EG_Chemistry-12_(22-07-2022) a

IUPAC convention, the reduction potential alone DG° = – nFE0cell is called as the electrode potential unless it is Here, DG° is the standard Gibbs free energy specifically mentioned. change. E0red = – E0oxid • Relationship between free energy change • It is not possible to determine the absolute and equilibrium constant: DG° = –2.303 RT log Kc value of electrode potential. For this a reference Cell potential: The potential difference between electrode [NHE or SHE] is required. The electrode the two electrodes of a galvanic cell. potential is only the difference of potentials between two electrodes that we can measure by Ecell = Ecathode – Eanode combining them to form a complete cell. SHE (Standard Hydrogen Electrode): • Standard electrode potential: The potential • It is used as a reference electrode to measure the difference developed between metal electrode standard electrode potential of the other metal and solution of ions of unit molarity (1 M) at 1 electrodes by assuming the standard electrode atm pressure and 25°C (298 K) is called standard potential of SHE as zero [E0(H+) = zero]. electrode potential. It is denoted by E°. • It consists of a platinum electrode (foil) coated • Electromotive force (emf) of a cell is the with finely divided platinum metal dipped in an difference between the electrode potentials of two acidic solution with 1 M H+ (aq) ions and pure half-cells and cause flow of current from electrode hydrogen gas (at 1 bar) is bubbled through the at higher potential to electrode at lower potential. solution. It is also the measure of free energy change. • Representation of SHE: H+(aq) |H2(g)| Pt(s) • Electrochemical series is the arrangement of electrodes in the increasing order of their standard reduction potentials. • Nernst equation: The relationship between the concentration of ions and electrode potential is given by Nernst equation. For a general reaction, aA + bB ne−→ cC + dD Ecell = E0cell – RT ln [C]c [D]d nF [A]a [B]b For an electrochemical cell, concentration of pure solids and liquids is taken as unity. • Nernst equation and KC: Standard Electrode Potentials at 298 K ∵ At equilibrium, Ecell = 0 Ions are present as aqueous species and H2O as liquid; 0.0591 gases and solids are shown by g and s. E0cell = n log KC at 298 K Reaction (Oxidised form +ne– → Reduced form) E/V 2.87 F2(g) + 2e– → 2F– 1.51 MnO4– + 8H+ + 5e– → Mn2+ + 4H2O 1.40 Au3+ + 3e– → Au(s) 1.33 Increasing strength 1.23 of oxidising agent 1.23 Increasing strength 1.09 of reducing agent 0.92 Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O O2(g) + 4H+ + 4e– → 2H2O 0.80 MnO2(s) + 4H+ + 2e– → Mn2+ + 2H2O Br2 + 2e– → 2Br– 2Hg2+ + 2e– → Hg22+ Ag+ + e– → Ag(s) 50 Chemistry-12

Increasing strength Fe3+ + e– → Fe2+ 0.77 of oxidising agentO2(g) + 2H+ + 2e– → H2O2 0.68 Increasing strengthI2 + 2e–→ 2I–0.54 of reducing agentCu2+ + 2e–→ Cu(s)0.34 AgCl(s) + e– → Ag(s) + Cl– 0.22 2H+ + 2e– → H2(g) 0.00 Pb2+ + 2e– → Pb(s) –0.13 Fe2+ + 2e– → Fe(s) –0.44 Zn2+ + 2e– → Zn(s) –0.76 2H2O + 2e– → H2(g) + 2OH–(aq) –0.83 Al3+ + 3e– → Al(s) –1.66 Na+ + e– → Na(s) –2.71 Li+ + e– → Li(s) –3.05 1. A negative E means that the redox couple is a stronger reducing agent than the H+/H2 couple. 2. A positive E means that the redox couple is a weaker reducing agent than the H+/H2 couple. ExErcisE 2.1 Multiple Choice Questions (MCQs) (1 Mark) (i) It does not participate in the cell reaction. 1. Which cell will measure standard electrode (ii) It provides surface either for oxidation or for potential of copper electrode? reduction reaction. (i) P(atq(.s,1) M| )H|2 (g,0.1 bar) | H+ (aq.,1 M) || Cu2+ (iii) It provides surface for conduction of electrons Cu (iv) It provides surface for redox reaction. (ii) Pt (s) |MH) |2 (g, 1 bar) | H+ (aq.,1 M) || Cu2+ 5. An electrochemical cell can behave like an (aq.,2 Cu electrolytic cell when ..................... (iii) Pt (s) |MH) |2 (g, 1 bar) | H+ (aq.,1 M) || Cu2+ (i) Ecell = 0 (ii) Ecell > Eext (aq.,1 Cu (iii) Eext > Ecell (iv) Ecell = Eext (iv) P(atq(.s,1) |MH) 2|(gC,u1 bar) | H+ (aq.,0.1 M) || Cu2+ Assertion-Reason Type Questions (1 Mark) 2. Which of the following statement is correct? Note: In the following questions a statement of assertion followed by a statement of reason is given. (i) eExCteelnl saivnedpDrorGperotfiecs.ell reaction both are Choose the correct answer out of the following choices. (ii) iEnCteelnl saivnedprDorpGerotifesc.ell reaction both are (i) Both assertion and reason are true and the reason is the correct explanation of assertion. (iii) EreCaeclltiisonanisinatnenesxitveenpsrivoepeprrtoypwerhtiyle. DrG of cell (ii) Both assertion and reason are true and the reason is not the correct explanation of assertion. (iv) EceClelllreisacatnionexisteannsiivnetepnrsoipveerptyrowpehritlye. DrG of (iii) Assertion is true but the reason is false. (iv) Assertion is false but the reason is true. 3. The difference between the electrode potentials of 6. Assertion: Cu is less reactive than hydrogen. two electrodes when no current is drawn through the cell is called .................... . Reason: E is negative Cu2+ / Cu 7. Afosrstehreticoenll:toEfCuelnl csthioonu.ld have a positive value (i) Cell potential (ii) Cell emf Reason: Ecathode < Eanode (iii) Potential difference 8. Assertion: Current stops flowing when ECell = 0. (iv) Cell voltage Reason: Equilibrium of the cell reaction is 4. Which of the following statement is not correct attained. about an inert electrode in a cell? eleCtroChemistry 51

9. Acosnsceernttiroanti:onEAofg+A/Agg+ increases with increase in 27. Consider the following diagram in which an ions. electrochemical cell is coupled with an electrolytic cell. What will be the polarity of electrodes ‘A’ and Reason: Equilibrium of the cell reaction is ‘B’ in the electrolytic cell? attained. 10. Assertion: Copper sulphate can be stored in zinc Zn plate Cu plate vessel. Reason: Zinc is less reactive than copper. Salt Bridge Very Short Answer Type Questions (1 Mark) 11. Out of the two cells i.e. (i) Galvanic cell, (ii) ZnSO4 solution CuSO4 solution Electrolytic cell which one is spontaneous? (A) Electrode (B) Electrode 12. What is the role of platinum wire in hydrogen electrode? Electrolytic 13. Can we store AgNO3 solution in a copper vessel Cell or not? 28. A galvanic cell has electrical potential of 1.1 V. If 14. Following two reaction occurs at cathode in the an opposing potential of 1.1 V is applied to this electrolytic solution of aqueous sodium chloride. cell, what will happen to the cell reaction and current flowing through the cell? Na+ + e– → Na(s) Ecoell = –2.71 V 2H2O(l) + 2e– → H2(g) + 2OH–(aq) 29. Which reference electrode is used to measure the Ecoell = – 0.83 V electrode potential of other electrodes? Which reaction takes place preferentially and why? 30. Name two metals which cannot displace hydrogen from dilute acids? 15. How can the reduction potential of an electrode be increased? Short Answer Type Questions-I (2 Marks) 16. wKhCyl,?NH4NO3 are suitable salts for a salt bridge, 31. (i) Can we store the solution of CuSO4 in a Zn vessel? 17. It is necessary to use a salt bridge in a galvanic cell, why? (ii) Can we store the solution of CuSO4 in a Ag vessel? 18. The standard reduction potential values of three metal cations X, Y and Z are 0.52, –3.03, –1.18 V 32. Consider the cell given below: respectively. What will be the order of reducing Cu|Cu2+||Cl–|Cl2(Pt) power of the corresponding metals. Write the reactions that occur at anode and cathode. 19. An electrochemical cell is made of aluminium and tin electrode with their standard reduction 33. Write the Nernst equation for the cell reaction potential –1.66 V and 0.14 V respectively. Select winhtehnecDonacneineltlracetilol.nHoofwZnw2+illiotnhseiEs cienllcbreeaasfefdec?ted the anode and cathode that represents the cell and write the cell reaction. Find out the emf of 34. Can we store copper sulphate in an iron vessel? the cell. Why or why not? 20. An unknown metal M displaces Ni from NiCl2 E0Cu2+/Cu = +0.34 V, E0Fe2+/Fe = –0.44 V. solution but cannot displace Mn from MnCl2 35. Write down the relationship between solution. Arrange metals M, Ni and Mn in correct order of reducing power. (i) Standard free energy change and emf of a cell. 21. Can absolute electrode potential of an electrode is measured? (ii) Standard free energy change and equilibrium constant. 22. Can Ecoell or DG° for a cell reaction ever be equal to zero? 36. Distinguish between EMF and potential difference. 37. With the help of the following possible reactions, 23. Under what condition is Ecell = 0 and DG = 0? 24. What does the negative sign in the expression arrange Mg, Zn, Cu and Ag in decreasing order of their reduction potentials. E°Zn2+/Zn = –0.76 V mean? 25. Depict the galvanic cell in which the cell reaction Cu + 2Ag+ → Cu2+ + Ag Mg + Zn2+ → Mg2+ + Zn is: [AI 2011 C] Zn + Cu2+ → Zn2+ + Cu Cu + 2Ag+ Æ 2Ag + Cu2+ 26. What is electrode potential? 52 Chemistry-12

38. Arrange the following in order of decreasing Determine DrG and E for the reaction. reducing nature: Mg, Na, Ag, Cu, Fe, Zn. [Delhi 2012] 39. State the factors that influence the cell potential Short Answer Type Questions-II (3 Marks) of the following cell 50. Calculate the potential for half cell reaction Mg(s)|Mg2+(aq) || Ag+(aq)|Ag(s) containing 0.1 M K2Cr2O7, 0.2 M Cr3+ (aq) and 1 × 10–4H+ (aq) 40. Calculate the potential of zinc electrode in which the zinc ion activity is 0.001 M. The half cell reaction is EoZn2+ /Zn = –0.76 V, R = 8.314 J K–1 mol–1, Cr2O72–(aq) + 14H+ + 6e– → 2Cr3+ + 2H2O(l) and standard electrode potential is given as F = 96500 coulomb. 41. Calculate the emf of the cell E° = 1.33 V [AI 2011] Cr|Cr3+(0.1 M) || Fe2+ (0.01 M)|Fe 51. A voltaic cell is set up at 25°C with the following half cells: Al3+ (0.001M) and Ni2+ (0.50 M). Al(s)| E°Cr3+/Cr = – 0.75V Al3+(0.001M||Ni2+(0.50M|Ni(s)) E°Fe2+/Fe = – 0.44V [Delhi 2009 C] Write an equation for the reaction that occurs 42. Write down the Nernst equation and calculate the emf of the following cell at 298K. when the cell generates an electric current and Cu(s) | Cu2+ (0.13 M) || Ag+ (10–4 M) | Ag(s) determine the cell potential. Given: E°Cu2+/Cu = +0.34 V and E°Ag+/Ag = +0.80 V (Given EoNi2+ /Ni = -0.250V, EoAl3+ /Al = –1.66 V). 43. Arrange the following metals in the order in which 52. Two half cell reactions of an electrochemical cell they displace each other from the solution of their are given as salts. MnO4– (aq) + 8H+ (aq) + 5e– → Mn2+ + 4H2O(l) E° = +0.151V Al, Cu, Fe, Mg and Zn. (NCERT) 44. Given the standard electrode potentials, Sn2+(aq) → Sn4+(aq) + 2e– E° = + 0.15 V K+/K = –2.93V, Ag+/Ag = 0.80 V, Construct the redox equation from the two half- Hg2+/Hg = 0.79 V cell reactions and predict if this reaction favours Mg2+/Mg = –2.37 V, Cr3+/Cr = – 0.74 V the formation of reactants or products as shown Arrange these metals in their increasing order of in the equation. [Delhi 2011, AI 2009] reducing power. (NCERT) [AI 2010] 53. Calculate the standard cell potentials of galvanic cell in which the following reaction take place: 45. Depict the galvanic cell in which the reaction 2Cr(s) + 3Cd2+ (aq) → 2Cr3+ (aq) + 3Cd Zn(s) + 2Ag+(aq) æÆ Zn2+(aq) + 2Ag(s) takes place. E 0 = – 0.40 V E 0 r3+ – 0.74 V Cd C Further show: +2 / Cd /Cr (i) Which of the electrode is negatively charged? Calculate the DrG and equilibrium constant of the reaction. (NCERT) [Delhi 2008 C] (ii) The carriers of the current in the cell. 54. Calculate the standard cell potentials of galvanic (iii) Individual reaction at reach electrode. cell in which the following reactions take place: (NCERT) [AI, Delhi 2008] Fe2+(aq) + Ag+ (aq) → Fe3+(aq) + Ag(s) 46. Write the Nernst equation and emf of the following E0Ag+/Ag = 0.80 V E0 = 0.77 V cells at 298 K. Fe3+ /Fe2+ Mg(s) | Mg2+(0.001 M) || Cu2+(0.0001 M) || Cu(s) Calculate the DrG and equilibrium constant of the reactions. (NCERT) (NCERT) 55. Write the Nernst equation and calculate emf of 47. Write the Nernst equation and emf of the following the following cell at 298 K. cells at 298 K. (P1t(bsa) r|)B| rP2t((ls))| Br–(0.010 M) || H+(0.030 M(N)C| EHR2(Tg)) Fe(s) | Fe2+(0.001 M) || H+(1M)|H2(g) (1 bar) || Pt(s) (NCERT) [Delhi 2013] 56. Consider the following cell reactions: 48. Write the Nernst equation and emf of the following Cu2+ + 2e– → Cu E° = 0.34 V cells at 298 K. Ag+ + 1e– → Ag E° = 0.80 V Sn(s) | Sn2+(0.050 M) || H+(0.020 M) | H2(g) (1 bar) (i) Construct a galvanic cell using the above | Pt (s) (NCERT) [AI 2011 C] data. 49. In the button cells widely used in watches and (ii) For what concentration of Ag+ ions will the other devices the following reaction takes place: emf of cell be zero at 25°C, if the concentration of Cu2+ ions is 0.1 M? [log 3.919 = 0.593]. Zn(s) + Ag2O(s) + H2O(l) æÆ Zn2+(aq) + 2Ag(s) + 2OH–(aq) eleCtroChemistry 53

Long Answer Type Questions (5 Marks) (v) Which electrode loses its mass? (vi) Which electrode gains mass? 57. In the galvanic cell: Cu|Cu2+ || Ag+|Ag, the (vii) Write the electrode reactions. electrons flow from Cu electrode to Ag electrode. (viii) Write the cell reaction. Answer the following questions. (ix) What is the function of salt bridge represented (i) What is anode? by the symbol ‘| |’? (ii) What is cathode? (x) Which metal, Cu or Ag has a greater tendency to lose electrons. (iii) What happens at anode – oxidation or reduction? (iv) What happens at cathode – reduction or oxidation? Answers 2.1 1. (iii) 2. (iii) 3. (ii) 19. In an electrochemical cell, the cathode must have 4. (iv) 5. (iii) 6. (iii) higher value of reduction potential. Therefore for 7. (iii) 8. (i) 9. (i) the cell: 10. (i) Al(s)|Al3+(aq) || Sn2+(aq) | Sn(s) 11. (i) Galvanic cell is spontaneous because its free energy change is–ve. (i.e. there is decrease in Ecoell = ESon2+ Sn − EoAl3+ Al free energy). = 0.14 – (–1.66) = +1.80 (ii) Electrolytic cell is non-spontaneous, because its free energy change is +ve (i.e. there is 20. EoMn2+ Mn = − 1.18 V − EoNi2+ Ni = 0.25 V increase in free energy). Since M replaces Ni but not Mn, hence the correct order of reducing power is: Mn > M > Ni. 12. Platinum being inert does not take part in reaction and also acts as the absorbent for hydrogen gas. 21. No, absolute electrode potential of an electrode cannot be measured. 13. It is clear from the values of E° of Ag and Cu that standard potential of Cu is less than Ag, therefore 22. No E° or DG° can never be equal to zero. Cu will reduce AgNO3. DG° = –nE°F. Cu + 2AgNO3 → Cu(NO3)2 + 2Ag Eoredox process = E° of reducing species ∵ E° is never zero, hence, DG° will also never be – E° of oxidising species equal to zero. 23. At equilibrium, i.e. when the cell is completely = 0.80 – 0.34 = 0.46 volts discharged Ecell = 0 and DG = 0. \\ DG = –nEF E° is positive, so the reaction will be spontaneous. \\ When E = 0, DG is also equal to zero. 14. Since the standard potential is greater for second 24. EZon2+ /Zn = –0.76 V process, therefore second process will take place Negative sign shows that the reducing power of at cathode to release hydrogen gas for the second Zn is greater than that of H2. reaction. Zn + H2SO4 → ZnSO4 + H2 15. Mn+ + ne– → M 25. Cu + 2Ag+ → 2Ag + Cu2+ Cu(s) | Cu2+ || Ag+ | Ag(s) EMn+ /M = 2.303 RT log 1 nF [Mn+ ] 26. Tendency of an electrode to gain or to lose electrons is called electrode potential. From the above equation it is clear that — reduction potential can be increased either by 27. (A) → Negative polarity increasing the temperature or by increasing the (B) → Positive polarity concentration of metal ions. 28. When Ecell = Eexternal 16. Electrolytes which produce cations and anions Then no current will flow through the circuit. having nearly the same transference number are suitable for a salt bridge. Cations and anions 29. Standard hydrogen electrode is used to measure should have same mobility in the salt bridge. the electrode potential of other electrodes. 17. Salt bridge is used to maintain the electrical 30. Cu, Ag neutrality in the cell and it also avoid the formation 31. (i) No, because Zn is more reactive than Cu. of liquid junction between the electrolytic solutions at cathode and anodic half cells. Therefore Zn will replace Cu from CuSO4. Zn + CuSO4 → ZnSO4 + Cu 18. More the negative value of reduction potential, (ii) Yes, because Ag is less reactive than Cu. So more will be the reducing power: Y > Z > X there will be no reaction. 54 Chemistry-12

32. The given cell is Cu(s) | Cu2+ || Cl– | Cl2(Pt) Cathode: Fe(20+.01 M) + 2e− → Fe(s) × 3 Anode: Cu(s) → Cu2+ + 2e– Cell reaction: Cathode: Cl2(g) + 2e– → 2Cl– 2Cr(s) + 3Fe(20+.01 M)  2Cr(30+.1 M) + 3Fe(s) n = 6 33. Daniell cell Zn(s) | Zn2+ || Cu2+ | Cu(s) Anode: Zn(s) → Zn2+ + 2e– Q= [Cr3+ ]2 [Fe]3 = (0.1)2 × 13 = 104 Cathode: Cu2+ + 2e– → Cu(s) [Cr]2 [Fe2+ ]3 12 × [0.01]3 Cell reaction Zn(s) + Cu2+ → Zn2+ + Cu(s) Q = [Zn2+ ] Using Nernst equation [Cu2+ ] E = E° – 0.059 log10 Q The Nernst equation for the cell is: n Ecell = Ecoell − 2.303RT log [Zn2+ ] = + 0.31 – 0.059 log10 104 = 0.270 V nF [Cu2+ ] 6 This equation is an indication that the cell 42. Ecoell = 0.80 – 0.34 = 0.46 V potential will decrease on decreasing [Zn2+]. Anode: Cu(s) → Cu2+ (0.13 M) + 2e– 34. No, we cannot store copper sulphate in an iron Cathode: 2Ag+(10–4 M) + 2e–→ 2Ag(s) vessel. This is because, iron being more reactive, Cell reaction: Cu(s)+2Ag+(10–4M)  reacts wanitdhCCuumSOet4aal ngdetrsedpelapcoesiCteud2:+ ion to give FeSO4 Cu2+ (0.13 M) + 2Ag(s), n = 2 Fe(s) + CuSO4 → FeSO4 + Cu(s) 35. (i) DG° =–nE°F Q = [Cu2+ ] [Ag]2 0.13 × 12 = 0.13×108 (ii) DG° = –2.303 RT log10 K [Cu] [Ag+ ]2 = 1 × (10−4 )2 36. Potential difference between the two electrodes E = 0.46 – 0.059 log10 (0.13 × 108) of a cell when no current is flowing through the 2 circuit is called EMF. On the other hand, potential difference between two ends of an electrode can also be measured even when current is flowing = 0.22 V through the electrode. 43. Cu < Fe < Zn < Al < Mg is increasing order of reactivity, increasing ability to displace less 37. Reactivity order from three reactions are Cu > Ag; reactive metals from their salt solution. Mg > Zn; Zn > Cu. 44. Ag < Hg < Cr < Mg < K is increasing order of Thus, overall reactivity order is reducing power. Mg > Zn > Cu > Ag 45. The galvanic cell is: 38. Na > Mg > Zn > Fe > Cu > Ag. Zn(s)|Zn2+(aq)||Ag+(aq)|Ag(s) 2.303 Mg2+  (i) Zn is negatively change 2F  39. E= Ecoell − log [Ag+ ]2 (ii) Ions carry current inside the cell and electrons carry current outside the cell. Cell potential will depend upon the concentration (iii) At anode Zn æÆ Zn2+ + 2e– of ions and temperature. 40. Half cell reaction || Zn2+ (0.001 M) | Zn(s) At cathode 2Ag+ + 2e– æÆ 2Ag(s) 46. E0 /Mg = – 2.37 V, E0 /Cu = + 0.34 V Mg2+ Cu+2 Zn2+ + 2e–  Zn(n = 2) Mg (s) æÆ Mg2(a+q) + 2e– Q= [Zn] = 1 = 1000 Cu(2a+q) + 2e– æÆ Cu(s) [Zn2+ ] 0.001 Mg (s) + Cu2+ (aq) æÆ Mg2+ (aq) + Cu (s) 0.059 E = E° – n log Q Ecell = E0cell – 0.0591 log 7Mg2+A 2 7Cu2+A 0.059 = –0.76 – 2 log (1000) = (EC0 u2+/Cu –E0Mg2+/Mg) – = –0.8485 V 0.0591 log 0.001 2 0.0001 41. Ecoell = EFo e2+ /Fe − ECo r3+ /Cr = + 0.34 V – (– 2.37 V) – 0.0591 log10 –0.44 – (–0.75) = +0.31 V 2 Anode: [Cr(s) → Cr3+ + 3e–] ×2 = 2.71 V – 0.02955 = 2.68 V eleCtroChemistry 55

47. E0H+/H2 = 0, E0Fe2+/Fe = – 0.44 V = 1.33 – 0.059 log10 (4 × 1055) = +0.78 Fe(s) æÆ Fe2+ + 2e– 6 2H+(aq) + 2e– æÆ H2(g) 51. Al(s)|Al3+(0.001 M)||Ni2+(0.50 M)|Ni(s) Fe(s) + 2H+(aq) æÆ Fe2+(aq) + H2(g) Ecoell = E°Ni2+ /Ni – EA°l3+/Al Ecell = E0cell – 0.0591 log 7Fe2+A = –0.25 – (–1.66) = +1.41 V 2 6H+@2 Cell reaction: 0.05091 1 –3 2 (1) 2 2Al(s) + 3Ni2+  2Al3+ + 3Ni(s) n = 6 = _E0H+ /H2 – E0 /Fe i – log Q= [Al3+ ]2 [0.001]2 10−6 [Ni2+ ]3 [0.5]3 0.125 Fe2+ = = = 60 – ]– 0.44 Vg@ – 0.0591 log 10–3 = 8 × 10–6 2 0.059 = + 0.44 V – 0.0591 × ]–3g E = E° – 6 log Q 2 = 0.44 + 0.1773 = 1.41 − 0.059 log (8 × 10–6) 2 6 = 0.44 + 0.089 = 0.529 = +1.46 V 48. E0H+/H2 = 0V, ES0n2+/Sn = – 0.14V Sn(s) æÆ Sn2+(aq) + 2e– 52. For spontaneous redox reactions, E° (reduced species) must be greater than E° (oxidised 2H+ + 2e– æÆ H2(g) species). Sn(s) + 2H+(aq) æÆ Sn2+(aq) + H2(g) This may be given as Ecell = E0cell – 0.0591 log [Sn2+]6H2@ [MnO4– (aq) + 8H+(aq) + 5e– → 2 [H+] 2 Mn2+ + 4H2O] × 2 [E° = +1.51 V] = (E0H+/H2 – ES0n2+/Sn) – 0.0591 log 0.05 × 1 [Sn2+ (aq) → Sn4+ (aq) + 2e–] × 5 2 (0.02) 2 [E° = + 0.15 V] = 0 – (– 0.14 V) – 0.0591 log 500 2MnO4– + 5Sn2+ + 16H+ → 2 4 2Mn2+ + 3Sn4+ + 8H2O = + 0.14 V – 0.0591 log 125 EC° ell = +1.51 – 0.15 = +1.36 V 2 Positive E° value indicates that the process is spontaneous. = + 0.14 V – 0.0591 × 2.0969 V 53. 2Cr(s) æÆ 2Cr3+(aq) + 6e– 2 3Cd2+(aq)+ 6e– æÆ 3Cd(s) = 0.14 V – 0.06 V = 0.08 V 2Cr(s) + 3Cd2+(aq) æÆ 2Cr2+(aq) + 3Cd(s) 49. Zn(s) æÆ Zn2+ + 2e– 2Ag+ + 2e– æÆ 2Ag(s) n=6 E0cell = E0 cathode – E0 anode = E0Ag+/Ag – E0Zn2+/Zn E0cell = E0 /Cr – E0 /Cr Cr2+ Cr3+ = –0.40 V – (–0.74 V) = +0.34 V = 0.80 V – (– 0.76 V) = 1.56 V DrG0 = –nE0F DrG0 = – nE0F = –6 × 0.34 V × 96500 C mol–1 = – 2 × 1.56 V × 96500 C = –196860 J mol–1 = – 301080 J mol–1 = – 301.080 kJ mol–1 = –196.860 J mol–1 log KC = n E0 = 6 × 0.34 V = 2.04 V 0.0591 0.0591 0.0591 [Cr3+ ]2 = 34.5177 50. Q= [Cr2O72− ] [H+ ]14 KC = Antilog 34.5177 = 3.294 × 1034 = [0.2]2 0.04 54. Fe2+(aq) æÆ Fe3+(aq) + e– [0.1] [10−4 ]14 = 0.1 × 10−56 Ag+(aq) + e– æÆ Ag(s) = 0.04 × 1057 = 4 × 1055 Fe2+(aq) + Ag+(aq) æÆ Fe3+(aq) + Ag(s) E = E° – 0.059 log10Q E0cell = E0Ag+ /Ag – E0 /Fe 6 Fe3+ = 0.80 V – (0.77 V) = 0.03 V 56 Chemistry-12

DrG0 = – nE0F 56. Since the reduction potential of Cu2+/Cu is less, so = – 1 × 0.03 V × 96500 C mol–1 copper electrode acts as anode and silver electrode acts as cathode. The galvanic cell is = – 2895 J mol–1 Cu|Cu2+ (C1)||Ag+ (C2)|Ag EMF = 0.8 – 0.34 = 0.46 V = – 2.895 kJ.mol–1 Using Nernst Equation; lop KC = n E0 = 13× 0.0 = 0.5076 0.0591 0.0591 Ecell = Ecoell 0.059 [Cu2+ ] KC = Antilog 0.5076 = 3.218 −n log [Ag+ ]2 55. E0H+/H2 = 0, E0Br2/Br– = 1.09 V 0 = 0.46 − 0.059 log [0.1] At anode: 2Br– æÆ Br2(l) + 2e– 2 [Ag+ ]2 At cathode: 2H+ + 2e– æÆH2(g) 2Br–(aq) + 2H+(aq) æÆ Br2(l) + H2(g) 15.59 = log(0.1) – log[Ag+]2 0.0591 1 16.59 = –2 log [Ag+] 2 6Br–@2 6H+@2 Ecell = E0cell – log [Ag+] = 5.070 × 10–9 mol L–1 = ( E0H+/H2 – E0Br2/Br– ) 57. (i) Anode is the electrode which attracts anions towards it. Here it is the electrode made up of Cu – 0.0591 log 1 (ii) Cathode is the electrode which attracts 2 (3 × 10–2)2 cations towards it. Here it is the electrode (10 –2) 2 made up of Ag. = (0 – 1.09 V) – 0.0591 log 108 (iii) oxidation, (iv) reduction, 2 9 (v) Cu, (vi) Ag 0.0591 = – 1.09 V – 2 (log 108 – log 9) (vii) Anode: Cu → Cu2+ + 2e– Cathode: Ag+ + e– → Ag = – 1.09 V – 0.0295 (8.000 – 0.9542) (viii) Cu + 2Ag+ → Cu2+ + 2Ag = – 1.09 V – 0.0295 × 7.0458 (ix) Its function is to complete the circuit. = – 1.09 V – 0.208 V = – 1.298 V (x) Cu Topic 2. Conductivity and its Variation for Electrolytic Solutions Substances which allow the passage of current Kohlrausch’s law: At infinite dilution, the molar conductivity of an electrolyte is the sum of the ionic by movement of electrons through their body are conductivities of the cations and anions, e.g., for L°HCl = λ°H + λ°Cl– . electronic conductors/metallic conductors. Application of Kohlrausch’s law are: Conductance (G): It is the ease of flow of electric • Determination of molar conductivities of weak current through the conductor. It is the reciprocal of electrolytes at infinite dilution. resistance (R). 1 • Determination of degree of dissociation (a) of an R electrolyte at a given dilution. G= , units ohm–1 or mho • Calculation of solubility of sparingly soluble salts. Specific conductivity (G): • It is the reciprocal of specific resistance. • Specific conductivity decreases on dilution. This is because concentration of ions per cc decreases upon dilution. S.No. Physical Unit (CGS) Unit (SI) 1. Conductance ohm–1 S (siemen) 2. Conductivity or specific conductance ohm–1 cm–1 S m–1 3. Cell constant cm–1 m–1 4. Equivalent conductance ohm–1 cm2 equiv–1 S m–1 (equiv L–1)–1 or S m2 (equiv–1) 5. Molar conductance ohm–1 cm2 mol–1 S m2 mol–1 or S m–1 (mol L–1)–1 cMonodlaurctcinognpdouwcetrivoiftayll(tLhme )i:onIts may be defined as the Mathematically Lm = k × V, produced by dissolving Lm = κ × 1000 or κ × 1000 one gram mole of an electrolyte placed between two C M large electrodes at one centimeter apart. eleCtroChemistry 57

where, V is the volume of solution in cm3 containing the sum of the individual contributions of the anion 1 gram mole of the electrolyte and C is the molar concentration. and cation of the electrolyte. 400 In general, if an electrolyte on dissociation gives CH3COOH cno+ncdautcitoinvistyainsdginv–enanbiyons then its limiting molar Λ o = ν+ λo+ + ν− λ o− m Λm/(S cm2 mol–1) Here, λco+orraensdpoλno−dianrge the limiting molar conductivities of the cation and anion, respectively. 200 Applications of Kohlrausch’s Law KCl (i) Calculation of molar conductivities of weak electrolyte at infinite dilution. For example molar conductivity of acetic acid at infinite dilution can be obtained from the knowledge of molar conductivities at infinite dilution of strong 0.2 0.4 c1/2/(mol/L)1/2 electrolyte like HCl, CH3COONa and NaCl as illustrated below. Graphic representation of the variation of Lm vs c Λom (CH3COOH) = λCo H3COOH− + λ o Units: Lm = κ × 1000 S cm−1 H+ C = mol cm−3 λ  = ohm–1 cm2 mol–1 or S cm2 mol–1 = o + λ oNa+  + CH3COO− Variation of conductivity and molar conductivity λ o + λ Cl−  – λoNa+ + λCo l−  with concentration: Conductivity decreases with H+   decrease in concentration, this is because the number of ions per unit volume that carries the current in the i.e., Λom(CH3COOH) = Λmo (CH3COONa) + solution decreases with dilution. Λ o − Λ om(NaCl) m(HCl) Minocloanrcceonntdruacttiiovnit.yT(hLims = k × V) increases with decrease is because the total volume ‘V’ (ii) Determination of Degree of Dissociation of of solution containing one mole of electrolyte also weak electrolytes: increases. It has been found that the decrease in k on dilution of a solution is more than compensated by the Degree of dissociation, (a) = Λm increase in its volume. Λ o m Limiting molar cwohnednuthceticvointcyen(Ltr°ma)ti:oTnhaeppvraolauceheosf (iii) Determination of Dissociation constant (K) molar conductivity of weak electrolytes: zero is known as limiting molar conductivity or molar K = cα2 conductivity at infinite dilution. 1−α It is possible to determine the molar conductivity at (iv) Determination of solubility of sparingly infinite dilution (L0m) in case of a strong electrolyte by soluble salts: ethxetrvaaplouleaotifomnoolfarthceoncduurcvteivoiftyLfmorvws eack . On contrary, electrolytes at Λ o = κ × 1000 = κ × 1000 m Molarity Solubility infinite dilution cannot be determined by extrapolation of the curve as the curve becomes almost parallel to the y-axis when concentration approaches to zero. or, Solubility = κ × 1000 Λ om Kohlrausch’s Law: It states that the limiting molar conductivity of an electrolyte can be represented as • Difference between metallic and electrolytic conductance. Metallic conductance Electrolytic conductance (i) Metals conduct electricity in solid state. (i) Electrolytes conduct electricity in molten state or in (ii) Electrons carry current. aqueous solution only. (iii) Conductance decreases with increase in temperature. (ii) Ions carry current (iv) Metals become hot when current is passed through (iii) Conductance increase with increase in temperature. (iv) Heat energy is used to dissociate electrolyte into ions. them. (v) No chemical change takes place. (v) Chemical change takes place. 58 Chemistry-12

ExErcisE 2.2 Multiple Choice Questions (MCQs) (1 Mark) Very Short Answer Type Questions (1 Mark) 1. Which of the following statements about limiting 7. How does the solution of electrolytes conduct molar conductivity is not correct? electricity? (i) It is also known as molar conductivity at 8. Define (i) conductivity and (ii) molar conductivity. infinite dilution. [AI 2015, 14 C] (ii) Its value for strong electrolytes can be 9. D e f i n e s p e c i f i c c o n d u c t a n c e a n d m o l a r determined graphically by extrapolation of conductance. [Delhi, 2014 C] Lm vs c curve. 10. List the factors affecting the conductance of an (iii) Its value for weak electrolytes can be electrolyte. determined using Kohlrausch’s law. 11. Draw a graph showing variation in molar (iv) None of these conductivity for weak and strong electrolytes. 2. Which of the following statements about solutions [Delhi 2008 C] of electrolytes is not correct? 12. Write the expression for molar conductivity of (i) Conductivity of solution depends upon size infinite dilution for Al2(SO4)3. of ions. 13. What is the effect of temperature on molar (ii) Conductivity depends upon viscosity of conductivity? solution. Short Answer Type Questions-I (2 Marks) (iii) Conductivity does not depend upon solvation 14. Explain Kohlrausch’s Law. of ions present in solution. [AI 2010, 2010 C, 2008 C] (iv) Conductivity of solution increases with 15. λ° for NaCl, HCl and rCeHsp3eCcOtiOveNlya,acraelc1u2l6a.t0e, 426.0 temperature and 91.05 cm2 mol–1 L° for CH3COOH. [Delhi 2010] 3. 0m (NH4 OH) is equal to .................... . 16. It is not possible to determine the conductance of (i) 0m (NH4 OH) + 0m (NH4 Cl) – (0HCl) weak electrolytes by extrapolation. Why? (ii) 0m (NH4 Cl) + 0m (NaOH) – (0NaCl) 17. Copper is conducting as such but copper sulphate is conducting only in molten state or in an aqueous (iii) 0m (NH4 Cl) + 0m (NaCl) – (0NaOH) solution. Explain why? (iv) 0m (NaOH) + 0m (NaCl) – 0 18. Solution of two electrolytes ‘A’ and ‘B’ are diluted. m (NH4 Cl) iTnhcereLaaseosf ‘B’ increases 1.5 times while that of ‘A’ 25 times. Which of the two is a strong Assertion-Reason Type Questions (1 Mark) electrolyte? Justify your answer. Note: In the following questions a statement of 19. In an aqueous solution how does the specific assertion followed by a statement of reason is given. conductivity of the electrolyte changes with Choose the correct answer out of the following choices. addition of water? (i) Both assertion and reason are true and the 20. dgWrrahadsytuioacnalllydly?il,uwtihonileththeaLtmofofCCHH3C3COOOON[CHaBiiSnnEccrr2ee0aa2ssee2ss] reason is the correct explanation of assertion. (ii) Both assertion and reason are true and the 21. 0.05 M NaOH solution offered a resistance of reason is not the correct explanation of assertion. 31.6 ohm in a conductivity cell at 298 K. If cell constant of the cell is 0.367 cm–1. Calculate the (iii) Assertion is true but the reason is false. molar conductivity of NaOH solution. (iv) Assertion is false but the reason is true. 4. Assertion: Conductivity of all electrolytes 22. The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm–1. Calculate its molar conductivity. decreases on dilution. (NCERT)[Delhi 2008] Reason: On dilution number of ions per unit 23. The resistance of a conductivity cell containing volume decreases. 0.001 M KCl solution at 298 K is 1500 W. What is the cell constant if conductivity of 0.001 M KCl 5. iAnscsreeratsieown:hLemn for weak electrolytes shows a sharp solution at 298 K is 0.146 × 10–3 S cm–1? the electrolytic solution is diluted. Reason: For weak electrolytes degree of (NCERT)[Delhi 2008] dissociation increases with dilution of solution. 6. Assertion: For measuring resistance of an ionic Short Answer Type Questions-II (3 Marks) solution an AC source is used. Reason: Concentration of ionic solution will 24. A conductivity cell was filled with 0.01 M change if DC source is used. solution of KCl which was known to have specific eleCtroChemistry 59

conductivity of 0.1413 ohm–1 m–1 at 298 K. Its of 0.052 M unknown electrolyte, the resistance was 96 ohm–1. Calculate the molar conductivity measured resistance at 298 K was 94.3 ohm. When the cell was filled with 0o.h0m2 .MCaAlcguNlaOte3 of the unknown electrolyte at this concentration solution, its resistance was 50.3 (specific conductance of 0.1 M KCl = 1.29 × 10–2 ohm–1 cm–1). (a) Cell constant, (b) The specific conductance of AgNO3 solution. 25. When a conductivity cell was filled with 0.1 M 26. Conductivity of 0.00241 M acetic acid is 7.896 × 10–5 S cm–1. Calculate its molar conductivity. If cm2 mol–1, what is KCl, it has a resistance of 85 ohm at 25°C. When L0 mdifsosrocaicaettiiocnaccoidnsitsa3n9t?0.5 S (NCERT) its the same cell was filled with aqueous solution Answers 2.2 1. (iv) 2. (iii) 3. (ii) 13. Molar conductivity increases with increase in temperature. 4. (i) 5. (i) 6. (i) 7. Electrolytes give ions in the solution and these 14. At infinite dilution each ion of the electrolyte ions conduct electricity by moving to their make a definite contribution towards conductivity respective electrodes i.e. cations towards cathode of the electrolyte and it is independent of the and anions towards anode. presence of other ions of the electrolyte. This is called Kohlrausch’s law of independent migration 8. (i) Conductivity is the reciprocal of the specific of ions. resistance represented by symbol k. For salt like KCl k= 1 ...(1) ρ λ∞m KCl = λ∞mK + + λ ∞ Cl − m S.I. unit of k is Sm–1 but often expressed as Scm–1. 15. L° CH3COOH = λ°(H+) + λ°(CH3COO–) = [λ°(H+) + λ°(Cl–)] + (ii) Molar conductivity is the conducting power of all the ions furnished by one mole of [λ°(Na+) + λ°(CH3COO–)] – [λ°(Na+) + λ°(Cl–)] an electrolyte in a solution of specified 426.0 + 91.05 – 126.0 = 391.05 cm2 mol–1 concentration. 16. The conductance of weak electrolytes increases Lm = 1000 κ on dilution experimentally. The curve meet the M axis at infinity. The curve is almost run parallel to Lm axis which cannot be extrapolated to cut the M = Molarity of solution axis. Therefore, conductance of weak electrolytes cannot be determined by extrapolation. Its unit is S cm2 mol–1. 9. Specific conductance is the conductance of a solution taken in a cell in which two electrodes are 1 cm apart and surface area of each one of them is 1 cm2. Equivalent conductance is the conductance of all Λm the ions furnished by one mole of an electrolyte in a solution of given concentration. C 10. Nature of electrolyte (strong or weak), valency 17. It is because Cu metal has free electrons whereas of its ions, speed of ions, concentration and in CuSO4 ions can move only in aqueous solution temperature. or in molten state. 11. 18. Electrolyte ‘B’ will be a strong electrolyte because it is completely ionised and on dilution the molar Weak electrolyte conductance will increase to a small extent due to increase in speed of ions only. Λm/(S cm2 mol–1) Strong electrolyte 19. Specific conductance of an electrolytic solution c1/2/(mol/L)1/2 decreases on dilution with water because the number of ions per unit volume decreases. ( ) ( )12. Al3+ λ ∞ Al2 (SO4 )3 = 2λ∞m + 3λ∞m SO42− m 60 Chemistry-12

20. iOnncrdeialusetisodnrLasmtiocfaCllHy 3dCuOeOtoHin(acrweaesaek electrolyte) 0.1413 = 1 × cell constant in degree of 94.3 ionisation. \\ cell constant = 13.32 m–1 CH3COOH + H2O  CH3COO–(aq) + H3O+ For, AgCl, k = 1 × 13.32 50.3 CHtboeHociwna3CucerOsveeeOairstNeiCasinHaint3schCtrereOoasnOpsgeeNseedalgercoiatfsrdioouclnayostlml.ey.pTolenhtederilelyufotirioeonnLimdseuodef = 0.2648 ohm–1 m–1 25. k = C × Cell constant 21. k= 1 × cell constant 1.29 × 10–2 = 1 × Cell constant R 85 = 1 × 0.367 = 0.0116 ohm–1 cm–1 Cell constant = 1.0965 cm–1 31.6 For unknown electrolyte, the cell constant will be As we know same 1000 1000 k = C × Cell constant M 0.05 Lm = k × = 0.0116 × = 1 × 1.0965 = 0.01142 ohm–1 cm–1 96 \\ Lm = 232 ohm–1 cm2 mol–1 22. 1000 1000 × 2.48 × 10–2 1000 1000 Lm = M κ = 0.2 Lm = k × M = 0.01142 × 0.052 = 124 S cm2 mol–1 = 219.61 ohm–1 cm2 mol–1 23. R = 1500 W, k = 0.146 × 10–3 S cm–1,  =? 26. k = 7.896 × 10–5 S cm–1, M = 0.00241 M a 1  Lm = 1000k = 1000 × 7.896 × 10–5 R a M 0.00241 We know: k = × 1  = 32.763 cm2 mol–1 1500 a 32.76 0.146 × 10–3 = × a= Lm = 390.5 = 0.0839  = 1500 × 0.146 × 10–3 L0m = 0.219 cm–1 \\ a Ca2 0.00241 × (0.0839)2 Ka = 1– α = 1 – 0.0839 24. For KCl, k = 1 × cell constant R = 1.85 × 10–5 Topic 3. Quantitative Aspects of Electrolysis Electrolysis: It is the process of decomposition of the substance deposited or liberated by passing of an electrolyte when electric current is passed 1A current for 1 sec (i.e. 1 coulomb, I × t = Q) through either its aqueous solution or molten state. In electrolysis both oxidation as well as reduction • Second Law: When the same quantity of process takes place. electricity is passed through different electrolytes, the amount of the substance deposited or liberated Faraday’s Law of Electrolysis: at the electrodes are directly proportional to their • First Law: The amount of the substance equivalent weight. deposited or liberated at cathode directly proportional to the quantity of electricity passed Mass of A = eq. wt. of A through electrolyte, i.e. Mass of B eq. wt. of B W µ I × t =I × t × Z = Q × Z • Equivalent weight: It is defined as atomic weight divided by valency or charge on ion. Where I is current in amp, t = time in sec. • Preferential discharge theory: If more than Q =quantity of charge (coulomb) one type of ion is attracted towards a particular electrode then the ion which has lower discharge Z is a constant known as electrochemical potential or requires lower energy will get equivalent of the substance. discharged on the electrode. When I = 1 amp, t = 1 sec then Q = 1 coulomb, Cations: K+ < Na+ < Ca2+ < Mg2+ < Al3+ < Zn2+ < and w = Z. H+ < Cu2+ < Ag+ < Au3+ Thus, electrochemical equivalent is the amount Anions: SO4 2– < NO3 – < Cl– < Br– < I– < OH– eleCtroChemistry 61

ExErcisE 2.3 Multiple Choice Questions (MCQs) (1 Mark) 12. o1Af0qAumgeiNonuOust3esasorlieunteisloeencptaorrfoalCytueseSedOleb4cytar1nodalymatiqpcuecereeolulcsus.rsWroelinulltttifhooner 1. The quantity of charge required to obtain one mole of aluminium from Al2O3 is .................... . (i) 1F (ii) 6F mass of copper and silver deposited on the cathode (iii) 3F (iv) 2F be the same or different? Explain your answer. 2. In the electrolysis of aqueous sodium chloride 13. Value of standard electrode potential for the solution which of the half cell reaction will occur oxidation of Cl– ions is more positive than that at anode? of water, even then in the electrolysis of aqueous (i) Na+ (aq) + e– Æ Na (s); EC6ell = –2.71V solution of sodium chloride. Why is Cl– oxidised (ii) 21H.223OV (l) Æ O2(g) + 4H+ (aq) + 4e–; EC6ell = at anode instead of water? [NCERT Exemplar] +Æe–12ÆCl212 14. How will the pH of brine solution (aq NaCl) be affected when it is electrolysed? (iii) H+ (aq) H2 (g); EC6ell = 0.00V [NCERT Exemplar] (iv) Cl– (aq) (g) + e–; EC6ell = 1.36V 15. How many moles of mercury will be produced cbuyrerleencttroofl2ysaimngpe1r.0esMforH3gh(NrsO. 3[)M2 osloalur tmioansswoifthHag Assertion-Reason Type Questions (1 Mark) Note: In the following questions a statement of = 200.6 g mol–1]. [AI 2011] assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. 16. How much electricity in terms of Faraday is required to produce: [NCERT] (i) Both assertion and reason are true and the (i) 20.0 g of Ca from molten CaCl2? reason is the correct explanation of assertion. (ii) 40.0 g of Al from molten Al2O3? (ii) Both assertion and reason are true and the [Delhi 2013 C] reason is not the correct explanation of assertion. 17. How much electricity is required in coulomb for (iii) Assertion is true but the reason is false. the oxidation of: (iv) Assertion is false but the reason is true. (i) 1 mol of H2O to O2? 3. Assertion : Electrolysis of NaCl solution gives (ii) 1 mol of FeO to Fe2O3? [NCERT] chlorine at anode instead of O2. Reason : Formation of oxygen at anode requires 18. pAlastoilnuutmioneleocf tNroid(NesOu3s)i2nigs electrolysed between overvoltage a current of 5 amperes 4. Assertion : In electrolysis, only reduction process for 20 minutes. What mass of Ni is deposited at takes place. the cathode? [NCERT] Reason : Metal ions from electrolyte reduced and 19. Predict the products of electrolysis in each of the deposited at cathode. following: (i) An aqueous solution of AgNO3 with silver electrodes. Very Short Answer Type Questions (1 Mark) (ii) An aqueous solution of AgNO3 with platinum 5. On passing 1 F electricity how much grams of electrodes. oxygen will be produced at STP? (iii) A dilute solution of H2SO4 with platinum 6. Although fluorine is a good conductor of electricity electrodes. but it cannot be obtained by electrolysis of HF. Why? (iv) An aqueous solution[DoeflChui C20l21w4 iCth] platinum electrodes. [NCERT] 7. What is meant by Faraday constant? 8. What amount of substance is deposited by 1 Short Answer Type Questions-II (3 Marks) coulomb of charge? What is it known as? 20. Silver is deposited on a metallic vessel of surface area 800 cm2 by passing current of 0.2 amp for Short Answer Type Questions-I (2 Marks) 3 hrs. Calculate the thickness of the silver metal 9. What are the products obtained when molten deposited. [Density of silver = 10.47 g/cm3, atomic NaCl and aqueous NaCl are electrolysed? mass of Ag = 107.92 amu] 10. In the electrolytic aqueous solution of NaCl, which among hydrogen or sodium is liberated at cathode, 21. How much charge is required for the following Explain. reductions: 11. On passing electricity in molten NaCl, one mole (i) 1 mol of Al3+ to Al? of sodium deposits whereas from molten AlCl3 1/3 mole of aluminium is deposited. Why? (ii) 1 mol of Cu2+ to Cu? (iii) 1 mol of MnO– 4 to Mn2+? [NCERT] 62 Chemistry-12

22. Three electrolytic cells A, B, C containing solutions of silver deposited at the cathode of cell B. How coof nZnneScOte4d, AingNsOer3ieasn.dACustSeOa4d,yrecsuprercetnivteolyf are long did the current flow? What mass of copper 1.5 and zinc were deposited? amperes was passed through them until 1.45 g [NCERT][HOTS; Delhi 2009, AI 2008] Answers 2.3 1. (iii) 2. (iv) 3. (i) of over potential of oxygen and high concentration of chloride ions. 4. (iv) 5. 1 Faraday charge liberates one gram equivalent of 14. Aqueous solution of brine contains Na+, Cl–, H+ the substance. Therefore, 8g of O2 will be liberated and OH– ions. Electrode processes are given below: by 1 F charge. 2H+ + 2e– → H2 (Cathode) 2Cl– → Cl2 + 2e– 6. On etlwecotrsoplyesciiseswpergeestenHt2wahnidchO2cannotoFxi2d. iTseheorne Since, H+ are consumed and the (Anode) are remaining solution will contain NaOH, therefore pH will anode viz. pHo2tOenatniadl F– from HF but due to higher increase. oxidation or lower reduction potential 15. Q = I × t = 2.0 × 3 × 60 × 60 = 21600 C of H2O, it will oxidise at anode to give O2. 7. Faraday constant = 96500 coulomb \\ Number of moles of Hg produced = Charge on one mole of electrons. = 21600 = 0.112 moles 2 × 96500 8. Amount deposited by 1 coulomb charge is equal 16. (i) Method -1: m = Z × Q to = Equivalent mass 20g = 40 ×Q 96500 2 × 96500 It is known electrochemical equivalent of the ⇒ Q = 96500 C = 1F substance. Method-2: Ca2+ + 2e– æÆ Ca(s) 9. Aqueous NaCl: From the above reaction, we can infer that 2H+ + 2e– → H2↑ 2F electricity will give 1 mole, i.e., 40g of 2Cl– → Cl2 + 2e– (Cathode) Ca (s). Therefore to produce 20g of Ca(s) from (Anode) CaCl2, electricity required is 1 Faraday. Molten NaCl: (ii) Al3+(aq) + 3e– æÆ Al(s) Na+ + e– → Na (Cathode) m= Z×Q 2Cl– → Cl2 + 2e– (Anode) 27 96500 10. Hydrogen, because standard reduction potential 40g = 3× × Q of sodium is less than that of hydrogen. Q = 4.2889 × 105 C 2H+ + 2e– → H2 11. As we know that one Faraday electric charge = 4.2885 × 105 96500 deposits 1 g equivalent of the substance Na+ + e– → Na = 4.44 F Al3+ + 3e– → Al 17. (i) 2H2O æÆ 4H+ + O2 + 4e– For sodium, equivalent mass For 2 moles ofofrH12mOoilte requires 4 moles of electrons. So of H2O it requires 2 = Atomic mass But for Al, equivalent mass moles of electrons. = Atomic mass of Al 2 Faraday = 2 × 96500 C = 193000 C 3 = 1.93 × 105 C is the electricity required. Hence 3 Faraday electric charge is required to deposit 1 mole of aluminium. (ii) Fe2+ æÆ Fe3+ + e– 1 Faraday = 96500 C = 9.65 × 104 C 12. According to Faraday’s second law of electrolysis, is the electricity required. when same amount of current is passed through 18. Ni2+ + 2e– æÆ Ni(s) CuSO4 and AgNO3 solutions then, m=Z×I×t Mass of Cu deposited Eq. wt. of Cu Mass of Ag deposited = Eq. wt. of Ag m= 2 58.9 × 5.0 × 20 × 60 × 96500 m = 1.83 g Mass of Ag and Cu will be different because the equivalent mass of Ag is different from Cu. 19. (i) At cathode: Ag+ + e– æÆ Ag(s) At anode: Ag(s) æÆ Ag+(aq) + e– 13. Cl– is oxidised at anode instead of water because eleCtroChemistry 63

(ii) Ag NO3(aq) æÆ Ag+ + NO3 – (ii) Cu2+ + 2e– æÆ Cu(s) H2O æÆ H+ + OH– At cathode: Ag+ + e– æÆ Ag(s) 2 Faraday or 2 × 96500 C At anode: 2OH– æÆ O2 + 2H+ + 4e– = 193000 = 1.93 × 105 C is required Or 2H2O æÆ 4H+ + O2 + 4e– (iii) H2SO4 æÆ 2H+ + SO4 2– (iii) 8H+ + MnO4 – + 5e– æÆ Mn2+ + 4H2O, H2O æÆ H+ + OH– 5 Faraday or 5 × 96500 C At cathode: 2H+ + 2e– æÆ H2(g) At anode: 2OH– æÆ O2 + 2H+ + 4e– = 4.825 × 105 C (iv) CuCl2(aq) æÆ Cu2+ + 2Cl– H2O æÆ H+ + OH– 22. m = Z × I × t At cathode: Cu2+ + 2e– æÆ Cu(s) 1.45 = 108 ×1.5 × t 96500 t= 1.45 × 96500 108 × 1.5 At anode: 2Cl– æÆ Cl2(g) + 2e– 20. Mass of silver deposited = 863.7 seconds. W= it Z WCu = WAg fi WCu = 1.45 96500 ZCu Z Ag 63.5 108 = 0.2 × 3 × 3600 × 107.92 21 96500 WCu = 1.45 × 63.5 = 0.426 g = 2.4156 g 108 × 2 Mass = Area × thickness × density WZn = WAg fi WZn = 1.45 ZZn Z Ag 65 108 2.4156 = 800 × x × 10.47 x = 2.88 × 10–4 cm 21 21. (i) Al3+ + 3e– æÆ Al(s) WZn = 1.45 × 65 = 0.436 g 3 Faraday or 3 × 96500 C charge is required 2 × 108 = 289500 C = 2.895 × 105 C Topic 4. Batteries, Fuel Cells, Electrochemical Aspects of Corrosion and Hydrogen Economy Primary cells Dry cell (Leclanche cell) In these cells the reaction occurs only once and the battery then becomes dead after use over a period of • Anode: Zn container time. It cannot be recharged and reused again. For • Cathode: Graphite rod surrounded by powdered example, Dry cell, Mercury cell MnO2 and carbon Dry Cell • Electrolyte: A moist paste of NH4Cl and ZnCl2 • Anode: Zn-rod 64 Chemistry-12 • At anode: Zn(s) → Zn2+(aq) + 2e– • Cathode: Carbon rod • At cathode: MnO2 + NH4+ + 1e– → MnO (OH) + NH3 • Overall reactions: Zn + 2MnO2 + 2H2O → Zn(OH)2 + 2MnO(OH) • Cell potential: Nearly 1.5 V Mercury cell • Anode: Zn-Hg amalgam • Cathode: A paste of HgO and carbon • Electrolyte: A paste of KOH and ZnO • Anode: Zn(Hg) + 2OH– → ZnO(s) + H2O + 2e– • Cathode: HgO(s) + H2O + 2e– → Hg(l) + 2OH– • Overall cell reaction: Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)

• Cell potential is 1.35 V Mercury Cell Nickel-Cadmium Cell Secondary batteries: Once used they can be Fuel cells recharged again by passing the current through it in opposite direction and thus they are rechargeable and Galvanic cells which converts the energy of combustion resauable, e.g. Lead storage battery, Nickel-cadmium of fuels like hydrogen, methane directly into electrical cell. energy. For example, Hydrogen – Oxygen Fuel cell Lead storage Battery Hydrogen and oxygen are bubbled through porous carbon electrode into the concentrated aqueous sodium • Anode: Pb plate hydroxide solution. To increase the rate of electrode • Cathode: Grid of lead packed with PbO2. reaction catalysts like palladium or platinum is used • Electrolyte: 38% solution of sulphuric acid (1.3 At Anode: 2H2(g) + 4OH–(aq) → 4H2O(l) + 4e– At Cathode: O2(g) + 2H2O(l) + 4e– → 4OH–(aq) g/ml) • At Anode: Pb(s) + SO42–(aq) → PbSO4(s) + 2e– • At →CaPthbSoOde4(:s)P+bO2H2(s2)O+(l)SO42– (aq) + 4H+(aq) + 2e– • Overall cell reaction: Pb(s) + PbO2(s) + 2H2SO4 → 2PbSO4(s) + 2H2O(l) • Used in automobiles and inverters. On recharging the battery, the above reaction is reversed. Fuel cell Advantages: 1. High efficiency 2. Continuous source of energy 3. Pollution free Electrochemical Theory of corrosion Lead Storage Battery • A particular spot made up of iron acts as anode Nickel-cadmium cell and oxidation takes place at this spot as follows. • Anode: Cd • 2Fe(s) → 2Fe2+ + 4e–, E°(Fe2+/Fe) = –0.44 V • Cathode: Ni(OH)3 These e– s move through the metal to another spot • Overall cell reaction: where oxygen is reduced in the presence of H+ ions Cd(s) + 2Ni(OH)3(s) → which come wfraotmerHo2rCfOro3mforomtheedrdauceidtiocdoixsisdoelus.tion CdO(s) + 2Ni (OH)2(s) + H2O(l) of CO2 and • This spot behaves as cathode where reduction • Nickel cadmium cell has a longer life than the lead storage cell but it is more expensive to takes place as follows: manufacture. O2(g) + 4H+(aq) + 4e– → 2H2O(l) E°H+|O2|H2O = 1.23 V eleCtroChemistry 65

O2 + Fe2+ Fe2+ 2H2O 4H+(aq)+ O2 2. Cathodic protection or sacrificial protection: 2e– 4e– In this method a more reactive metal like Mg (or) Fe2O3 Fe2+ Zn are made as sacrificial anode and are connected Fe to the iron pipe or tank. Fe 2e– 3. Painting and using anti-rust solutions are some other methods to prevent corrosion of iron. Oxidation : Fe(s) → Fe2+(aq) + 2e– Reduction : O2(g) + 4H+(aq) + 4e– → 2H2O(l) Hydrogen Economy: Atomospheric oxidation: 2Fe2+(aq) + 2H2O(l) • Hydrogen provides an ideal alternative to fossil fuels as its combustion results in water only. + ½O2(g) → Fe2O3(s) + 4H+(aq) • Hydrogen production must come from splitting The overall cell reaction is: water using solar energy. Therefore, hydrogen 2Fe + O2 + 4H+ → 2Fe2+ + 2H2O, E°(cell) = 1.67 V can be used as non-polluting renewable source of energy. Method of Preventing corrosion • Both the production of hydrogen by electrolysis 1. Galvanisation: The process of coating zinc metal of water and by combustion in a fuel cell will be over iron metal. important in the future. Both the technologies are based on electrochemical principles. ExErcisE 2.4 Multiple Choice Questions (MCQs) (1 Mark) 6. How is the presence of CO2, SO2, NO2 etc. in water due to pollution affects the process of corrosion? 1. While charging the lead storage battery .................... . 7. Unlike dry cell, the mercury cell has a constant cell potential throughout its useful life. Why? (i) PbSO4 anode is reduced to Pb. [NCERT Exemplar] (ii) PbSO4 cathode is reduced to Pb. (iii) PbSO4 cathode is oxidised to Pb. Short Answer Type Questions-I (2 Marks) (iv) PbSO4 anode is oxidised to PbO2. 2. Which metal coating is done over iron during 8. What advantages do the fuel cell have over galvanisation? primary and secondary batteries? [NCERT Exemplar] (i) Mg (ii) Mn 9. Write the cell reaction of a lead storage battery, when it is discharged. How does the density of the (iii) Zn (iv) Ga electrolyte change when the battery is discharged? Assertion-Reason Type Questions (1 Mark) [AI 2011C][NCERT Exemplar] Note: In the following questions a statement of 10. Which cell was used in Apollo space programme assertion followed by a statement of reason is given. and what is its role in this programme? Choose the correct answer out of the following choices. 11. From the given cells: Lead storage cells, Mercury (i) Both assertion and reason are true and the cell, Fuel cell and Dry cell. reason is the correct explanation of assertion. Answer the following: (ii) Both assertion and reason are true and the reason is not the correct explanation of assertion. (i) Which cell is used in hearing aids? (iii) Assertion is true but the reason is false. (ii) Which cell was used in Apollo space programme? (iv) Assertion is false but the reason is true. (iii) Which cell is used in automobiles and 3. Assertion : Mercury cell does not give steady inverters? potential. (iv) Which cell does not have long life? Reason : In the cell reaction, ions are not involved in solution. [Delhi 2016] 4. Assertion : Fuels like H2, CH4 are used in fuel Short Answer Type Questions-II (3 Marks) cells. 12. Define and give one example of each of the Reason : Catalysts like Pd or Pt are used to following: increase the rate of electrode reaction. (i) Primary cell Very Short Answer Type Questions (1 Mark) (ii) Secondary cell 5. What is the principle of fuel cell? (iii) Fuel cell [Delhi 2008, 2014; Foreign 2008] 66 Chemistry-12

Answers 2.4 1. (iii) 2. (iii) 3. (iv) Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l) 4. (ii) Idniscthhaisrgcinelgl,thHe2dSeOn4sitiys used as electrolyte. On of the electrolytic solution 5. It is designed to convert the heat of combustion of a fuel directly to electricity. decreases. It is because H2SO4 is consumed and H2O is produced. 6. These oxides are acidic in nature. When they 10. In Apollo space programme hydrogen-oxygen fuel combine with water they form oxyacids H2O + CO2→ H2CO3 cell was used. It was used not only to provide H2O + SO2→ H2SO3 2NO2 + H2O → HNO3 + HNO2 electricity but also the drinking water for the These acids act as electrolytes for the micro cells astronauts. developed on the surface of the iron and facilitate 11. (i) Mercury cell the corrosion process. (ii) Fuel cell (H2–O2 fuel cell) 7. Electrolyte is not consumed in the cell process of (iii) Lead storage Battery mercury cell, hence it will deliver the current at constant potential throughout its life. (iv) Dry cell 12. (i) Primary cell: Irreversible and non rechargeable, e.g. dry cell. 8. Fuel cell is an environment friendly cell. It does not require recharging as required by the (ii) Secondary cell: Reversible and rechargeable, secondary batteries and never exhausted like e.g. lead storage battery. primary cells until reactants are supplied to it. In future, vehicles will be driven by fuel cells. (ii) Fuel cell: Galvanic cells are designed to convert energy of combustion of fuels like 9. The overall reaction and the discharge of lead hydrogen, methane, etc. into electricity. For storage battery takes place as: e.g. H2–O2 fuel cell. Case based questions 1. Case Study A semiempirical model with five parameters suitable for both pure and mixed solvents systems was proposed based on the influence of the electrolyte concentration and temperature on the number of free ions and ion mobility: K = (P1T + P2 ) mn exp  − P3m  . The conductivities  T − P4  obfinNaaryClsoalnvdenCtsaCwle2 rien mpreoapsyulreendea ctatrebmonpaetrea−tuHr2eOs varying from 283.15 to 333.15 K. The feasibility and accuracy of the proposed model were verified by the experimental data and the (k, m, T) data of 28 electrolyte solutions from the literature. The results showed that in a wide temperature (Reference: Weitao Zhang, Xia Chen, Yan Wang, Lianying Wu, and Yangdong Hu. Experimental and concentration range, the proposed model and Modeling of Conductivity for Electrolyte Solution Systems. ACS Omega 2020,5,22465- can fit the experimental data well for both pure 22474) The following questions are multiple choice and mixed solvents systems. The parameters questions. Choose the most appropriate answer: in the proposed model are related to the type of (i) What is the effect of change in temperature solvent, and the value of n is closely relevant to the on conductivity of electrolyte? dielectric constant of the solvent. In a solvent with (a) Conductivity increases with increase in a high dielectric constant, the value of n is close temperature. to 1. In addition, compared with the conductivity model reported in the literature, the proposed model can obtain good accuracy with fewer parameters, which is of great value to practical engineering applications. eleCtroChemistry 67

(b) Conductivity decreases with increase in (d) Conductivity decreases, molar conductivity temperature. increases (c) Conductivity increases initially and (iv) What is the correct unit of molar conductivity then becomes constant on increasing of an electrolyte solution? temperature. (a) S cm–1 (b) S cm (d) No change in conductivity on change in (c) S cm–1mol–1 (d) S cm2 mol–1 temperature. 2. Case Study The development of advanced energy storage (ii) What are the factors that influence devices is at the forefront of research geared conductivity of electrolytes? towards a sustainable future. Nano-structured (a) Number of free ions materials are advantageous in offering huge (b) Ion mobility (c) Both (a) and (b) surface to volume ratios, favourable transport (d) Number of free electrons features, and attractive physicochemical (iii) In the proposed model, the value of ’n’ is properties. closely related to which property of the Nanostructured Mo-based electrode materials solvent? aidnriccelhuasdtlicunodgnieimddeosflyo(brMdieomnXpu2r,moXvoex=mideSen,stS(Mien)o, Oraexnc,dh2ao≤rxgxyes≤aabl3tl)es, (a) Molar mass (b) Charge (c) Size (d) Dielectric constant lithium/sodium-ion batteries, Mg batteries, and OR supercapacitors. Mo-based compounds including How conductivity and molar conductivity M= oFOe,2,CMo,oON3i,, CMao,OM3–ny ,(0Zn<, y< 1), CMdM; xoxO= y1(,My of a weak electrolyte vary with increase in Mg, or concentration? Ls=tiaM4t;eosxOa2=,ndL3,ie2yxMh=oiOb8i3t),,reiMtccho. Scph2o,esmsMeiosssStremy2.,uT(ltMhipeoylOea2v)ra2ePlev2neOrc7ye, (a) Conductivity decreases, molar conductivity decreases (b) Conductivity increases, molar conductivity attractive candidates for efficient electrochemical increases energy storage systems because of their unique (c) Conductivity increases, molar conductivity physicochemical properties, such as conductivity, decreases mechanical and thermal stability and cyclability. Mo-based materials with rich chemistry Supercapacitor MMooOS22 AMMooOO34 Mg battery MMxMo6oS68S8 MoS2 MoO3 Na-ion battery MMooSSe22 Li-ion battery LLiixMMooOO22 MLniMM22MMooOSooO232O38NMaxoMOo3O-x 3 (MMoAoOMOM32o.)oOn2OPH2422OO7 Multivalent Molybdenum III IV V VI (Reference: Xianluo Hu, Wei Zhang, Xiaoxiao (a) Assertion and reason both are correct Liu, Yueni Mei and Yunhui Huang, 2015, statements and reason is correct explanation Nanostructured Mo-based electrode materials for for assertion. electrochemical energy storage, Chemical Society Reviews, 44(8), 2376-2404) (b) Assertion and reason both are correct In these questions, a statement of assertion statements but reason is not correct followed by a statement of reason is given. explanation for assertion. Choose the correct answer out of the following choices. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. 68 Chemistry-12

(i) Assertion: Nanostructured materials are Reason: (MoO2)2P2O7 is used in advantageous in energy storage devices. supercapacitors. (iv) Assertion: MoS2 is a dichalconide with Mo Reason: These materials offer low surface in oxidation state of +4. area to volume ratio. Reason: MoS2 can be used in supercapacitors, Mg batteries and Na-ion batteries but not in (ii) Assertion: Mo-based nanostructured Li-ion battery. material have unique physicochemical properties which helps in making efficient OR energy storage devices. Assertion : MMoxOy where, M = Li, x = 1 and y = 4 has attractive physicochemical Reason: Conductivity, stability (mechanical properties. and thermal) and cyclability are such Reason: LiMoO2 is used in Li-ion batteries. physicochemical properties. (iii) Assertion: (MoO2)2P2O7 has Mo in oxidation state of +6. Answers 1. (i) (a) (ii) (c) (iii) (d) OR (c) (iv) (d) 2. (i) (c) (ii) (b) (iii) (c) (iv) (c) OR (d) Analogy based questions 2. Complete the following analogy 1. Which of the following analogies is correct? Nickel-Cadmium Cell : A :: Lead storage Battery (i) Galvanic cell : Daniel cell :: Voltaic cell : :B Electrolysis of NaCl (ii) Conductance ; Sm–1 :: Cell Constant : m–1 (i) A : Longer life, B : Shorter life (iii) Degree of dissociation : Λm :: (ii) A : Cheaper, B : Expensive Λ 0m (iii) A : Rechargeable, B : Non-Rechargeable (iv) A : Zinc Anode, B : Lead Anode Dissociation Constant : cα2 1− α 1. (iii) Answers (iv) Molar conductivity : Decreases with dilution 2. (i) :: Conductivity : Increases with dilution Multiple type questions 1. Match the terms given in Column I with the units given in Column II. Column I Column II A. Lm (1) S cm–1 B. Ecell (2) m–1 C. k (3) S cm2 mol–1 D. G* (4) V Code: (i) A (4) B (3) C (2) D (1) (ii) A (3) B (4) C (1) D (2) B (2) C (3) D (4) (iii) A (2) B (1) C (4) D (3) (iv) A (1) 2. Match the items of Column I and Column II. Column I Column II A. Lechlanche cell (1) cell reaction 2H2 + O2 Æ 2H2O B. Ni–Cd cell (2) does not involve any ion in solution and is used in hearing aids. eleCtroChemistry 69

C. Fuel cell (3) rechargeable (4) reaction at anode, Zn Æ Zn2+ + 2e– D. Mercury cell (ii) A (1) B (4) C (3) D (2) Code: (iv) A (4) B (1) C (2) D (3) (i) A (4) B (3) C (2) D (1) Answers (iii) A (3) B (4) C (1) D (2) 1. (ii) 2. (i) Quick revision notes • Element with lowest value of reduction potential • An electrochemical cell is a cell which either produces electric current as a result of redox is the best reducing agent whereas the one with reaction or facilitate a chemical reaction (redox reaction) through the introduction of electrical higher value of reduction potential is the best energy. oxidising agent. • Electrolyte is an important component of electrochemical cell. • If the given electrode when coupled with SHE • In galvanic cell, the chemical energy of a (Standard hydrogen electrode) acts as anode, spontaneous redox reaction is converted into electrical energy. we give negative sign to its standard reduction potential, e.g. E0Zn2+/Zn = – 0.76 V. • In electrolytic cell, electrical energy is used to • If the given electrode acts as cathode when carry out a non-spontaneous redox reaction. coupled with SHE, then we give positive sign to • The standard electrode potential of any electrode its reduction potential, e.g. E0Cu2+/Cu = + 0.34 V. dipped in a suitable solution is defined with • In the cell representation, anode is on the left respect to the standard electrode potential of hydrogen electrode, taken as zero. hand side and cathode is on the RHS. • The standard potential of cell can be obtained • Pt is an inert electrode used in SHE and as an by difference in standard reduction potential of cathode and anode, electrode in solution containing ions only, e.g. Pt(s)/Fe3+/Fe2+. • The conductivity (k) of an electrolytic solution depends upon the concentration of electrolyte, E0 = E0 E0 nature of solvent and temperature. cell cathode anode • – where SRP stands for • wMhoelnar‘Cc’oins dmuoclatirvitiytyin(Lmmo)lisL–d1eafinnded‘ka’ sin1S00cm0 k–1/C. (SRP) (SRP) standard reduction potential. • Conductivity decreases but Lm increases with decrease in concentration. • Cathode is +ve and anode is –ve in Galvanic cell but in an electrolytic cell anode is +ve, cathode • oLfmsitnrocrnegaesleescstlroowlyltyews iwthhidleecirnecarseeaisnescovnecreynsttreaetpiolny is –ve. • The standard potential of the cells and equilibrium for weak electrolytes in very dilute solution. constant are related to standard Gibbs energy as: • Kohlrausch found that molar conductivity at Dr G0 = – nE0cellF infinite dilution for an electrolyte is the sum of Dr G0 = – 2.303 RT log Kc • Concentration dependence of the potential of the contribution of the cation as well anions. It the electrodes and the cells are given by Nernst is known as law of independent migration of ions equation as: and has many applications. For a general reaction, aA + bB ne−→ cC + dD • Electrolysis is a process of breaking down of an electrolyte into simpler substances by passing Ecell = E0cell – RT ln [C]c [D]d electricity through molten state or in aqueous nF [A]a [B]b solution. • The quantitative aspects of electrolysis are • Electrochemical series is a series of elements governed by Faradays laws of electrolysis. in which elements are arranged in increasing or decreasing order of their standard reduction • The products of electrolysis are decided by potential. preferential discharge theory as well as reduction potentials of electrodes. 70 Chemistry-12

• Ions conduct electricity through the solution but • Primary cells are non-rechargeable while oxidation and reduction of ions take place at the secondary cells are rechargeable. electrodes in electrochemical cells. • Corrosion of metals is an electrochemical • Batteries and fuel cells are very useful forms of phenomenon. galvanic cells. • Electrochemical principles are also relevant to the Hydrogen economy. Important FormulaE 1. E0cell = E0cathode – Ea0node 9. κ = 1 = 1 c l m = G l ρ R A A 2. Ecell = E0cell – 2.303 RT log [Anode ion] nF [Cathode ion] 10. Lm = Specific conductance = κ Molarity C 3. E0cell = 0.059 log Kc (At 298 K) n 11. Lm = κ × 1000 C 4. DG° = – nFE0cell K c 5. DG° = – RTlnK 12. a= m o 6. R= l = 1 l K m A κ A ρ 13. Ka = Ca2 1–a 7. Conductance: G = 1 R 14. Kohlrausch’s Law Λ0m = v+ λ+0 + v– λ0– 8. Conductivity = Conductance × Cell constant 15. Faraday constant = 96487 C mol–1 Common Errors Errors CorrECtIons (i) Students make mistakes in identification of (i) For electrochemical cells, use the concept of anode and cathode. LOAN-Left-Oxidation-Anode-Negative while for cathode it will be opposite of this or clarity (ii) Students generally write wrong Nernst equation. in terms of E° values and overall reaction need to be given. (ii) Whenever ‘+’ sign is used in the Nernst equation then, the log term should be log[reactant]/ [product] Ecell =Ec°ell + 2.303RT log [reactants] nF [products] Ecell =Ec°ell – 2.303RT log [products] nF [reactants] (iii) Students mistakenly made wrong substitution (iii) Practice on many cell reactions. in the equations and formulae. (iv) Students mistakenly identify wrong products of (iv) If electrodes are inert, they do not involve in electrolysis reaction or E° values to be stressed upon. (v) Students generally have confusion in terms (v) Proper understand the concept in terms of like conductance, conductivity and molar definitions and symbols. conductivity. (vi) Students have confusion in determining the number (vi) Students have to write the overall reaction (if not of electrons gained or lost during a redox reaction given) and find the difference in total charge of or in a given cell representation. each reacting species. eleCtroChemistry 71

rEvIsIon Chart Electrochemistry: A branch of science that deals with  Electrochemical cells (Galvanic or Voltaic cells): the study of production of electricity from spontaneous Devices which converts chemical energy into electrical chemical reactions, or the use of electrical energy to bring energy. about non-spontaneous chemical reactions.  Based on spontaneous redox reactions.  Electrolytic cells: Devices which convert electrical energy into chemical energy.  Two electrodes, namely cathode and anode containing  Used to carry out non spontaneous chemical reactions. different electrolytes.  Both the electrodes are dipped in the same container, containing an electrolyte.  A salt bridge/porous pot acts as a set up for the  No salt bridge is used in such cells. transference of electrons. Nernst Equation:  For a reduction reaction: Mn+(aq) + ne– Æ M(s) Ecell = Ecoell − 2.303RT log 1 ] nF [Mn+  Electrolysis is a process of decomposition of electrolyte = Eocell − 0.0n591log 1 ] (at 298 K) [Mn+ on passing electricity through it.  Applications of Nernst Equation:  Farday’s Law of Electrolysis: (a) To calculate electrode potential of cell : Ist Law: Amount of substance deposited on the aA + bB ne−→ cC + dD electrode is directly proportional to the quantity of Ecell = 0.0n591log [C]c [D]d (at 298 K) electricity passed i.e., w ∝ Q or w = ZQ = Z × I × t, where Eocell − [A]a [B]b Z is electrochemical equivalent of the substances. (b) To calculate equilibrium constant: 2nd Law: The amount of different substances liberated At equilibrium, Ecell = 0 \\ Eocell = 0.0n591log Kc at 298 K during electrolysis is proportional to their chemical (c) Relation between cell potential and Gibbs Energy equivalent weights, i.e. w1 = E1 , where E Æ equivalent weight. w2 E2 change: DG° = – nF E0cell = –2.303 RT log Kc  Corrosion (eg. Rusting of iron) is an electrochemical  Limiting molar conductivity: Molar conductivity at process, involving slow eating away of metals in the presence of air and moisture. infinite dilution (or when concentration approaches zero).  Barrier method, sacrificial method and anti-rust solutions can be used to prevent (rusting) corrosion of  Molar conductivity: Conductivity power of all the ions iron metal. produced by dissolving 1g mole of an electrolyte placed  Primary cells: Once used will get exhausted.  Secondary cell: Can be recharge again, once used. between two electrodes at 1 cm apart.  Fuel cells: Convert energy of combustion of fuels, like H2, CH4, etc. into electrical energy.  Lm = κ ×1000 C  Degree of dissociation =(a)(1=C−αΛΛα2 0cmm)  Strong electrolytes: Ionises completely in aqueous solutions.  Kc for an electrolyte Kc  Weak electrolytes: Ionises only partially in aqueous where C = conc. of ions. solutions. Kohlrausch’s Law: It states that limiting molar conducting of an electrolyte is equal to the sum of the individual contributions by cation and anion, respectively. It is used to determine molar conductivity of weak electrolytes. 72 Chemistry-12

ChaptEr trEnd—Based on past Years’ CBsE Exams ➣ It has been observed from this chapter that the weightage of topics ‘Conductivity and its Variation for Electrolytic Solutions’(especially, numericals based on Lm = κ × 1000 , a = Λ c , K = cα2 ) and‘Electrochemical Cells and Electrode C m 1− α 0 Λ m Potentials’(especially, numericals based on Nernst equation) are maximum. Hence, these are the most important topics. ➣ From this chapter , generally 3 marks questions were asked from the topic ‘Electrochemical Cells and Electrode Potentials’. ➣ Most of the 1, 2 and 5 marks questions from this chapter belong to the topic ‘Conductivity and its Variation for Electrolytic Solutions’. Questions For practice Very Short Answer Types Questions (1 Mark) 1. Why does the conductivity of a solution decreases with dilution? 2. Suggest a way to determine the Lm0 value of water. Short Answer Types Questions-I (2 Marks) 3. Calculate the conductivity of 1.5 M solution of an electrolyte whose molar conductivity is 138.9 S cm2 mol–1. 4. Resistance of a conductivity cell filled with 0.1 mol L–1 KCl solution is 100 W. If the resistance of the same cell when filled with 0.02 mol L–1 KCl solution is 520 W, calculate the conductivity and molar conductivity of 0.02 mol L–1 KCl solution. The conductivity of 0.1 mol L–1 KCl solution is 1.29 S/m. Short Answer Types Questions-II (3 Marks) 5. Calculate E0cell and DrG° for the following reaction of 25°C. A2+(aq) + B+(aq) → A3+(aq) + B(s) [Given: KC = 1010, 1F = 96500 C mol–1] [AI 2015] 6. Estimate the minimum potential difference needed to reduce Al2O3 at 500°C. The Gibb’s free energy change for decomposition reaction: 2 4 3 Al2 O3 → 3 Al + O2 is 960 kJ mol–1. (F = 96500 C) [Delhi 2014(C)] [HOTS] Long Answer Types Questions (5 Marks) 7. Consider the given figure and answer the following questions: (i) What is the direction of flow of electrons? Zinc Plate (ii) Which is anode and which is cathode? Salt bridge Silver Plate Zn2+(aq) Ag+(aq) (iii) What will happen if salt bridge is removed? (iv) How will concentration of Zn2+ and Ag+ ions be affected when cell functions? (v) How will concentration of these ions be affected when the cell becomes dead? [Delhi 2017(C)] Answers 3. 0.20835 S cm–1 4. 0.248 Sm–1, 0.0124 Sm2mol–2 5. Ec°ell = 0.59, DrG0 = –57.058 kJ mol–1 6. E0 =– 2.48 V eleCtroChemistry 73

assIgnmEnt Time: 45 Minutes M.M.: 25 Multiple Choice Questions (MCQs) (1 Mark) 1. Which of the following statements about solutions of electrolytes is not correct? (i) Conductivity of solution depends upon size of ions. (ii) Conductivity depends upon viscosity of solution. (iii) Conductivity does not depend upon solvation of ions present in solution. (iv) Conductivity of solution increases with temperature 2. The cell constant of a conductivity cell .................... . (i) changes with change of electrolyte. (ii) changes with change of concentration of electrolyte. (iii) changes with temperature of electrolyte. (iv) remains constant for a cell Assertion Reason Type Questions (1 Mark) In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (i) Assertion and reason both are correct statements and reason is correct explanation for Assertion. (ii) Assertion and reason both are correct statements but reason is not correct explanation for Assertion. (iii) Assertion is correct statement but reason is wrong statement. (iv) Assertion is wrong statement but reason is correct statement. 3. Assertion: For measuring resistance of an ionic solution an AC source is used. Reason: Concentration of ionic solution will change if DC source is used. 4. Assertion: Conductivity of all electrolytes decreases on dilution. Reason: On dilution number of ions per unit volume decreases. Very Short Answer Types Questions (1 Mark) 5. Can you store copper sulphate solution in a zinc pot? Short Answer Types Questions-I (2 Marks) 6. Calculate the equilibrium constant of the reaction: Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s) E(cell) = 0.46 V. 7. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10. 8. The standard electrode potential for Daniell cell is 1.1 V. Calculate the standard Gibbs energy for the reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) 9. The conductivity of 0.001028 mol L–1 acetic acid is 4.95 × 10–5 S cm–1. Calculate its dissociation constant if Lm0 for acetic acid is 390.5 S cm2 mol–1. Short Answer Types Questions-II (3 Marks) 10. Represent the cell in which the following reaction takes place: Mg(s) + 2Ag+(0.000 1M) → Mg2+(0.130 M) + 2Ag(s) Calculate its E(cell) if E(cell) = 3.17 V. 11. The electrical resistance of a column of 0.05 mol L–1 NaOH solution of diameter 1 cm and length 50 cm is 5.55 × 103 ohm. Calculate its resistivity, conductivity and molar conductivity. 12. The molar conductivity of 0.025 mol L–1 methanoic acid is 46.1 S cm2 mol–1. Calculate its degree of dissociation and dissociation constant. Given λ0(H+) = 349.6 S cm2 mol–1 and λ0(HCOO–) = 54.6 S cm2 mol–1. 13. If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire? 1. (iii) 2. (iv) Answers 3. (i) 4. (i) 5. No 6. KC = 3.89 × 1015 7. –0.591V 8. D°G = –212.3 kJ mol–1 10–21W39–..1c2Km.a2–4=1,×1L.17m07=2×22e13l00e–cW5tr–mo1ncomsl L2m–1ol–1 10. r = 87.21 ohm cm, k = 12. E(cell)= 2.96V 11. 1.15 × a = 0.114, Ka = 3.67 × 10–4 mol L–1 74 Chemistry-12

3 Chemical Kinetics Topics covered 3.1 Rate of a Chemical Reaction 3.2 Factors Influencing Rate of a Reaction 3 .3 Integrated Rate Equations and Pseudo First Order Reaction 3.4 Effect of Temperature on Reaction Rate and Collision Theory C hapter map Topic 1. Rate of a Chemical Reaction If D[A] represents the change in concentration Rate of a Reaction and Dt the time taken for the change, then The rate of a reaction tells us the speed at which it occurs. Rate of Reaction = ∆ [A] ∆t Let us consider a simple reaction:A → B The concentration of reactants decreases as the Since the concentration of the reactant decreases, time passes. Suppose the concentration of A at therefore D[A] is a negative quantity. We know that one mole of A produces one mole of B. triamteeatt1 =wh[Aic]h1 and caotntciemnetrta2 t=io[nA]o2f. Therefore, the the A has changed Therefore, the rate of reaction is also given as: may be expressed as −∆ [A] ∆ [B] ∆t ∆t Change in concentration of A = [A]2 − [A]1 Rate of Reaction = = Time taken for the change (t2 − t1 ) 75

• Rate of chemical reaction may be defined as the • At a fixed temperature, the rate of a given reaction depends on the concentration of reactants. change in the concentration of a reactant (or • An expression which shows how the reaction rate product) in a given time interval. is related to the concentration is called the rate law or rate equation. The rate of a reaction is proportional to the concentration of the reactants. As the reaction • The power (exponent) of concentration in the rate progresses, the concentration of the reactants law is usually a small whole number integer (1, changes and the rate of reaction also changes. 2, 3) or a fractional number. Thus, the rate of reaction can be calculated by • The proportionality constant k is called the rate constant for the reaction. dividing the total concentration change by the • Usually, the power of concentration in the rate total time of change is the average rate, or law are different from the coefficients in balanced chemical equation. Average reaction rate • Rate law for a reaction must be determined by = Total change of concentration experiments. It cannot be written merely by Time needed for the change looking at the equation with a background of our knowledge of law of mass action. The units of reaction rates may be: How Fast are Chemical Reactions ã mole/litre sec or mol L–1 s–1 • The rate of chemical reactions vary from the ã mole/litre min or mol L–1 min–1 slowest with reaction time of the order of millions of years to the fastest with reaction times of the ã mole/litre hour or mol L–1 h–1 order of 10–12 seconds. • The average rate is obtained over a large time Modern techniques, mainly using lasers, are now available to study extremely fast reactions. The interval during which the rate of reaction changes range of reaction times observed cover as many significantly. So a better way to estimate the rate as 26 orders of magnitude: 1 picosecond = 10–12 of a reaction is to make the time interval as small second to 1 million years = 3 × 1013 seconds. Some as possible. examples are: • If the interval be infintesimally small (i.e. as Dt approaches zero), the rate is referred to as the instantaneous rate and is written as Rate = d [c] dt where [c], is the concentration at time t. Reaction Reaction time (half life) 1. First step in photosynthesis 10–12 s 2. H+ (aq) + OH– (aq) → H2O (l) 10–10 s 3. NO2 (g) + NO2 (g) → N2O4 (g) 10–6 s 4. Fe3+ + SCN– → Fe (SCN)3 5 × 10–3 s 5. H2CO3 → H2O + CO2 60 seconds 6. N2O5 → 2NO2 + 1/2O2 (at 55°C) 89 mins. 7. Sucrose + Water → Glucose + Fructose 200 mins. 8. [Cr(H2O)6]3+ (aq) + F– (aq) → [Cr(H2O)5F]2+ (aq) + H2O (l) 1000 hrs. 9. CH3COOC6H5 (aq) + H2O (l) → CH3COOH (aq) + C6H5OH (aq) 1000 days 10. [Cr(H2O)6]3+ (aq) + I– (aq) → [Cr(H2O)5I]2+ + H2O (l) 2000 years • Consider example: CH3COOC2H5 H2 O (in excess) • The reaction seems to be of second-order but Hydrolysis in practice it is found to be first-order as the cisonacpesneturadtoi-ofnirsotf-oHr2dOerisreiancltaiorng.e excess. Thus, it CH3COOH + C2H5OH Here, Rate = k[CH3COOC2H5] 76 Chemistry-12

ExERCisE 3.1 Multiple Choice Questions (MCQs) (1 Mark) [B] 1. A graph of volume of hydrogen released vs time (ii) Concentration for the reaction between zinc and dil. HCl is given in the figure. On the basis of this mark the correct option. V5 Time [A] V4 [B] V3 V2 (iii) Concentration V1 O 20 30 40 50 Time [A] [A] (i) Average rate upto 40 seconds is V3 – V2 40 V3 – V2 (ii) Average rate upto 40 seconds is 40 – 30 Concentration (iii) Average rate upto 40 seconds is V3 (iv) 40 4V03 – 2V01 (iv) Average rate upto 40 seconds is – [B] Time 2. Consider the graph given in above question. Very Short Answer Type Questions (1 Mark) Which of the following options does not show instantaneous rate of reaction at 40th second? 5. What do you mean by Chemical Kinetics? (i) V5 − V2 (ii) V3 − V2 6. What is rate? 50 − 30 40 − 30 7. Explain the difference between average rate and (iii) V3 − V2 (iv) V3 − V1 instantaneous rate of a chemical reaction. 40 − 30 40 − 20 [NCERT Exemplar][AI 2010 C] 3. Which of the following expressions is correct for 8. How can you determine the rate law of the the rate of reaction given below? following reaction: 5Br– (aq) + BrO3– (aq) + 6H+ (aq) → 2NO(g) + O2(g) → 2NO2(g) 3Br2 (aq) + 3H2O (l) 9. Why does the rate of any reaction generally (i) ∆[Br− ] = 5 ∆[H+ ] (ii) ∆[Br− ] = 6 ∆[H+ ] decreases during the course of reaction? ∆t ∆t ∆t 5 ∆t [NCERT Exemplar] (iii) ∆[Br− ] = 5 ∆[H+ ] (iv) ∆[Br− ] = 6 ∆[H+ ] 10. State a condition under which a bimolecular ∆t 6 ∆t ∆t ∆t reaction kinetically becomes a first order reaction. 4. Consider the reaction A Æ B. The concentration [NCERT Exemplar] of both the reactants and the products varies 11. For which type of reactions order and molecularity have the same value? [NCERT Exemplar] exponentially with time. Which of the following 12. For zero order reaction will the molecularity be figures correctly describes the change in equal to zero? Explain. [NCERT Exemplar] concentration of reactants and products with 13. Why can’t molecularity of any reaction be equal to zero? time? [B] 14. Why rate of reaction does not remain constant throughout? [Delhi 2011] Concentration (i) 15. For a reaction A → B, the rate of reaction −d[A] d[B] [A] can be denoted by dt or + dt .What is the Time significance of plus and minus signs? ChemiCal KinetiCs 77

16. Express the rate of reaction, 2N 2cOh5ange →of (i) N2 + 3H2 → 2NH3; − d[H2 ] =? N2N2O2O5,4(i+i) Oco2nucesnintrgat(iio) ncochnacenngteroaftiOo2n. dt 17. raFanotdretYoh2fearreneadaccwttiioiotnhn:itnXh2ete+arpm2pYse2aorfadnics→eapo2pfXepYarro2a,dwnucrceittXeofYthX2.e2 (ii) H2 +  1 O2 → H2O; − d[O2 ] = ?  2 dt 18. What is the usefulness of initial rate method? (iii) 2NO + Cl2 → 2NOCl; d[NOCl] = ? dt 19. For a chemical reaction: 2A + B → C, the rate of formation of C is x mol L–1 h–1. What is the rate (iv) CO + NO2 → CO2 + NO; − d[NO2 ] = ? dt of disappearance of A and B? 29. Following reaction takes place in one step: 20. Define rate of a reaction. What are its units? 2NO (g) + O2 (g) → 2NO2 (g) [Delhi, AI 2010] How will the rate of above reaction changes if Short Answer Type Questions-I (2 Marks) the volume of the reaction vessel is diminished 21. The study of kinetics has been divided into two to one third of its original volume? Will there be parts, what are these two parts? any change in the order of the reaction with the 22. For the reaction: reduction in volume? [NCERT Exemplar] 4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O(g) Short Answer Type Questions-II (3 Marks) If the rate expression in terms of disappearance 30. List out the experimental techniques used to of NH3 is − ∆ (NH3 ) . Write the rate expression in measure the rate of reaction. ∆t terms of concentrations of O2 and H2O. 31. Thermodynamic feasibility of the reaction alone 23. Why can we not determine the order of reaction by cannot decide the rate of the reaction. Explain taking into consideration the balanced chemical with the help of one example. equation? [NCERT Exemplar] 24. Why molecularity is applicable only for elementary 32. On the basis of rates how will you classify reactions and order is applicable for elementary chemical reactions? Give example of each. as well as complex reactions?[NCERT Exemplar] 33. For an elementary reaction, 2A + B → 3C, the rate of appearance for C at time t is 25. Define the rate of a reaction. For the reaction 1.3 × 10–4 mol L–1 s–1. Calculate at this time: X → Y, the rate of reaction can be denoted by, (i) rate of reaction dX dY − dt or + dt (ii) rate of disappearance of A State the significance of the plus and minus signs 34. Decomposition of N2O5 is expressed by the equation: in these expressions. 1 26. The rate expression for the reaction, A + B2 → N2O5 → 2NO2 + 2 O2 dx Products is dt = k [A] [B2]2. If the concentration If during certain time interval the rate of mdaneicdno–Om1,2pwdohusairttiinowgnitllohfebeNsat2hmOe5eriiasntt1ee.ro8vfa×flo?r1m0–a3tmionololfitNreO–12 of both rAeaacntidonBi2sacrheatnrgipedle.d,[NbyCEwRhaTtEfaxcetmorpltahre] 35. In hydrogenation reaction at 25°C, it is observed rate of that hydrogen gas pressure falls from 2 atm to 1.2 atm in 50 min. Calculate the rate of reaction 27. oFforretahcetiroenacctaionnb, e2Nex2Opr5essed→in4NthOre2 e+ wOa2,ytsheasr:ate in moles per second (R = 0.0821 litre atm Kelvin–1 mol–1) − d[N2O5 ] = k [N2O5 ], + d[NO2 ] = dt dt k′ [N2O5 ], + d[O2 ] = k″ [N2O5 ] 36. From the concentrations of R at different times dt gives below, calculate the average rate of the reaction: R → P during different intervals of What is the relationship between k and k′ and time. between k and k′′? 28. For each of the following reactions express the Time (sec) 0 5 10 20 30 given rate of change of the concentration of the reactants or products in terms of the rate of [R] × 103 mol L–1 160 80 40 10 2.5 change of concentration of the other reactants or products in that reaction. 37. The following reaction was carried out in a closed vessel: 78 Chemistry-12

2N2O5(g) 4NO2(g) + O2(g) Calculate: It was fo und that the concentr aftivioenseocfoNndOs2. (i) the rate of reaction. increases by 2 × 10–2 mol L–1 in (ii) the rate of change in concentration of N2O5. Answers 3.1 1. (iii) 2. (ii) reactant can be individually determined. It avoids the complications caused by the presence 3. (iii) 4. (ii) of products. 5. The branch of chemistry which is concerned with 19. Rate of disappearance of reaction rates and the sequence of steps by which A = –2x mol L–1 h–1; the reaction occurs is called chemical kinetics. Rate of disappearance of 6. Rate is defined as the change in a particular B = –x mol L–1 h–1 quantity per unit time interval. Mathematically, 20. Change in the concentration of the reacting Rate = Change in quantity species per unit interval time is called the rate of Time taken for the change reaction. Its units is mol L–1 time –1. 7. The rate measured over a long time interval 21. The first part addresses the rate of reactions and during which rate of reaction changes significantly the second part is concerned with collision theory is called average rate and the rate measured for of reactions and explanation of the mechanism of an infinitely small time interval is called the reactions. instantaneous rate. 22. For the given reaction 8. Rate law and order of reaction can be determined by initial rate method. − 1 ∆(NH3 ) = − 1 ∆(O2 ) 4 ∆t 7 ∆t 9. Rate of reaction decreases with passage of time due to decrease in the concentration of reactants. = 1 ∆ (NO2 ) = + 1 ∆(H2O) +4 6 ∆t 10. Let us consider a biomolecular reaction ∆t A + B → Product \\ 4 ∆(O2 ) = + 2 ∆(H2O) = − ∆(NH3 ) Rate = k [A] [B] −7 ∆t 3 ∆t ∆t When the concentration of [B] taken in large excess than rate law will become 23. Order of reaction is an experimentally determined quantity, it does not depend on the balanced Rate = k′ [A], where k′ = k [B] chemical equation written to describe the The order reaction will then be equal to one. reaction. For example 11. Elementary reactions have the same value of KClO3 (s) + 6FeSO4 (aq) + 3H2SO4 (aq) → order and molecularity. KCl + 3Fe2 (SO4)3 + 3H2O 12. Molecularity of reaction can never be zero. This It is actually a second order reaction, which is not is because there is no possibility that no reactant illustrated from the balanced chemical equation. molecule is involved in the formation of a product without collisions with other reactant molecules 24. A complex reaction involves several elementary and undergoing a change to form product reactions. Molecularity of each step is theoretically molecules. calculated. Thus, there is no physical significance of the molecularity of overall reaction. Order of 13. Molecularity is the number of molecules involved reaction is however, determined from the slowest in the elementary process. Its value is at least one. step of the reaction. Thus, zero molecularity is not possible. 25. Change in the concentration of the reacting 14. Rate depends upon the concentration of the species per unit interval time is called the rate of reactants which keeps on changing. reaction. Its units is mol L–1 time –1. (–) sign shows the consumption of reactants and (+) sign shows 15. (–) sign indicates decrease in the concentration of the formation of products. A. (+) sign indicates increase in the concentration 26. Rate = k[A] [B2]2 ...(i) of B, i.e. product formation. New Rate = k[3A] [3B2]2 ...(ii) From (i) and (ii) 16. − d[N2O5 ] ; + d[O2 ] 2 dt dt New rate = 27 × Rate 17. Rate = − d[X2 ] = − d[Y2 ] =+ d[XY2 ] 27. 2N2O5(g) → 4NO2(g) + O2(g) dt 2dt 2 dt 1 d[N2O5 ] = + 1 d[NO2 ] = + d[O2 ] 18. This can be applied when the number of reactants −2 dt 4 dt dt are more than one. Order with respect to each 2k = 4k′ = k′′ ChemiCal KinetiCs 79

28. (i) − d[H2 ] = − 3 d[N2 ] = + 3 d[NH3 ] (i) Rate of reaction: dt dt 2 dt = 1 d(C) = + 1 × 1.3 × 10−4 +3 dt 3 (ii) d[O2 ] 1 d[H2 ] 1 d[H2O] − dt = −2 dt = 2 dt = 4.33 × 10–5 mol L–1 s–1 d[NOCl] d[NO] 2 d[Cl2 ] (ii) Rate of disappearance of A: dt dt dt (iii) = − = − = d(A) = 2 d(C) − dt 3 dt d[NO2 ] d[CO] d[CO2 ] d[NO] (iv) − dt = − dt = dt = dt = 2 × 1.3 × 10–4 3 29. Rate = k[NO]2 [O2] If volume of the vessel is diminished to one third, = 8.66 × 10–5 mol L–1 s–1 the concentration will tripled. 34. The rate expression for the decomposition of N2O5 is: New rate = k[3NO]2[3O2] = 33 rate \\ Rate will increase by 27 times. − ∆[N2O5 ] = 1 ∆[NO2 ] = 2∆[O2 ] Order will not get affected by the volume change. ∆t 2 ∆t ∆t 30. (i) By measuring the volume of the gaseous So ∆[NO2 ] = 2 ∆[N2O5 ] product ∆t ∆t (ii) By measuring the change in pressure = 2 × 1.8 × 10–3 (iii) By pH measurements = 3.6 × 10–3 mol L–1 min–1 (iv) By measuring the optical rotation and ∆[O2 ] = 1 ∆[N2O5 ] = 1 × 1.8 × 10−3 ∆t 2 ∆t 2 (v) By volumetric analysis method = 0.9 × 10–3 mol L–1 mol–1 31. Thermodynamic feasibility of a reaction depends on Gibbs free energy, i.e. DG which must be always Rate is always positive, hence – ∆(N2O5 ) is taken negative for a spontaneous process. Kinetic as positive. ∆t feasibility depends on the activation energy of reaction. Lesser the activation energy, greater 35. Rate = dP = 2 − 1.2 will be the feasibility of reaction. dt 50 × 60 i.e. Diamond → Graphite, = 2.666 × 10–4 atm s–1 DG = –ve PV = nRT, P =  n RT  V  The process is thermodynamically feasible but it is very slow due to its high activation energy. P = CRT 32. On the basis of rates, chemical reactions are dP =  dC RT classified into three categories. dt  dt  (i) Very fast or instantaneous reactions: These or  dC =  dP  1 are very fast  dt   dt  RT Eg.: AgNO3 + NaCl → AgCl + NaNO3 As rate (in mol L–1 s–1) =  dC HCl + NaOH → NaCl + H2O  dt  (ii) Very slow reactions: These are extremely slow  dP  1 and occur at a slow rate.  dt  RT Eg.: 2H2 + O2 250°C→ 2H2O = = 2.666 × 10−4 0.0821 × 298 Pb(CH3COO)2 AtmHos2pShere → PbS = 1.09 × 10–5 mol L–1 s–1 White Black (iii) Moderate reactions: Their speed is in between 36. Average rate reaction between 0 – 5 sec: the above two extremes. = −[{R2 } − {R1 }] Eg.: 2H2O2 → 2H2O + O2 [t2 − t1 ] CH3COOC2H5 + NaOH → CH3COONa + C2H5OH = −[80 × 103 − 160 × 103 ] 5 33. The given reaction is: = 16 × 103 mol L–1 sec–1 2A + B → 3C Average rate of reaction between 5 – 10 seconds: ⇒ 1 d(A) d(B) = + 1 d(C) − 2 dt = − dt 3 dt = – {[R2 ] − [R1 ]} [t2 − t1 ] = rate the reaction 80 Chemistry-12

= – [40 × 103 − 80 × 103 ] = 1 ∆(NO2 ) 5 4 ∆t = 8 × 103 mol L–1 s–1 = 1 # 2 # 10−2 4 5 37. 2N2O5 (g) → 4 NO2 (g) + O2 (g) 'N2 O5 ' (NO2) = 10–3 mol L–1 s–1 1 't 1 't =  2  4 (ii) −∆(N2O5 ) = + 1 × ∆(NO2 ) ∆t 2 ∆t = Rate of reaction: ⇒ −∆ (N2O5 ) = + 1 ∆(NO2 ) = + 1 × 2 × 10−2 ∆t 2 ∆t 2 5 (i) Rate of reaction: = 2 × 10–3 mol L–1 s–1 Topic 2. Factors Influencing Rate of a Reaction • The factors can be summarised as follows: ã In heterogeneous reactions, the reaction rate also depends on the surface area of ã Concentration or Pressure: The greater reactants. With increase in surface area, the concentration of reactants or greater the the reaction rate increases. pressure (in the case of gaseous reactions), greater is the reaction rate. Order of a Reaction: The order of a reaction is defined as the sum of the power of concentration of ã Catalyst (positive catalysts): It increases the reacting species in the rate law, the rate of reaction. For a reaction ã Temperature: An increase in temperature increases the reaction rate to a large extent 2A + B → Products for several reactions. The rate law is ã Intensity of radiations: In photochemical reactions, the rate of the primary step that Rate = k[A]m [B]n involves light absorption depends only upon The order of such a reaction is (m + n). the intensity of the absorbed light and not on Order of a reaction can also be defined with respect factors like concentration and temperature. to a single reactant. Thus, the reaction order with For example, Cl2 hv→ 2 C: l respect to ‘A’ is m and with respect to ‘B’ it is n. The overall order of reaction (m + n) may ranges Rate = kIabs from zero, 1, 2, 3 ... so on and it can be fractional too. Examples of Reaction Order: Rate law Reaction order Rate = k [N2O5] 1—(first order) reaction Rate = k [H2] [I2] Rate = k [NO2]2 1 + 1 = 2—(second order) reaction Rate = k [H2] [NO]2 2 (second order)—reaction Rate = k [CHCl3] [Cl2]1/2 1 + 2 = 3 (third order)—reaction 1+ 1 =1 1 (one and half order), fractional order reaction 2 2 The unit of rate constant ‘k’ for nth order of reaction • For example, the rate law for the reaction: is given by unit of k = [conc]n–1 [Time]–1 ( ) ( ) ( )2NH3 g P1tc1a3t0alKyst→ N2 g + 3H2 g Zero Order Reaction: • A reactant whose concentration does not affect At high pressure on a hot platinum surface it is the reaction rate is not included in the rate law. given as: • In effect, the concentration of such a reactant has Rate = k[NH3]0 = k the power 0. Thus, [A]0 = 1. • Here the rate does not depend on [NH3], so this • A zero order reaction is the one whose rate is is not included in the rate law and the order of independent of the concentration of the reacting reaction is understood to be zero. species. ChemiCal KinetiCs 81

Molecularity of a Reaction Difference between Order and Molecularity Chemical reactions may be classified into two types: • Molecularity is a theoretically deduced number, • Elementary reaction is a simple reaction which while order of a reaction is an experimentally determined value. occurs in a single step. • Order and molecularity are not necessarily • Complex reaction is the one which occurs in two equal. For complex reactions they are generally or more steps. different. (i) Molecularity of an elementary reaction is • The molecules or atoms participating in a reaction (i.e. molecularity) can be in whole number values defined as the number of reactant molecules only. On the other hand, the order of a reaction being an experimental value can be a fractional involved in a reaction. The elementary value too. reactions having molecularity 1, 2 and 3 Effect of Light on Rate of Reaction are called unimolecular, bimolecular and • Some reactions proceed by absorption of light radiations. The reactant molecules absorb photons trimolecular reactions respectively. Examples of light and gets excited. These excited molecules then carry on the reaction. These reactions are are: called photochemical reactions and they differ from thermal reactions in the following aspects. Br2 → 2Br (molecularity = 1) ã In thermal reactions, the energy of activation 2HI → H2 + I2 (molecularity = 2) is acquired by heat. In photochemical reactions energy is gained by the absorption CH3COOC2H5 + H2O → CH3COOH + C2H5OH of photons of visible or ultraviolet radiations. (molecularity = 2) ã The rate of thermal reactions depends on the 2NO + O2 → 2NO2 (molecularity = 3) temperature while the rate of photo-chemical (ii) Complex reactions occur in a series of reactions are independent of temperature. However, their rate does depends on the steps. The stepwise sequence of elementary intensity of radiation used. reactions that converts reactants to products ã All thermal reactions proceeds with a decrease in free energy. But certain photochemical is called the mechanism of the reaction. reactions proceed with an increase in free In any mechanism, some of the steps will be energy, e.g., photosynthesis in plants. This is so because the energy of reaction gets fast, while others will be slow. converted into free chemical energy of the products. Slowest step is the rate determining step of ã In comparison with thermal reactions, the reaction. photochemical reactions are selective. The absorbed photon of radiation excites an For example: The decomposition of N2O5 appropriate bond or group of atoms which occurs in the following steps: becomes the site for the reaction. Step 1 2N2O5 → 2NO2 + 2NO3 • Some of the examples where light is used for (Slow) initiating chemical reactions are: Step 2 NO2 + NO3 → NO + NO2 + O2 ã Photosynthesis (Slowest) ã Vision (A compound, retinal absorbs the light Step 3 NO + NO3 → 2NO2 falling on the eye and undergoes a chemical (Fast) change. This is the first step visualisation). Step 1 + 2 +3 2N2O5 → 4NO2 + O2 ã Photography Each elementary reaction has its own molecularity ã Photochemical synthesis of compounds equal to the number of molecules or atoms participating ã Sterilisation of water (exposure of water to U.V radiations) in it. It is meaningless to give molecularity of the overall reaction because it is made up of several elementary reactions, each perhaps with a different molecularity. At best it could be thought of as, the number of molecules or atoms taking part in the rate-detemining step. Thus, step 2 in the above example is rate determining and has molecularity ‘2’, which could also be considered as the molecularity of the decomposition reaction of N2O5. 82 Chemistry-12

ExERCisE 3.2 Multiple Choice Questions (MCQs) (1 Mark) 7. The rate law for the decomposition of N2O5 is: Rate = k [N2O5] 1. The role of a catalyst is to change .................... . (i) gibbs energy of reaction. What is the significance of ‘k’ in this equation? (ii) enthalpy of reaction. 8. Give the name of two important factors which affects the rate of a reaction. (iii) activation energy of reaction. 9. The reaction, A + 2B → product, the rate law (iv) equilibrium constant is given by: 2. In the presence of a catalyst, the heat evolved or Rate = k [A]1/2 [B]2 absorbed during the reaction .................... . What is the order of the reaction? (i) increases. (ii) decreases. [NCERT][AI 2013][CBSE 2022] (iii) remains unchanged. 10. The specific reaction rate of a reaction is 6.2 × 10–3 (iv) may increase or decrease. mol L–1 s–1. What is the order of the reaction? 3. Which of the following statements is not correct [AI 2011] about order of a reaction. 11. What are the units of rate constant for a first order (i) The order of a reaction can be a fractional reaction? number. 12. State a condition in which a bimolecular reaction (ii) Order of a reaction is experimentally may be kinetically of first order. determined quantity. [NCERT Exemplar] (iii) The order of a reaction is always equal to the sum of the stoichiometric coefficients of 13. What are the units of rate constant for a zero order reaction? reactants in the balanced chemical equation 14. For which type of reactions order and molecularity for a reaction. is same. [NCERT Exemplar] (iv) The order of a reaction is the sum of 15. What is the difference between rate law and law the powers of molar concentration of the of mass action? reactants in the rate law expression 16. Define specific reaction rate or rate constant. Assertion-Reason Type Questions (1 Mark) 17. For a reaction: A → B, the rate increases 2.25 times when the concentration of A is increased by Note: In the following questions a statement of assertion followed by a statement of reason is given. 1.5 times. What is the order of the reaction? Choose the correct answer out of the following choices. 18. Can molecularity of zero order reaction be zero? (i) Both assertion and reason are correct and the reason is correct explanation of assertion. 19. For a reaction, A + H2O → B Rate ∝ [A] (ii) Both assertion and reason are correct but reason does not explain assertion. What is the molecularity and order of reaction? (iii) Assertion is correct but reason is incorrect. 20. The rate law for the reaction is, (iv) Assertion is incorrect but reason is correct. Rate = k [NO]2 [Cl2] to NO, Cl2 and 4. Assertion: Order of the reaction can be zero or Write down the order with respect fractional. overall order of the reaction. Reason: We cannot determine order from 21. Give the rate law expression of the reaction, balanced chemical equation. A → C, whose rate increases four times when the concentration of A is doubled. 5. Assertion: Order and molecularity are same. 22. The reaction A + B + C → Products is found Reason: Order is determined experimentally to obey the rate law, and molecularity is the sum of the stoichiometric coefficient of rate determining elementary step. What iddsxtth=e k [A]2 [B]3/2 [C]–1/2. reaction? overall order of the Very Short Answer Type Questions (1 Mark) 23. Why is the probability of reaction with molecularity 6. Consider the following reaction, higher than three very rare? 2NO(g) + 2 H2(g) → N2(g) + 2H2O(g) [NCERT Exemplar] The rate law for the reaction is first order wrt to Short Answer Type Questions-I (2 Marks) Hla2wanfodr second order wrt NO. Write down the rate this reaction. 24. Identify the order of reaction from the following unit of rate constant: L mol–1 s–1. [Foreign 2010] ChemiCal KinetiCs 83

25. (i) The reaction, A + B → 2C; obeys the rate 37. The rate expression for the reaction, equation, Rate = k[A]1/2 [B]3/2 dx 2NO + O2 → 2NO2 is dt = k [NO]2 [O2]. What is the order of the reaction? [NCERT] (ii) The reaction A + B → C has zero order. If two experiments involving this reaction are Write the rate equation. carried out at the same temperature, but in second 26. The rate constant of reaction is 3 × 102 h–1. What experiment the initial concentration of NO is is the order of the reaction? [NCERT] doubled while the initial concentration of O2 is 27. bFeocrotmheesr,etahcrtieoent2imXes wh→enX2t,htehecornacteenotfrraetaioctnioonf halved, the initial rate in the second experiment will be ...... times that of the first. X is increased by 27 times. What is the order of 38. Possible mechanism for the reaction, the reaction? 2NO + 2H2 → N2 + 2H2O is: 28. Rate of reaction is given by the equation. (i) 2NO  N2O2 Rate = k[A]2 [B] (ii) N2O2 + H2 slow→ N2O + H2O (iii) N2O + H2 fast→ N2 + H2O What are the units for the rate and rate constant What is the rate law for the reaction? What is the for the reaction? order of the reaction? 29. The rate law for the reaction, Ester + H2O → Acid + Alcohol, is dx = k [Ester] [H2O]0 39. When ammonia is treated with O2 at elevated dt temperatures, the rate of disappearance of What would be the effect on the rate if: ammonia is found to be 3.5 × 10–2 mol dm–3 s–1 (i) concentration of the ester is doubled, during a measured time interval. Calculate the (ii) concentration of H2O is doubled? rate of appearance of nitric oxide and water in the 30. (i) Define molecularity of a reaction. same time interval. [NCERT] (ii) What is activation energy of a reaction? 40. Write down the units for rate of reaction for different orders of reactions. 31. What is the difference between order of a reaction and its molecularity? Short Answer Type Questions-II (3 Marks) [CBSE 2022][NCERT][Delhi 2014] 41. Mention the factors that affect the rate of a chemical reaction. 32. Nitric oxide, NO, reacts with oxygen to produce nitrogen dioxide: [Delhi 2010C; AI 2008, Delhi, Foreign] 2NO (g) + O2 (g) → 2NO2 (g) 42. The reaction, 2N2O5 → 4NO2 + O2 was studied The rate law for this reaction is, and the following data were collected: Rate = k[NO]2 [O2] Expt (m[No2lOL5–]1) Rate of disappearance Propose a mechanism for the reaction. No. 33. Discuss the effect of catalyst on the activation of N2O5 (mol L–1 mol –1) energy. of C 34. A reaction is second order with respect to a 1. 1.13 × 10–2 34 × 10–5 reactant. How is the rate of reaction affected if the concentration of the reactant is reduced to 2. 1.84 × 10–2 25 × 10–5 half? What is the unit of rate constant for such a 3. 0.62 × 10–2 18 × 10–5 reaction? [AI 2011] 35. For the reaction, 2A + B → Products, doubling Determine (i) order of the reaction, (ii) the rate the initial concentrations of both the reactants, law and (iii) rate constant for the reaction. the rate increases by 8 times and on doubling [AI 2008C] the concentration of B alone, the rate increases 43. A reaction is first order wrt ‘A’ and second order wrt ‘B’, by 2 times. Write the rate expression for such a (i) Write the differential equation for the rate reaction. of reaction. 36. Write rate expression for the following reaction if (ii) How is the rate affected if the concentration the order of reaction is first, second and zero with of B is tripled? respect to A, B and C respectively: (iii) H o w i s t h e r a t e a f f e c t e d w h e n t h e concentration of both A and B are doubled? 2A + B + C → D + 2E [Delhi 2010] Also, what will be the effect on the velocity of reaction if the concentrations of A, B and C are doubled? 84 Chemistry-12

Long Answer Type Questions (5 Marks) (b) Rate constant [Delhi 2015C] 44. (i) Define the following: (ii) What is the significance of rate constant in (a) Law of mass action the rate expression? Answers 3.2 1. (iii) 2. (iii) 3. (iii) 25. (i) Order of the reaction = 1 + 3 = 2 4. (ii) 5. (iv) 6. Rate = k[H2] [NO]2 2 2 7. k is the specific rate constant. k becomes equal to the rate of the reaction when the concentration of (ii) Rate law expression for this reaction: N2O5 is unity. k¢ [A]° [B]° 8. (i) Concentration of the reactants 26. k = 3 × 102 h–1, (ii) Temperature Unit of rate constant: 9. Overall order of reaction = 1 +2= 5 =  L n − 1 × Time−1 10. Zero order 2 2  mol  11. Time–1 =  L  °h−1 =  L n−1 × h−1  mol   mol  12. When one of the reactants is taken in large excess. 13. mol L–1 time–1 On comparing power of ‘L’, we get 14. For an elementary reaction, order and molecularity n–1=0 have same value. or n = 1, i.e. first order reaction For example, 27. 2X → X2 CH3COOC2H5 + NaOH → CH3COONa + C2H5OH Let the order with respect to X be a, then Here, Order = 2, Molecularity = 2. Rate = k [X]a ...(a) 15. Rate law is based on experimental data of a ...(b) reaction whereas law of mass action is theoretical, \\ 3 × Rate = k [27 X]a which is based on balanced chemical equation for a reaction. From (a) and (b), a = 1 3 16. Specific reaction rate (or rate constant) is the rate of a reaction when the concentration for each \\ Order of reaction = 1 reactant is taken to be unity. 3 17. Two 28. Rate = k [A]2 [B] 18. No, molecularity is never zero. Order = 2 + 1 = 3 19. Molecularity = 2; order = 1 Unit of K =  L n−1 × sec−1  mol  20. Order with respect to NO = 2; order with respect to Cl2 = 1; overall order = 2 + 1 = 3. = L2 mol–2 s–1 (Q n = 3) 21. Rate = k[A]2 unit of rate = conc time–1 22. Order = 2+ 3 − 1 = 3 = mol L–1 s–1 2 2 23. Probability of simultaneous collision of more 29. (i) Rate is doubled. than three molecules is very rare, therefore (ii) Rate remains unaffected. molecularity greater than three is also rare. 30. (i) Molecularity of a reaction is the number of reactant molecules or ions taking part in a 24. For an nth order reaction, single step of the reaction.  1 n − 1 Unit of rate constant, k =  conc  × s−1 (ii) It is the minimum energy required by the reactant molecules to form an activated Unit of k =  1 n − 1 s−1 complex so as to form products in a reaction.  mol L−1  × 31. Order of a Reaction Molecularity ⇒ k = Ln–1 mol1 – n s–1 It is the sum of the powers It is the number of of the concentration of reacting species taking Given, unit of k = L mol–1 s–1 = Ln–1 mol1 – n s–1 the reactants in the rate part in an elementary law expression. reaction. Comparing the power of ‘L’ we get 1=n–1 It can be determined It is a theoretical n=2 experimentally. concept. \\ This reaction will be of second order. ChemiCal KinetiCs 85

It may have an integral It is always an integer; 39. Consider the following equation: value or a fractional 1, 2 or 3 but it can never value or it may also be be zero or fraction. 4NH3 + 5O2 → 4NO + 6H2O zero. The rate of reaction can be given as: It is applicable for It is applicable only for 1 d[NH3 ] = 1 d[NO] = 1 d[H2O] elementary as well as elementary reactions. −4 dt 4 dt 6 dt for complex reactions. d[NO] = − d[NH3 ] dt dt 32. The proposed mechanism for the reaction is: = 3.5 × 10–2 mol dm–3 s–1 NO + O2 (Fa∆st)→ NO3 ...(i) d[H2O] −6 d[NH3 ] NO3 + NO k1 2NO2 ...(ii) dt = 4 dt ]Slowg 6 Here, step (ii) is the slowest, therefore = 4 × 3.5 × 10–2 Rate = k [NO3] [NO] = 5.25 × 10–2 mol dm–3 s–1 Let [NO3] = k1[NO] [O2] Rate = k1k [NO]2 [O2] = k≤[NO]2 [O2], 40. Zero order reaction – mol L–1 s–1 [(conc)1 (time)–1] where k≤ = k1k 33. Positive catalyst lowers the activation energy First order reaction – s–1 [(conc)0 (time)–1] while negative catalyst increases it. Second order reaction – L mol–1 s–1 [(conc)–1 (time)–1] 41. (i) Nature of reactants 34. Rate = k[A]2 (ii) Concentration of reactants Rate will become 1 th, if the concentration is reduced to half. 4 (iii) Temperature of the system (iv) Presence of catalyst For second order reaction, (v) Surface area Unit of k = [conc]n–1 [time]–1 (vi) Intensity of the radiation, in case of = [mol L–1]2–1 s–1 photochemical reactions = mol L–1 s–1 42. (i) Let the rate law for the reaction be 35. Rate = k[A]2 [B] Rate = k[N2O5]x Rate will increase by 8 times, if concen- tration \\ 34 × 10–5 = k[1.13 × 10–2]x , and of both A and B are doubled. Moreover, rate will be doubled if the concentration of reactant B is 25 × 10–5 = k[0.84 × 10–2]x doubled. or 34 × 10−5 = [1.13 × 10−2 ]x 36. Rate = k[A]1 [B]2 [C]0 25 × 10−5 [0.84 × 10−2 ]x Rate will increase by eight times if concentrations of A, B and C are doubled. 1.36 = [1.345]x x= 1 \\ Order of the reaction is 1. 37. r1 = k[NO]2 [O2 ] (ii) Rate law: Rate = k[N2O5] r2 k[2NO]2  1 O2  (iii) k = Rate = 34 × 10−5  2  [N2O5 ] 1.13 × 10−2 \\ r1 = 1 = 30.08 × 10–3 min–1 r2 2 43. (i) Rate = k[A] [B]2 or r2 = 2r1, (ii) Rate will increase by 32 or 9 times i.e. rate will become double the value of r1. 38. Slowest step is the rate determining step. (iii) Rate will increase by 2 × 22, i.e. 8 times \\ Rate = k[N2O2] [H2] …(i) 44. (i) (a) At a given temperature the rate of is reaction at a particular instant is Npr2eOse2 nsthionutlhde be eliminated bec ause it is not proportional to the product of the active overall reaction masses of the reactants taking part in the reaction raised to the power equal to the From (a) : [N2O2] = k1[NO]2 ...(ii) numbers of their respective molecules in From (i) and (ii) the stoichiometric equation describing the reaction. Rate = kk1[NO]2 [H2] = k′[NO]2 [H2][k′= kk1] Rate = [A] [B] Order = 2 + 1 = 3 or Rate = k′ [CA] × [CB] 86 Chemistry-12

(b) Rate constant of a reaction may be (ii) Rate constant represent the rate of the defined as the rate of reaction when the reaction when the concentration of all the concentration of each reactant is taken to reactants is taken to be unity. Its units for be unity. That is why rate constant is also reactions of different order is different. called specific reaction rate. Topic 3. Integrated Rate Equations and Pseudo First Order Reaction Zero Order Reactions 2.303 22.8 k= 1200 log 8.2 = 8.523 × 10–4 sec–1 A→P • Rate of a reaction at time t The average value of k = 8.445 × 10–4 sec–1 − d [A] Rate = dt = k[A]0 The value of k being fairly constant, indicates that the reaction is a first order reaction. − d [A] \\ Rate = dt =k or d[A] = – kdt On integrating both sides, we get [A] = – kt + [A0] or k= [A0] − [A] t First Order Reactions A → Products • Rate of the reaction at time t = −d(C0 − x) = dx Second Order Reactions = k [A]′ = kC dt dt • Consider the general reaction: where the concentration of A at that instant is C. Assume C0 to be the initial concentration of A and A → Products ‘x’ to be the change in concentration in time t. where the order is 2 with respect to A or A + B • The integrated expression for the rate constant → Products is given as: where the order is 1 each with respect to A and k= 1 ln C0 = 2.303 log10 C0 B, respectively. The total order is therefore t C t C being 2. In the later case, assume that the initial concentration of A and B is the same. Rate of the  −k  reaction after time, t.  2.303 log C= t + log C0 = −d[A] = k [A]2 in the first case dt • A plot of log C versus ‘t’ gives a straight line with or k [A] [B] in the second case. a negative slope =  −k   2.303 In both the cases, the integrated expression for the rate constant is. • For a 1st order reaction, the half life period: k= 1x or 1x t1/2 = 2.303 log10 2 = 0.693 tC0 C tC0 ⋅ (C0 − x) k k (Where C = C0 – x) i.e. t1/2 is independent of the initial concentration. KIlMlunstOr4astoiolunt:iAonsoaltudtiiofnfeoref nHt2Oti2mwehienntetirtvraaltsedgaavgeaitnhset or k = 1 1 − 1  , i.e. 1 = (k)t + 1 . following results: t  C0  C C0  C  t (in secs) 0 600 1200 Therefore, a plot of slC1opeve=rsku.s t gives a straight line with a positive Volume of KMnO4 (in c.c.) 22.8 13.8 8.2 • For a 2nd order reaction: A →Products, the half Show that the decomposition of H2O2 is a first order period; t1/2 = 1 , reaction. C0 k Solution: Here, where C0 is the initial concentration of A. Illustration-1: Hydrolysis of ethyl acetate by NaOH k= 2.303 log 22.8 = 8.368 × 10–4 sec–1 using equal concentrations of the reactants was 600 13.8 ChemiCal KinetiCs 87

studied by titrating 25 mL of the reaction mixture at 100 seconds. How long will it take for the decomposition different time intervals against a standard solution to be 90% complete? of an acid. From the data given below, establish that this is a second order reaction: Solution: For a first order reaction, t (in minutes) 0 5 15 25 k= 0.693 = 0.693 = 6.93 × 10–3 sec–1 t1 / 2 100 acid used (in mL) 16 10.24 6.13 4.32 k= 2.303 log C0 t Ct Solution: k= 1 x tC0 (C0 − x) For a 90% decomposition, Ct = 0.1 C0 k= 1 5.76 = 0.0070 6.93 × 10–3 = 2.303 log C0 = 2.303 ×1 16 × 5 10.24 t 0.1 C0 t k= 1 9.87 = 0.0067 \\ t= 2.303 secs = 332 secs. 16 × 15 6.13 6.93 × 10−3 k = 1 11.68 = 0.0068 Pseudo First Order Reactions 16 × 25 4.32 • A reaction in which one of the reactants is present in large excess and shows an order different from The value of K being fairly constant indicates that this the actual order. reaction is of the second order. • The experimental order which is not the actual one is referred to as the preudo order. 1 • Since for elementary reactions molecularity C and order are identical, therefore pseudo-order reactions may also be called pseudo molecular t reactions. • In general for a nth order reaction of the type • For example, if in a reaction A + B → Products A → Products, where the rate, dx = k[A]n, Then by the rate law: dt Rate = k[A] [B] then where k is the rate constant, • If ‘B’ is present in excess, then its concentration remains practically constant during the course of k= 1 1 − 1 the reaction. Thus, the rate law can be rewritten   as t(n − 1)  Cn−1 C0n−1  • This is the general expression applicable to all Rate = k′[A] values of n, except 1. • For a nth order reaction, the half-period is where the new rate constant k′ = k [B]. The actual t1/2 = 1 1) ⋅ 2n−1 − 1 , i.e. (except n = 1) order of the reaction is second order but in practice k(n − C0n−1 it will be of first order. Therefore, the reaction is cto1/n2 ciesntirnatvieonr s)ne–1l.y p r o p o r t i o n a l t o ( i n i t i a l said to have a pseudo first order. An example of such a reaction is: Illustration-2: The half-life of a substance in a CH3COOC2H5 + H2 O → CH3COOH + C2H5OH first order reaction is 15 minutes, calculate the rate constant. (excess) If the potential energy of the products is less Solution: For a first order reaction than that of the reactants, the energy obtained t1/2 = 0.693 in going from the activated complex to products k will be more than the activation energy (Ea). Thus, Pfourtkti,nwget1h/2a=v1e5 min in the expression and solving such a reaction will always be exothermic. On the other hand, if the potential energy of the products k= 0.693 0.693 = 4.62 × 10–2 min–1 is greater than that of the reactants, then energy t1/2 = 15 min released in going from the activated complex to Illustration-3: The time required for 50% products will be less than the activation energy decomposition of reactant in a first order reaction is and the reaction will be endothermic in nature. 88 Chemistry-12

A... B...C A... B...C AB + C Potential Energy Potential Energy Ea Ea A + BC DE DE A – B + C Products Reaction Coordinate Reaction Coordinate Exothermic Reaction Endothermic Reaction ExERCisE 3.3 Multiple Choice Questions (MCQs) (1 Mark) 1. Consider a first order gas phase decomposition (iii) In k reaction given below: 1/T A(g) → B(g) + C(g) The initial pressure of the system before decomposition of tAhewsaysstpei.mAifntecrrelaapsesde of time ‘t’, total pressure of by x units and became ‘pt’. (iv) In k The rate constant k for the reaction is given as. (i) k = 2.303 log pi 1/T t pi − x 3. Rate law for the reaction A + 2B → C is found to (ii) k = 2.303 log pi be t 2 pi − pt Rate = k [A][B] (iii) k = 2.303 log pi t 2 pi + pt Concentration of reactant ‘B’ is doubled, keeping the concentration of ‘A’ constant, the value of rate (iv) k = 2.303 log pi t 2 pi + x constant will .................... . 2. According to Arrhenius equation rate constant (i) the same (ii) doubled k is equal to k = Ae–Ea/RT. which of the following (iii) quadrupled (iv) halved options represents the graph of ln k vs1/T? 4. A first order reaction is 50% completed in 1.26 × 104 s. How much time would it take for 100% completion? (i) In k (i) 1.26 × 1015 s (ii) 2.52 × 1014 s 1/T (iii) 2.52 × 1028 s (iv) infinite 5. Compounds ‘A’ and ‘B’ react according to the following chemical equation. A (g) + 2B (g) → 2C (g) (ii) In k Concentration of either ‘A’ or ‘B’ were changed keeping the concentrations of one of the reactants 1/T constant and rates were measured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate equations for this reaction. Experiment Initial concentration of [A]/ Initial concentration of [B]/ Initial rate of formation of mol L–1 mol L–1 [C]/mol L–1 s–1 1. 0.30 0.30 0.10 2. 0.30 0.60 0.40 3. 0.60 0.30 0.20 ChemiCal KinetiCs 89

(i) Rate = k[A]2 [B] 14. Give the relation between rate constant and half- (ii) Rate = k[A] [B]2 life period for a first order reaction. (iii) Rate = k[A] [B] (iv) Rate = k[A]2 [B]0 15. Give an example of a pseudo first order reaction. 16. What is half-life period of a reaction? Assertion-Reason Type Questions (1 Mark) 17. Give one example of a reaction with fractional order. Note: In the following questions a statement of assertion followed by a statement of reason is given. 18. What is rate determining step of a reaction? Choose the correct answer out of the following choices. 19. For (ag)r,etahcetiroant:eNlaOw2 (egx)p+reCsOsio(ng)is: → CO2(g) + (i) Both assertion and reason are correct and the NO reason is correct explanation of assertion. Rate = k [NO2]2 (ii) Both assertion and reason are correct but reason How many molecules are involved in the slowest does not explain assertion. step? (iii) Assertion is correct but reason is incorrect. (iv) Assertion is incorrect but reason is correct. 20. The gas phase decomposition of A follows the rate law, r = k[A]n. What are the units of its rate 6. Assertion: If in a zero order reaction, the constant if concentration is taken in terms of concentration of the reactant is doubled, the half- life period is also doubled. pressure? Reason: For a zero order reaction, the rate of reaction is independent of initial concentration. 21. The reaction, A + B → C has zero order. What is the rate of the reaction? [NCERT Exemplar] 7. Assertion: Half-life period of a reaction of first order is independent of initial concentration. 22. Write the expression showing the initial Reason: Half-life period for a first order reaction concentration and concentration at time t in the is t½ = 2.303 × log 2/K. exponential form for first order reaction. Very Short Answer Type Questions (1 Mark) 23. Give an example of pseudo order reaction. 24. What is lifetime of a reaction? Short Answer Type Questions-I (2 Marks) 8. 50% of a reaction is completed in 16 minutes. 25. The rate constant for a reaction of zero order in A What fraction of the reaction would occur in 32 is 0.003 mol L–1 s–1. How long will it take for initial minutes? concentration of A to fall from 0.1 M to 0.075 M? 9. Fraotretlhaewrehaacstiboenen2Ade+teBrm+iCned to→beAR2Bate+ C, the [Delhi, Foreign 2010] = k[A] 26. rTehaectdioencowmitphosaitiroanteocfoNns2tOa5n(tg)ofis5a×fi1r0s–t4 order s–1 at [B]2 250°C. It the value of k is 2.0 × 10–6 mol–2 L2 s–1 for the reaction, determine the initial rate with [A] = 0.2 2N2O5(g) → 4NO2(g) + O2(g) mol L–1, [B] = 0.1 mol L–1, [C] = 0.5 mol L–1. If the initial ccoonncecenntrtartaitoinonofaNfte2Or 52ism0i.n2u5teMs.. Calcu late its [NCERT] Calculate half life for the decomposition of 10. NFoOr(gth),etrheeacetxipoenriNmOe2n(gta)+llyCOde(tge)rmine→d CraOte2(gla) w+ expression at 400 K is rate = k[NO2]2 N2O5 (g). [Delhi 2008] What mechanism can be proposed for the above reaction? 27. The rate constant for a 1st order reaction is 60 s–1. How much time will it take to reduce the concentration of the reactant to 1/16th of its initial 11. Nitric oxide reacts with H2 to give N2 and H2O as value? [NCERT] per the reaction: 28. Find the two t5h.4ir8d×lif1e0(–t124/3s)–o1f. a first order reaction in which k= 2NO + 2H2 → N2 + 2H2O The kinetics of this reaction is explained by the 29. The first order reaction takes 69.3 minutes for following steps: 50% completion. Calculate the time required for (i) 2NO + H2 Slow→ N2 + H2O2 80% completion of this reaction. (ii) H2O2 + H2 Fast → 2H2O What is the predicted rate law? 30. The first order decomposition reaction takes 40 minutes for 30% decomposition. Calculate t1/2. 12. Write the rate equation for the reaction 2A + B [NCERT][Delhi 2013, AI 2008] → C if the order of the reaction is zero. 31. A first order reaction has specific rate constant [NCERT Exemplar] of 10–3 sec–1. How much time will it take for 10 g 13. Derive an expression to calculate time required of it to reduce to half the quantity. for completion of zero order reaction. [NCERT] 32. For the first order reaction, calculate the ratio between time taken to complete 3/4th of the 90 Chemistry-12

reaction and the time taken to complete half of 45. The decomposition of an aqueous solution of ammonium nitrite was studied. The volume of the reaction. [NCERT] nitrogen gas collected at different intervals of time as follows: 33. Archaeological remains of a wooden artifact was found to contain 2% 14C activity of what was expected in freshly cut wood. Find out the age of Time (min) 10 15 20 ∞ 34. the wood (t1/2 for 14C = 5760 yrs.). Vol. of N2 (mL) 6.25 9.00 11.40 35.05 The amount of 1o46fCthisaottoppreesiennat piece of wood was found to be 1/5 in a fresh piece of From the above data prove that the reaction is of the first order. wood. Calculate the age of wood. (Half-life of 14C 46. Optical rotation of sucrose in 1 N HCl at various = 5577 years). times intervals was found as shown below: 35. The half life period of radon is 3.8 days. After how Time (sec) 0 7.18 18.0 27.05 ∞ many days will only 1/20 of the radon sample be left over? Rotation (deg) +24.09 +21.7 +17.7 +15.0 –10.74 36. 131I has a half life period of 13.3 hrs. After 79.8 Show that the inversion of sucrose is a first order hour, what fraction of 131I will remain? reaction. 37. A piece of wood was found to have 14C/12C ratio as 47. The rate law for the decomposition of gaseous 0.6 times to that in a living plant. Calculate the N2O5, period when the plant died (Half life of 14C = 5760 1 years). N2O5 → 2NO2 + 2 O2 is observed to be 38. The possible mechanism for the reaction: 2H2 (g) + 2NO (g) → N2(g) + 2H2O is Rate = − d[N2O5 ] = k[N2O5] (i) 2NO  N2O2 dt (ii) N2O2 + H2 → N2O + H2O (Slow) (iii) N2O + H2 → N2 + H2O (Fast) A reaction mechanism which has been suggested What is (a) rate law of the reaction? to be consistent with this is: (b) the order of the reaction? N2O5 K NO2 + NO3 (fast) 39. In a reaction if the concentration of A is tripled, NO2 + NO3 k1 NO2 + NO + O2 (slow) the rate of reaction becomes twenty seven times. NO + NO3 k2 → 2NO2 What is the order of the reaction? [NCERT] (fast) 40. For reaction A + B → Products, the rate law Show that the mechanism is consistent with the is – Rate = k [A] [B]3/2. Can the reaction be an observed law. elementary reaction? Explain. 48. Write the overall reaction and the rate law [NCERT Exemplar] equation from the following elementary steps: 41. Rate of a reaction is given by the equation: Rate NO + Br2  NOBr2 = k[A]2 [B] NOBr2 + NO → 2NOBr 49. A first order reaction is 15% completed in 20 What are the units for the rate and rate constant minutes. How long will it take for its 60% for the reaction? completion. 42. In a first order reaction the concentration of the 50. oTrhdeerthreeramctaiolndewciotmh paorsaittieoncoonfsHtaCnOt o2Hf 2i.s4 a first reactant is reduced to one-fourth of its initial × 10–3 value in 50 seconds. Calculate the rate constant of the reaction. s–1 at a certain temperature. Calculate how long 43. For a certain first order reaction, it take 5 minutes will it take for three-fourth of the initial quantity for the initial concentration of 0.6 mol L–1 to become 0.2 mol L–1. What is the rate constant for of HCO2H to decompose? 51. Draw a schematic graph showing how the rate of a first order reaction changes with the change in this reaction? [log 3 = 0.4771] [AI 2011 C] the concentration of the reactant. [NCERT] 44. bTyhetictraatatilnygticitdaetcdomiffpeorseinttioinntoefrHva2Ols2wwiaths studied and the following data were obtained: KMnO4 Short Answer Type Questions-II (3 Marks) Time (sec) 0 600 1200 52. The decomposition of a compound is found to follow the first order rate law. It take 15 minutes Vol. of KMnO4 (mL) 22.8 13.8 8.3 for 20% of the original material to react. Calculate: Calculate the velocity constant for the reaction, (i) the specific rate constant assuming it to be a first order reaction. (ii) the time in which 10% of the original material remains unreacted. ChemiCal KinetiCs 91

(iii) the time it take for the next 20% of reactant (ii) DfirisstsoocridaetriorneaocftSioOn2Cwli2thinrathtee gas phase is a left to react after first 15 minutes. constant equal to 2.3 × 10–5 sec–1 at 600 K. Calculate the 53. A first order reaction is 15% completed in p20er0cemnitnaugteeosfoSfOth2Celr2etahcattiownoutilmd ere. main after 20 minutes. How long will it take to be 60% completed? 63. Consider the following data for reaction A + B → Product 54. Half life of a radioactive element is 10 seconds. Calculate its average life. In how much time 75% of the radioactive element will decay? Exp. Initial conc. Initial conc. Initial rate No. of A (mol L–1) of B (mol L–1) (mol L–1 s–1) 55. For a general reaction A → B, plot of 1 0.10 1.00 2.1 × 10–2 concentration of A v/s time is 2 0.20 1.00 8.4 × 10–3 given in the figure. Answer the following question on the 3 0.20 2.00 8.4 × 10–3 basis of this graph: Determine the order of reaction wrt. to A and wrt. to B and overall order of the reaction. (i) What is the order of the reaction? Long Answer Type Questions (5 Marks) (ii) What is the slope of the 64. oTrhdeerreraecaticotnio:nSOw2iCthl2 half →lifeSOof2 3+.1C5l2×is1a04fisresct curve? (iii) What is the unit of rate constant? adtec5o7m5pKos.eWd ohnathepaetricnegntaatg5e75ofKSfOo2rC9l02 would be minutes? [Delhi 2014] 56. Find two-third life (t2/3) of a first order reaction in 65. The rate otuf dreecowmapsomsiteiaonsuorfeHd2Ob2yatt a particular which: tempera itrating its k = 5.48 × 10–1 sec–1 (log 3 = 0.4771; log 2 = 0.3010) solution with acidic KMnO4. Following results were obtained: 57. A reaction that is first order with respect to reactant ‘A’ has a rate constant 6 min–1. If we start Time (min) 0 10 20 with A = 5 mol L–1, when would A reach the value KMnO4 (mL) 22.8 13.8 8.3 of 0.05 mol L–1? (i) Show that the reaction is of first order. 58. Identify the reaction order from each of the (ii) Calculate the rate constant. following rate constants: 66. mtThheeeafdsiruescrtoeomdrdpaeotrsidrtaiifotfneerloaefnwNt.2TOtih5meinecoCinnCctlee4nrsvtoralaulsttiioaonnreofofglNliov2weOns5 below: (i) 6.5 × 10–3 mol L–1 s–1 (ii) 7.0 × 10–4 mol–1 L s–1 (iii) 2.25 × 10–2 mol–2 L2 s–1 (iv) 5.0 × 10–5 s–1 (v) 4.5 × 10–6 atm–1 s–1 Time (sec) 0 80 160 410 600 1130 1720 (vi) 4.0 × 10–6 atm–1 s–1 N2O5 5.5 5.0 4.8 4.0 3.4 2.4 1.6 59. Write the rate law for a first order reaction and Calculate the rate constant at t = 410 sec and at justify the statement the half-life period of such a t = 1130 sec. reaction is independent of the initial concentration 67. Methyl acetate was hydrolysed in the presence of dil. HCl and the mixture is withdrawn at of reactants. [Delhi 2009C] 60. Methyl acetate is hydrolysed with 1 N HCl at 25°C. different time intervals and titrated against 5 mL of the reaction mixture is withdrawn at; standard NaOH solution. The following results various time intervals and is quickly titrated with were obtained: 0.185 N NaOH. The volumes of NaOH consumed are as follows: Time (min) 0 25 40 61 800 Time (sec) 339 1242 2745 4546 ∞ Volume of alkali 19.24 24.20 26.60 29.50 42.1 (in mL) VNaOH (mL) 26.34 27.80 29.70 31.81 31.81 Prove the reaction of first order. Show that the hydrolysis reaction is of first order. 61. Calculate the ratio of the catalysed and uncatalysed 68. At certain temperature the half life period for the catalytic decomposition of NH3 were found as rate constant at 20°C, if the energy of activation given below: of a catalysed reaction is 20 kJ mol–1 and for the uncatalysed reaction is 75 kJ mol–1. Pressure (Pascals) 6667 13333 26666 62. (i) How is half-life period related to the initial 1.0 Half life period (in hours) 3.52 1.92 concentration for a second order reaction? Calculate the order of the reaction. 92 Chemistry-12

69. For the reaction in which A and B forms C, (b) Calculate the value of rate constant for the following results were obtained from three the reaction. experiments: (ii) The following data were obtained during the first order thermal decomposition of SO2Cl2 Exp. Initial Conc. (mol L–1) Initial rate at a constant volume: No. AB mol L–1 s–1 SO2Cl2(g) → SO2(s) + Cl2(g) 1 0.03 0.03 0.3 × 10–4 Experiment Total Pressure/ 2 0.06 0.06 1.2 × 10–4 No. Time/sec atm 3 0.06 0.09 2.7 × 10–4 1. 0 0.5 2. 100 0.6 What is the rate equation of the reaction and what is the value of rate constant? Calculate the rate of the reaction when total 70. Derive the integrated rate equation for the rate pressure is 0.65 atm. constant of a first order reaction. What would be the unit of rate constant if the concentration is [Delhi 2014, Foreign 2012][HOTS] expressed in mol L–1 and time in seconds? 73. For the following reactions, state the order with [Delhi, AI 2008] respect to each reactant and the overall order: 71. With the help of an example explain what is meant (i) 2NO(g) + 2H2(g) → N2(g) + 2H2O(g) by pseudo first order reaction? Rate = k [NO]2 [H2] [NCERT Exemplar] [AI 2011] (ii) 2N2O(g) → 2N2(g) + O2(g) Rate = k [N2O] 72. (i) Consider the reaction: 2A + B → C + D Following results were obtained in (iii) CHCl3 + Cl2 → CCl4 + HCl experiments designed to study the rate of Rate = k [CHCl3] [Cl2]1/2 reaction: (iv) H2O2 + 3I– + 2H+ → 2H2O + I3– Experiment Initial Initial Initial Rate = k [H2O2] [I–] No. conc. of conc. of rate of [A] mol/L–1 [B] mol/L–1 formation (v) CH3CHO → CH4 + CO of [D] Rate = k [CH3CHO]3/2 (mol/min) 74. If concentrations are expressed in mol L–1 and time in seconds, what are the units for the rate 1. 0.10 0.10 1.5 × 10–3 constant for reactions that are: 2. 0.20 0.20 3.0 × 10–3 (i) zero order (ii) first order 3. 0.20 0.40 6.0 × 10–3 (iii) second order (iv) of the order 1 1 2 (a) Write the order of the reaction wrt A and B and the rate law. (v) of order n? [NCERT] Answers 3.3 1. (ii) 2. (i) 3. (ii) 4. (iv) 12. 2A + B → C Rate = k [A]0 [B]0 5. (ii) 6. (ii) 7. (i) 3 13. Kinetic equation for a zero order reaction is 8. The time required for 4 th completion of the 1st [R] = [R]° – kt order reaction is twice of its half life. For completion of the reaction, [R] = 0 \\ t3/4 = 2 × t1/2 = 2 × 16 = 32 min 9. Rate = k[A] [B]2 = 2 × 10–6 [0.2] [0.1]2 \\ 0 = [R]°– kt ⇒ t = [R]°/k = 4 × 10–9 mol L–1 s–1 14. t1/2 = 0.693 k 10. Step 1: NO2 (g) + NO2(g) Slow→ NO(g) + NO3(g) 15. CH3COOC2H5 + H2 O H+→ Step 2: NO3 (g) + CO(g) Fast → CO2(g) + NO2(g) 16. i.e. hCalHf 3oCfOtOheHr+eaCc2tHio5nOHis Slowest step is the rate determining step (excess) Time in which 50%, completed. So, Rate = k [NO2]2 11. Slowest step is the rate determining step, 17. CHCl3 + Cl2 → CCl4 + HCl Experimental rate expression is therefore rate will be determined from the first step. Rate = k[CHCl3][Cl2]1/2 \\ Overall order of reaction Thus, the rate law would be: 1 1 = 1+ 2 = 1 2 Rate = k[NO]2 [H2] ChemiCal KinetiCs 93

18. The slowest step of the reaction is the rate 0.693 = 2.303 log  100  determining step. 69.3 t  100 − 80  19. Two 20. (atm)1 – n time–1 t = 161 minutes 21. Rate = k 30. For a first order reaction 22. [A] =[A0]e–kt 0.693 2.303  a  wcohnecreent[rAa0t]ioins initial concentration and [A] is t1 / 2 t log  − x  at time t and k is the rate constant. = a 23. CH3COOC2H5 + H2O H+→ a = 100, x = 30, t = 40 min CH3COOH + C2H5OH By putting these values in the above equation: 24. Time for the complete consumption of reactants. 0.693 = 2.303 log  100  t1 / 2 40  100 − 30  25. The zero order reaction is ⇒ t1/2 = 77.7 minutes x = kt, 0.693 0.693 k 10−3 k = 0.003 mol L–1 s–1 ...(a) 31. t1 / 2 = = = 693 sec x = 0.1 M – 0.075 M = 0.025 M \\ From (a) 0.025 = 0.003 × t 32. 0.693 = 2.303 log  1  t1 / 2 t3 / 4  − 3 t= 0.025 = 8.3 s  1 4  0.003 26. Given k = 5 × 10–4 s–1 0.693 = 2.303 × 0.602 t1 / 2 t3 / 4 Half life, t1/2 = 0.693 = 0.693 = 1386 sec k 5 × 10−4 t3/4 2 t3/4 = 2 × t1/2 or t1/2 = 1 For the first order reaction k= 2.303 log  A°  = 5 × 10–4 33. We know l = 2.303 log10  N0  t  A° − x  t  N  = 2.303 log  0.25  = 5 × 10–4 0.693 = 2.303 log10 100  2 × 60  0.25 − x  5760 t  32  On solving for x, we get, t = 9472 yrs. x = 0.015 34. t= 2.303 × t1/2 log  N0  0.693  N  Thus, concentration of N2O5 after 2 minutes = 0.25 – 0.015 = 0.235 M N0 2.303 27. For a 1st order reaction N = 5 ⇒ t = 0693 × 5577 log 5 k= 2.303 log  C0  = 12954.5 years t  Ct  35. For radioactive decay 60 = 2.303 log 1 0.693 = 2.303 log  N0  t 1/16  t1 / 2 t  N  t= 2.303 × 1.20 0.693 = 2.303 log 1 60 3.8 t 1 = 2.303 × 0.02 s–1 = 0.046 s–1 20 2.303  C0  t = 16.4 days 28. k= t log  Ct  36. Number of half lives, i.e.  n= Time = 79.8 = 6 2.303  1  Half life 13.3 5.48 × 10–14 = t log  −  1 2  n  3  As we know N = N0 1   2 t = 2 × 1013 sec–1 N  1 6 1 29. For a first order reaction N0  2 64 = = 0.693 = 2.303 log  a a  \\ Remaining fraction = 1 t1 / 2 t  − x  64 94 Chemistry-12

37. In radiocarbon dating k 2.303 log  V0  2.303 log  22.8  t  Vt  600  13.8  0.693 2.303 log  N0  = = t1 / 214 C t(age)  N  = = 8.4 × 10–4 s–1 0.693 = 2.303 log 1 45. [A]0 = volume of nitrogen at ∞ = 35.05 5760 t(age) 0.6 [A]t = V – Vt ∞ t = 4246.6 years When t = 10, 38. (i) Step 1: 2NO (g)  N2O2 (g) (fast) k = 2.303 log 35.05 = 0.0196 Step 2: N2O2(g) + H2(g) → 10 (35.05 − 6.25) N2O(g) + H2O(g) (slow) Step 3: N2O(g) + H2(g) → Similarly, N2(g) + H2O(g) (Fast) Rate is determined from step 2, as it is the when t = 15, k = 0.0197 s–1 slow step. when t = 20, k = 0.0196 s–1 A nearly constant value of k indicates that the \\ Rate = k [N2O2] [H2] ...(i) reaction is a first order reaction. Npr2eOs2ensht oinultdhbeyoveelirmalilnraetaecdtiboenc.ause it is not 46. Initial concentration, [A]0 = r0 – r ∞ At time t, From step 1, we get [A]t = rt – r ∞ When t = 7.18 s N2O2  K= k= 2.303 log d 24.09 –]–10.74g n [NO]2 7.18 21.7 –]–10.74g [N2O2] = K [NO]2 ...(ii) = 2.303 log c 34.83 m= 0.0093 s–1 \\ From (i) and (ii) 7.18 32.44 Rate = k K [NO]2 [H2] or k′ [NO]2 [H2] When t = 18.0 s (ii) Order = 3 k= 2.303 log d 24.09 –]–10.74g n 18 17.7 –]–10.74g 39. Rate = k [A] ...(i) k= 2.303 log c 34.83 m= 0.011 s–1 27 × Rate = k [A]a ...(ii) 18 28.44 Dividing (ii) by (i): a = 3 Since, ‘k’ was nearly the same at different time \\ Order of reaction, a = 3 intervals, t. Thus, the reaction is a first order reaction. 40. Rate = k [A] [B]3/2 47. r = k1[NO2] [NO3], Order of reaction = 1 + 3 = 5 2 2 [NO2 ] [NO3 ] K[N2O5 ] K= [N2O5 ] or [NO3 ] = [NO2 ] Order of elementary reaction cannot be fractional. 41. Unit of rate = mol L–1 s–1 r = k1K[N2O5] = k[N2O5] Thus the mecahnism is consistent with the Order = 2 + 1 = 3 observed law and it is a first order reaction. Unit of k =  L 3−1 sec–1 = L2 mol–2 s–1 48. Adding both the steps given in the question, we  mol  get the overall reaction as: 42. k = 2.303 log  a a  2NO + Br2 → 2NOBr t  − x  The second step is slow and therefore, is the rate when (a – x) = a , t = 50 sec determining step. 4 So, r = k′[NOBr2] [NO] k= 2.303 log  a a 4  = 2.303 × 0.602 K= [NOBr2 ] 50  /  50 [NO] [Br2 ] = 2.77 × 10–2 sec–1 [NOBr2] = K[NO] [Br2] r = k′ K[NO]2 [Br2] 43. For first order reaction: r = k[NO]2 [Br2] [k = k′K] k = 2.303 log  a  = 2.303 log  0.6  2.303 a t  a− x  5  0.2 t − 49. k= log = 0.2197 min–1 (a x) 44. For a the decomposition of H2O2 k= 2.303 log  100  20  100 − 15  ChemiCal KinetiCs 95

= 8.127 × 10–3 min–1 Given: N0 = 100, N = 100 – 75 = 25 k = 2.303 log  a a  l= 2.303 log  N0  20  − x  t  N  8.127 × 10−3 = 2.303 log  100  0.693 = 2.303 log 100  t  100 − 60  10 t  25  t = 112.78 min t = 20 seconds 50. k = 2.303 log R0 55. (i) Zero order t Rt (ii) Rate constant, k 2.4 × 10–3 = 2.303 log 1 (iii) Unit of rate constant, k = mol L–1 sec–1 t / 1 4 56. t1 / 2 = 0.693 = 0.693 = 1.264 sec k 5.48 × 10−1 t = 577.6 sec 51. We know, 0.693 = 2.303 log a t1 / 2 t − a x ⇒ 0.693 = 2.303 log a 1.264 t2 / 3 a − 2 a 3 ⇒ t2/3 = 2.005 sec 52. (i) For the 1st order reaction 57. For a first order reaction: k= 2.303  a  ...(a) k = 2.303 log10  a  ...(i) t  − x  t  a− x  log a Given: k = 6 min–1, a = 100, x = 20, t = 15 a = 5 mol L–1, k= 2.303 log  100  (a – x) = 0.05 mol L–1 15 100 − 20  From (i), = 0.0149 min–1 6 = 2.303 log  5  t  0.05  (ii) a = 100, (a – x) = 10 Now, 0.0149 = 2.303 log 100  ⇒ t = 0.768 min t  10  t =154.6 minutes 58. Unit of rate constant is used to determine the order of the reaction. (iii) a= 80, x = 80 × 20 = 16 (i)  L  n−1 sec–1 = mol L–1 s–1 100  mol (a – x) = 64 Comparing the power of L or mol. Now, 0.0149 = 2.303 log  80  n – 1 = –1 t  64  n = 0, zero order \\ t = 15 minutes Similarly, the order of the reactions are calculated and found to be 53. For 15% completion, (ii) Second order (iii) Third order k= 2.303 log  a  t  a − x  (iv) First order (v) Second order 2.303  100  (vi) Second order 20  100 − 15 = log 59. For a first order reaction: = 8.127 × 10–3 min–1 k= 2.303 . log [R0] t [R] For 60% completion At half life, 8.127 × 10–3 = 2.303 log  100  R = ; R0 E t  100 − 60  2 2.303 ⇒ t = 112.7 min \\ At t = t1/2, k = t1/2 log [2] 54. Average life = t1/2 × 1.44 or t1/2 = 2.303 log 2 = 0.693 = 10 × 1.44 = 14.4 sec k k 96 Chemistry-12

60. [ester]t = (V – Vt) 100 = Antilog (0.0515) ∞ 100 − x On applying, k = 2.303 log [ester]t1 100 = 1.1259 (t2 − t1 ) [ester]t2 100 − x The value of k is found to be 1.27 × 10–4 s–1 in the x = 11.2% first case, 1.19 × 10–4 s–1 in the second case and 1.24 × 10–4 s–1 in the third case. 65. Let us suppose reaction is of first order A nearly uniform value of k indicates the reaction k= 2.303 log  V0  , then follows the first order. t  V2  61. Lowering of activation energy, At 10 min, DE = 75 – 20 = 55 kJ mol–1 k= 2.303 log 22.8 10 13.8 kp = Antilog  ∆E  ...(i) ka  2.303RT  = 0.05022 min–1 twhheeprereksep nacnedankda are the rate constants in At 20 min, in the absence of catalyst, respectively. k= 2.303 log 22.8 20 8.3 kp \\ ka = Antilog  55 × 1000  = 0.05053 min–1  2.303 × 8.314 × 293  Since the values of rate constant, are nearly = 6.4 × 109 the same at two different instants, hence our assumption is correct and the reaction follows \\ Rate will increase by 6.4 × 109 times. first order kinetics. 62. (i) t1 / 2 ∝ 1 The average value of rate constant, k α n −1 = 0.05022 + 0.05053 (ii) k = 2.303 log  a a  2 t  − x  = 0.05037 min–1 2.303 100 2.3 × 10−5 = 200 × 60 log (a − x) 66. According the first order kinetics equation: (a – x) = 75.88% k= 2.303 log10  a a  t  − x  Here, (a – x) = Percentage of the remaining concentration k= 2.303 log (N2O5 )0 t (N2O5 )t 63. Rate = k [A]a [B]b At 410 sec, 2.1 × 10–3 = k[0.1]a [1]b ...(a) 2.303 5.5 8.4 × 10–3 = k[0.2]a [1]b ...(b) k= 410 log 4.0 8.4 × 10–3 = k[0.2]a [2]b ...(c) Dividing (b) by (a), 4 = (2)a = 7.768 × 10–4 sec–1 a=2 At 1130 sec, Dividing (c) by (b) k= 2.303 log 5.5 1130 2.4 1 = [2]b b=0 = 7.41 × 10–4 sec \\ Overall order = 2 + 0 = 2 67. As we know 64. For a first order reaction 2.303  V∞ − V0  t  V∞ − Vt  0.693 = 2.303 log  a  ...(a) k= log t1 / 2 t  − x  a At t = 25 min, Given: t1/2 = 3.15 × 104 sec, t = 90 × 60 sec k=  2.303 log 42.1 − 19.24 a = 100, x = ?  25  42.1 − 24.20 By putting these values in (a) 0.693 = 2.303 log  100 x  = 2.303 log  22.86 3.15 × 104 5400  100 −  25  17.9  log 100 x = 0.0515 = 9.765 × 10–3 min–1 100 − At t = 40 min, ChemiCal KinetiCs 97

k= 2.303 log  42.1 − 19.24  \\ ln[R0] = – k × 0 + I 40  42.1 − 26.6  or ln[R0] = I R = 2.303 log 22.86 Thus, ln < R0 F = – kt 40 15.50 Taking antilog on both sides, we get = 9.615 × 10–3 min [R] = [R0] e–kt Values of rate constant are nearly the same at Or [R] = e–kt different time intervals, hence the reaction follows first order kinetics. [R0] 2.303 [R0] (t1/2 )1  a2  n−1 \\ k= t log [R] (t1/2 )2  a1  68. = Unit of rate constant ‘k’ for first order reaction is s–1 where the value of n is the order of the reaction. 71. A reaction which may appear to be first order reaction, but in actual is not truly a first order From the given data reaction is a pseudo first order reaction. For 3.52  13333 n−1 1.92  6667  illustration, Consider the hydrolysis of an ester as = (a ∝ initial pressure) described by a chemical equation written below: Taking log on both sides CH3COOC2H5 + H2 O H+ log 3.52 = (n – 1) log 2 (in excess) CH3COOH + C2H5OH 1.92 Here, water is taken in large excess; 0.2632 = 0.3010 × (n – 1) So, the rate law equation, n = 1.88 ª 2 Rate, R = k[CH3COOC2H5] [H2O] Thus, this reaction follows second order kinetics. practically becomes: 69. Let the rate equation be Rate, R = k¢[CH3COOC2H5], Rate = k [A]x × [B]y where k¢ = k[H2O] [Since H2O is taken in large excess] From experiment (1), Another example is: 0.3 × 10–4 = k(0.03)x [0.03]y ...(a) From experiment (2), C12 H22 O11 + H2 O H+ C6 H12 O6 + C6 H12 O6 (cane sugar) (in excess) 1.2 × 10–4 = k[0.06]x [0.06]y ...(b) (Glu cos e) (Fructose) On dividing equation (b) from equation (a) Like the hydrolysis of ester it is also pseudo first 1.2 × 10−4 = (0.06)x (0.06) y order reaction. 0.3 × 10−4 (0.03)x (0.03) y 72. (i) (a) The initial rate is = k[A]0p [B]0q 4 = (2)x × (2)y ...(c) Comparing and substituting the values we ...(d) get the experimental results Similarly from experiment (1) and (3) (Rate)1 = k(0.1)p (0.1)q (2)x × (3)y =9 = 1.5 × 10–3 ...(i) ...(ii) Solving equations (c) and (d) (Rate)2 = k(0.20)p (0.20)q ...(iii) = 3.0 × 10–3 x = 0, y = 2 (Rate)3 = k(0.20)p (0.40)q \\ Rate equation, gives: = 6.0 × 10–3 Rate = k[B]2 Dividing eq. (iii) by (ii), we get Considering equation (a) again, k = 0.3 × 10−4 (0.4)q = 6.0  103 = 2 (0.03)2 (0.2)q 3.0  103 ⇒ k = 3.33 × 10–2 mol L–1 s–1 70. For a first order reaction, R → P 2q = 21 d [R] Rate = – dt = k[R] or q = 1 = Order wrt B. or d [R] = – k dt Similarly comparing the equations of [R] experiments I and II: On integrating above equation, we get (0.2) p ×2 = 3.0 × 10−3 (0.1) p 1.5 × 10−3 lnR = – kt + I, where I = constant of integration 2p × 2 = 2 When, t = 0, R = R0 = initial concentration of the or 2p = 1 reactant or p = 0 = Order wrt A. 98 Chemistry-12

Rate law is: (iii) Order wrt CHCl3 = 1, Order wrt Cl2 = ½ ; Overall order = 1½ \\ Rate = k[A][B] or k [B] (b) Also, k= Rate (iv) Order wrt [H2O2] = 1, 5B? Order wrt I–= 1; = 3 × 10–3 = 15 × 10–3min–1 Order wrt H+ = 0; Overall order = 2 0.2 3 (v) Order wrt CH3CHO= 2 ; (ii) We have, 3 k= 2.303 log Pi Overall order = 2 or 1½ t Pt Sum of the power of concentration terms = 2.303 Pi t log (2Pi – Pt) involved in the rate law is equal to the order = 2.303 0.5 of the reaction. 100 0.5 – log 2 × 0.6 74. wUhneitreofnk==or;dmLeorloEnf–r1esa–1ction. = 2.23 × 10–3 s–1 (i) For zero order reaction, n = 0, So unit of rate constant is mol L–1 s–1. When pressure, Pt = 0.65 atm, i.e. (ii) For first order reaction, n = 1, So unit of rate Pi + P = 0.65 atm constant is s–1. or P = 0.15 atm (iii) For second order reaction, n = 2, So unit of Now, PSO2Cl2 = Pi – P =0.50 – 0.15 = 0.35 atm rate constant is L mol–1 s–1. ⇒ Rate = k × PSO2Cl2 1 3 3 (iv) Forr1 2 o 2 order reaction, n = 2 , So, unit = 2.23 ×10–3s–1×0.35 atm = 7.805 × 10–4 atm s–1 of rate constant is L1/2 mol–1/2 s–1. 73. (i) Order wrt NO = 2, order wrt H2 = 1; Overall (v) For a nth order reaction, n = n, So unit of rate order = 3 constant is ; L n–1 s –1 (ii) Order wrt N2O = 1; Overall order = 1 mol E Topic 4. Effect of Temperature on Reaction Rate and Collision Theory • Increase in temperature increases the rate of Kelvin. Taking log on both sides of the Arrhenius reaction. equation, it can be put in a more useful form: The change of rate with temperature is expressed ln k = – Ea + ln A by the change in the specific rate constant, k. RT For every reaction, k increases with temperature. • Arrhenius equation is valuable because it can be In several reactions, a 10°C rise in temperature used to calculate the minimum energy necessary approximately doubles the reaction rate. to bring about a reaction between the colliding The ratio of the specific constant for a reaction at emxopleercuimleesn, tia.el.,vtahleueacotfivtahteionraetenecrognys, tEana,ti,f the two temperatures separated by 10°C (generally known. k is at 25°C and 35°C) is known as the temperature coefficient. Energy of Activation for a Reaction Temperature coefficient = k at (t + 10°C) • For reactants to get converted into products, k at t°C there is an energy barrier, which has to be overcome. Only those encounters in which = k35°C reactant molecules have sufficient energy to cross k25°C this barrier, would result in products. • The temperature coefficient for most of the • It is of tremendous consequence to the existence of life that such energy barriers exist. Otherwise, reactions varies between 2 and 3. The quantitative all the fuels would have been consumed by the oxygen in the air. Infact life would not be possible relation between k and T is given by Arrhenius on earth without the presence of these barriers. equation • Every reaction feasible from consideration of k = Ae–Ea/RT free energy need not necessarily take place. Many may be prevented from occurring by a high ‘A’ is a constant which is experimentally tdheetegramsinceodn,stEaantisatnhde activation energy, R is T is the temperature in ChemiCal KinetiCs 99