C1-Allens Made Chemistry Theory {PART-1}

["SOLVED EXAMPLESEx .1The volume of a closed reaction vessel in which the equilibrium :2SO (g) + O (g) 22\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 2SO (g) sets is halved, Now -3(A) the rates of forward and backward reactions will remain the same.(B) the equilibrium will not shift.(C) the equilibrium will shift to the left.(D) the rate of forward reaction will become double that of reverse reaction and the equilibrium will shift to the right.Sol.( D )In the reaction2SO (g) + O (g) 22\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 2SO (g)3In this reaction three moles (or volumes) of reactants are converted into two moles (or volumes) of productsi.e. there is a decrease in volume and so if the volume of the reaction vessel is halved the equilibrium will beshifted to the right i.e. more product will be formed and the rate of forward reaction will increase i.e. doublethat of reverse reaction.Ex .2The equilibrium constant of the reaction A (g) + B (g) 22\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 2AB (g) at 100\u00b0C is 50. If one litre flaskcontaining one mole of A is connected to a 3 litre flask containing two moles of B the number of moles of22AB formed at 373 K will be -(A) 1.886(B) 2.317(C) 0.943(D) 18.86Sol.( A )The equilibrium is represented as :A (g) + B (g) 22\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 2AB (g)Initial concentration1 2 0Moles at equilibrium1\u2013x 2\u2013x2xTotal volume = 1 + 3 = 4 litres[A ] = 21 x4 \uf02d, [B ] = 22x4 \uf02d and [AB] = 2x4K = 222[AB][A ][B ] = 22x4501 x2x44\uf0e6\uf0f6\uf0e7\uf0f7\uf0e8\uf0f8\uf03d\uf02d \uf0f6\uf0e6\uf02d \uf0f6\uf0e6\uf0e7\uf0f7\uf0e7\uf0f7\uf0e8\uf0f8\uf0e8\uf0f8On solving we get 23x \u2013 75 x + 50 = 02\uf05c x = 2.31 or 0.943, since x can't be more than 1so, x = 0.943\uf05c moles of AB formed = 2 \u00d7 0.943 = 1.886Ex .3H (g) + I (g) 22\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 2HI (g)When 92 g of I and 1g of H are heated at equilibrium at 450\u00b0C, the equilibrium mixture contained 1.9 g of22I . How many moles of I and HI are present at equilibrium.22(A) 0.0075 & 0.147 moles(B) 0.0050 & 0.147 moles(C) 0.0075 & 0.7094 moles(D) 0.0052 & 0.347 molesSol.( C )moles of I taken =292254= 0.3622","moles of H taken =212 = 0.5moles of I remaining =21.9254 = 0.0075moles of I used = 0.3622 \u2013 0.0075 = 0.35472moles of H used = 0.35472moles of H remaining = 0.5 \u2013 0.3547 = 0.14532moles of HI formed = 0.3547 \u00d7 2 = 0.7094At equilibriummoles of I = 0.0075 moles2moles of HI = 0.7094 molesEx .4When 1.0 mole of N and 3.0 moles of H was heated in a vessel at 873 K and a pressure of 3.55 atm. 30%22of N is converted into NH at equilibrium. Find the value of K for the reaction.23P(A) 3.1 \u00d7 10 atm\u20132 \u20132(B) 4.1 \u00d7 10 atm\u20132 \u20132(C) 5.1 \u00d7 10 atm\u20132 \u20132(D) 6.1 \u00d7 10 atm\u20132 \u20132Sol.( C )N (g)2+3H (g)2\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 2NH (g)31 mole3 moles 0Initial moles1\u20130.33.0 \u2013 0.9 0.6 molesat equilibrium= 0.7 moles =2.1 molesTotal no. of moles at equilibrium = 3.4K = P230.63.553.40.72.13.553.553.43.4\uf0e6\uf0f6\uf0b4\uf0e7\uf0f7\uf0e8\uf0f8\uf0e6\uf0f6\uf0e6\uf0f6\uf0b4\uf0b4\uf0e7\uf0f7\uf0e7\uf0f7\uf0e8\uf0f8\uf0e8\uf0f8= 5.1 \u00d7 10 atm\u20132\u20132Ex .52SO (g) + O (g) 22\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 2SO (g)3 If the partial pressure of SO , O and SO are 0.559, 0.101 and 0.331 atm respectively. What would be the223partial pressure of O gas, to get equal moles of SO and SO .223(A) 0.188 atm(B) 0.288 atm(C) 0.388 atm(D) 0.488 atmSol.( B )2SO (g) + O (g) 22\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 2SO (g)3 K = P3222SO2SOO[P][P] [P ]= 22(0.331)(0.559) (0.101)= 3.47If SO and SO have same number of moles, their partial pressure will be equal and2332SOSOPP\uf03d\uf05c2 O P= 10.288atm3.47\uf03dEx .6A (g) and B (g) at initial partial pressure of 98.4 and 41.3 torr, respectively were allowed to react at 400 K.22At equilibrium the total pressure was 110.5 torr. Calculate the value of K for the following reaction atP400 K. 2A (g) + B (g) 22\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 2 A B (g)2(A) 124(B) 134(C) 154(D) 174Sol.( B )","The given reaction is,2A (g)2+B (g)2\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf0882A B (g)2Initial pressure (torr)98.441.3 0At equilibrium98.4\u2013x41.3\u2013x2 xTotal pressure at equilibrium = 110.5 torr.(98.4 \u2013 x) + (41.3 \u2013 x2) + x = 110.5\uf05c x = 58.4 torr (760 torr = 1 atm)P(A B)2 = 58.4 torr = 7.68 \u00d7 10 atm\u20132P(A )2 = 98.4 \u2013 58.4 = 40 torr = 5.26 \u00d7 10 atm\u20132P(B )2 = 41.3 \u2013 58.42 = 12.1 torr = 1.59 \u00d7 10 atm\u201322222A BP2ABPKPP\uf03d\uf0b4 = 2 222(7.68 10 )(5.26 10 )(1.59 10 )\uf02d\uf02d\uf02d\uf0b4\uf0b4\uf0b4 = 134Ex .7K for the reaction N + 3H P22\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 2NH at 400\u00b0C is 3.28 \u00d7 10 . Calculate K .3\u20134 c(A) 0.3 mole litre\u201322(B) 0.4 mole litre\u201322(C) 1.0 mole litre\u201322(D) 0.6 mole litre\u201322Sol.( C )N + 3H 22\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 2NH3\uf044n = \u20132 and K = K (RT)PC\uf044 n3.28 \u00d7 10 = K (0.0821 \u00d7 673)\u20134c\u20132and K = 1.0 mole litre .c\u201322Ex .80.96 g of HI were heated to attain equilibrium 2 HI \uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 H + I . The reaction mixture on titration requires2215.7 mL of N\/20 hypo. Calculate % dissociation of HI.(A) 18.9%(B) 19.9%(C) 10.46%(D) 21.9%Sol.( C )2HI \uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 H2 + I2Initial moles0.961280 0 = 7.5 \u00d7 10\u20133Moles at equilibrium (7.5 \u00d7 10 \u2013 x)\u20133x\/2 x\/2Now Meq. of I formed at equilibrium = Meq. of hypo used2W1000 = 15.7 \u00d7 E\uf0b4120 or WEof I = 0.785 \u00d7 102\u20133\uf05c Moles of I formed at equilibrium = 230.785 102\uf02d\uf0b4= 0.3925 \u00d7 10\u20133or x2= 0.3925 \u00d7 10 or x = 0.785 \u00d7 10\u20133 \u20133\uf05c degree of dissociation of HI = moles dissociatedinitial moles = \u20133x7.5 10\uf0b4\uf061 = 3\u201330.785 107.5 10\uf02d\uf0b4\uf0b4= 0.1046 = 10.46%","Ex .9A mixture of H and I in molecular proportion of 2 : 3 was heated at 444\u00b0C till the reaction22H + I 22\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 2HI reached equilibrium state. Calculate the percentage of iodine converted into HI.(K at 444\u00b0C is 0.02)C(A) 3.38 %(B) 4.38%(C) 5.38%(D) 6.38%Sol.( C ) H +2 I2 \uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf0882HIInitial moles 2 3 0Equi.conc.2xv\uf02d3xv\uf02d2xvK = C24x(2x)(3x)\uf02d\uf02d = 0.02199 x + 5x \u2013 6 = 02 x = 0.1615Out of 3 moles, 0.1615 moles I is converted into HI.2\uf05c Percentage of I converted to HI = 20.1615 1003\uf0b4 = 5.38%Ex. 10 The equilibrium composition for the reaction is :PCl3+Cl2\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088PCl50.200.050.40 moles\/litreIf 0.25 moles of Cl is added at same temperature. Find equilibrium concentration of PCl (K = 20)25C(A) 0.48 moles\/litre(B) 0.38 moles\/litre(C) 0.56 moles\/litre(D) 1.20 moles\/litreSol.( A )PCl3+Cl2\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088PCl50.200.050.40 moles\/litreIf 0.25 moles of Cl is added then at equilibrium [Let V = 1L]20.20 \u2013 x0.30\u2013x0.40 +x20 = 0.40x(0.20x)(0.30x)\uf02b\uf02d\uf02d or x = 0.08[PCl ] = 0.4 + 0.08 = 0.48 moles\/litre5Ex. 11 The equilibrium constant K, for the reaction N + 3H 22\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 2NH is 1.64 \u00d7 10 atm at 300\u00b0C. What will3\u20134 \u20132be the equilibrium constant at 400\u00b0C, if heat of reaction in this temperature range is \u2013 105185.8 Joules.(A) 0.64 \u00d7 10 atm\u20135\u20132(B) 6.4 \u00d7 10 atm\u20133\u20132(C) 0.64 \u00d7 10 atm\u20133\u20132(D) 0.64 \u00d7 10 atm\u20131\u20132Sol.( A )1 p K= 1.64 \u00d7 10 atm , \u20134\u201322 p K = ?T = 300 + 273 = 573 K1T = 400 + 273 = 673 K2\uf044H = \u2013105185.8 JoulesR = 8.314 J\/K\/moleApplying equation12 pplog Klog K \uf02d=H2.303 R\uf0442112TTT T\uf0e6\uf0f6\uf02d\uf0e7\uf0f7\uf0e8\uf0f82 p log K\u2013 log 1.64 \u00d7 10 = \u2013\u20134 105185.82.303 8.314\uf0b4673573673 573\uf02d\uf0e6\uf0f6\uf0e7\uf0f7\uf0b4\uf0e8\uf0f8or K2 p= 0.64 \u00d7 10 atm\u20135\u20132","Ex. 12 In an experiment at 500 K, the concentration of different species are [NH ] = 0.105 mol dm ,3\u20133[N ] = 1.10 mol dm and [H ] = 1.50 mol dm then find the followings :-2\u201332\u20133 (a) values of K and K for the reactionCP N + 3H 22\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 2NH3(b) value of K for the reaction -c 2NH 3\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 N + 3H22Sol.(a)For the reaction N + 3H 22\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 2NH3K = C23322[NH ][N ][H ][NH ] = 0.105 mol dm , [N ] = 1.10 mol dm and3\u201332\u20133[H ] = 1.50 mol dm2\u20133K =C3 233 3(0.105 mol dm )(1.10 mol dm ) (1.50 mol dm )\uf02d\uf02d\uf02d\uf0b4= 2.97 \u00d7 10 mol dm\u20133 \u201326Now K = K \u00d7 (RT)PC\uf044 n\uf044n = \u20132,R = 0.082 atm dm K mol , T = 500 K3 \u20131 \u20131 \uf05cK = (2.97 \u00d7 10 mol dm ) \u00d7 [(0.082 atm dm K mol ) \u00d7 (500 K)]P\u20133 \u20133 63 \u20131 \u20131\u20132 = 1.76 \u00d7 10 atm\u20136 \u20132(b)The equilibrium constant K for the reverse reaction is related to the equilibrium constant K for theCCforward reaction as :K' = C c1K = \uf02d\uf0b43\u20132612.97 10 mol dm = 3.37 \u00d7 10 mol dm\u20132 2 \u20136Ex. 13 The equlibrium pressure of NH CN (s) 4\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 NH (g) + HCN (g) is 0.298 atm. Calculate K . If NH CN (s)3P4is allowed to decompose in presence of NH at 0.50 atm then calculate partial pressure of HCN at equilib-3rium.Sol.NH CN (s) 4\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 NH (g) + HCN (g)3Pressure at equilibrium \u2013P P\uf05c Total pressure at equilibrium = 2P = 0.298 atm\uf05c P = 0.149 atm\uf05c K = P3 NHHCNPP\uf0b4 = 0.149 \u00d7 0.149 = 0.0222 atm2If dissociation is made in presence of NH at 0.5 atm3NH CN (s) 4\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088NH (g) + HCN (g)3Initial pressure \u2013 0.50 0Pressure at equli. \u2013 (0.50+P') P'Also K = P' (0.50 + P')Por 0.0222 = P' (0.50 + P')\uf05c P' = 0.1656 atm","Ex. 14 The value of K for the reaction,CN + 2O 22\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 2NO at a certain temperature is 900. Calculate the value of equilibrium constant for2 (i) 2NO 2\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 N + 2O22(ii) \u00bd N + O 22\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 NO2Sol.Equilibrium constant (K ) for the reactionCN + 2O 22\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 2NO is2 K = C 22222[NO ]900[N ][O ]\uf03d(i) For the reaction 2NO 2\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 N + 2O , K' =22C22222[N ][O ][NO ]=c1K K' =C1900= 0.0011 mole litre\u20131(ii) For the reaction \u00bd N + O 22\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 NO2 K\\\" = C 2\u00bd22[NO ][N ] [O ] = C K K\\\" = C 900 = 30 lit mol\u00bd\u2013\u00bdEx. 15 Ice melts slowly at higher altitude, why ?Sol.According to Le Chatelier principle, the melting of ice is favoured at high pressure because the forwardreaction ice \uf0be\uf0ae water shows a decrease in volume. At higher altitude atmospheric pressure being low andthus ice melts slowly.Ex. 16 Both metals Mg and Fe can reduce copper from a solution having Cu ion according to equilibria.+2 Mg (s) + Cu+2 \uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 Mg + Cu (s) ; K = 5 \u00d7 10+21 90Fe (s) + Cu +2\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 Fe + Cu (s) ; K = 2 \u00d7 10+2 226Which metal will remove cupric ion from the solution to a greater extent.Sol.Since K > K , the product in the first reaction is much more favoured than in the second one. Mg thus12removes more Cu from solution than Fe does.+2Ex. 17 The equilibrium constant K for Y (g) C\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 Z (g) is 1.1. Which gas has molar concentration greaterthan 1.Sol.For Y (g) \uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 Z (g)K = C[Z][Y]= 1.1if Z = 1 ;[Y] = 0.91Case I0.9 < [Y] 1 only Z = 1\uf03cCase II[Y] 1 both [Y] and [Z] > 1\uf03e","Ex. 18 When S in the form of S is heated 800 K, the initial pressure of 1 atmosphere falls by 30 % at equilibrium.8This is because of conversion of some S to S . Calculate the K for reaction.82PSol.S (g) 8\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 4 S (g)2Initial pressure1 atm 0Equilibrium pressure(1\u20130.30) 4\u00d7 0.30= 0.70 atm = 1.2 atmNow, K = P284SSPP= 4(1.2)0.70= 2.9622 atm3Ex. 19 A vessel at 1000 K contains CO with a pressure of 0.6 atm. some of the CO is converted into CO on22addition of graphite. Calculate the value of K, if total pressure at equilibrium is 0.9 atm.Sol.CO (g) + C (s) 2\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 2 CO (g)Initial pressure0.6 atm \u2013 0Equilibrium pressure(0.6\u2013x)atm 2x atmFrom question, (0.6 \u2013 x) + 2x = 0.9 hence, x = 0.3 atm.Now, K = P22COCOPP= 2(2x)(0.6x) \uf02d= 1.2 atm.","","1 .CLASSIFICATION OF REACTIONS : [IN TERMS OF RATES](i)There are certain reactions which are too slow e.g. rusting of iron, weathering of rocks.(ii)Instantaneous reactions i.e. too fast e.g. Detonation of explosives, acid-base neutralization,precipitation of AgCl by NaCl and AgNO .3(iii)Neither too fast nor too slow e.g. combination of H and Cl in presence of light, hydrolysis of ethyl22acetate catalysed by acid, decomposition of azomethane2 .RATE OF REACTION :The change in concentration of either reactant or product per unit time.Formula : v = \u00b1 dcdtdc = change in concentration in a small interval dt.[\u2013] sign is used when we refer to reactant concentration.[+] sign is used when we refer to product concentration.\uf075Example :N + 3H 22\uf0ae 2NH3(i)Rate of formation of ammonia = +3 d[NH ]dt(ii)Rate of disappearance of nitrogen = \u20132 d [N ]dt(iii)Rate of disappearance of hydrogen = \u20132 d [H ]dtRate of reaction = + 1 d[NH ]23dt = \u2013 2 d [N ]dt = \u2013 132 d [H ]dtThus, Rate of reaction = \u2013 2 d [N ]dt = 1 d[NH ]23dtor rate of formation of ammonia = Twice the rate of disappearance of nitrogeni.e.3 d[NH ]dt = 232 d [H ]\u2013dt\uf0e9\uf0f9\uf0ea\uf0fa\uf0eb\uf0fb3 .AVERAGE VELOCITY OF REACTION :Change in the concentration of reactants or products per unit time is called average reaction velocity. If c\uf044is the change in the concentration of reactants and product in t time, then\uf044Average velocity = \u00b1 ct\uf044\uf044orAverage velocity = (\u2013)Change in the concentration of reactantsTime(+ )Change in the concentration of productsTimeCHEMICAL KINETICS","CProductReactantTimeUnit of average velocity = Unit of concentrationUnit of time = gram molelitre\u00d7 Second = gram mole litre second\u20131\u201314 .INSTANTANEOUS RATE OF THE REACTION :The rate of reaction determined at specified concentration or specified time BTimeCis called instantaneous rate.The instantaneous rate of the reaction can be determined by measuringconcentration of reactant or product at a instant of time and plotting con-centration versus time.The instantaneous rate at any time is determined by the slope of the tan-gent at a point on the time-concentration curve corresponding to the speci-fied time. The slope of the tangent at a point is the limiting value of ct\uf044\uf044.t0cLimt\uf044 \uf0ae\uf044\uf044=dcdtIn terms of the concentration of reactant, the rate of the reaction = \u2013dcdt. The \u2013sign indicates that theeconcentration of reactant decreases with time.In terms of the concentration of product, the rate of the reaction = +dcdt. The +sign indicates that theeconcentration of product increases with time. In the reaction if at a time the concentration of product is x andtat time t + dt, the concentration becomes x + dx then the reaction rate =dxdt.\uf075For example the rate of reaction : N + 3H 22\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 2NH3in terms of the concentrations of N , H and NH can be expressed as :223\u2013 2 d[N ]dt, \u20131 d[H ]32dt, + 1 d[NH ]23dtEx .1What should be (a) the rate of disappearance of B and (b) the rate of formation of C, if the rateof disappearance of A for the reaction A + B \uf0ae 2C is 10 mole\/litre\/second at a particular\u20132temperature ?Sol.(a) Rate of disappearance of A = Rate of disappearance of B = 10 mole\/litre\/second\u20132(b) Rate of disappearance of A = 12 \u00d7 Rate of formation of CRate of formation of C = 2 \u00d7 Rate of disappearance of A = 2 \u00d7 10 mole\/litre\/second\u20132","Ex.2A gaseous reaction : 2A(g) + B(g) \uf0ae 2C(g),Show a decrease in pressure from 120 mm to 100 mm in 10 minutes. The rate of appearance of C is -[A] 2 mm\/min[B] 4 mm\/min[C] 10 mm\/min[D] 12 mm\/min. Ans. [B]Sol.Suppose 2p is the pressure of C after 10 min.Fall in pressure of A = 2p ; Fall in pressure of B = pTotal fall in pressure = (2p + p) \u2013 2p = p = 20 mmPressure of C = 2p = 40 mmRate of appearance of C = 40\/10 = 4 mm\/minEx.3The term dxdt in the rate expression refers to the -[A] concentration of the reactants[B] increase in concentration of the reactants[C] instantaneous rate of the reaction[D] average rate of the reactionAns. [C]Sol.It is expression for instantaneous rateEx.4Which of the following expression can be used to describe the instantaneous rate of the reaction ?2A + B\uf020\uf0ae A B2[A] \u2013 1 d[A]2 dt[B] \u2013d [A]dt[C] 21 d[A B]2dt[D] \u20131 d[A] d[B].2 dtdtAns. [A]Sol.The instantaneous rate of the reaction can be expressed by any of the following expressions1 d[A]\u20132 dt or \u2013d[B]dt or 2d[A B]dtEx.5Which of the following will react at the highest rate ?[A] 1 mol of A and 1 mol of B in a 1 L vessel [B] 2 mol of A and 2 mol of B in a 2 L vessel[C] 3 mol of A and 3 mol of B in a 3 L vessel [D] All would react at the same rateAns. [D]Sol.Since all have same conc. of reactants, all would react at same rate.5 .FACTORS AFFECTING THE RATE OF REACTION :(i)Concentration : Law of mass action enunciates that greater is the conc. of the reactants, the morerapidly the reaction proceeds.(ii)Pressure (Gaseous reaction) : On increasing the pressure, volume decreases and conc. increasesand hence the rate increases.(iii)Temperature : It is generally observed that rise in temperature increases the reaction rate.(i v)Nature of the reactants : The rate depends upon specific bonds involved and hence on the natureof reactants.gs \uf03e \uf03e \uf06c( v )Surface area of the reactants : In heterogeneous reactions, more powdered is the form ofreactants, more is the velocity. [as more active centres are provided](v i)Catalyst : Affects the rate immensely.Ex .6For the reaction :4NH (g) + 5O (g) 32\uf0ae 4NO(g) + 6H O(g)2Given : d[NO]dt= 3.6 \u00d7 10 mol L s\u20133\u20131\u20131Calculate : (i) rate of disappearance of ammonia (ii) rate of formation of water","Sol.From the eqn. it is clear thatRate = \u2013 143 d[NH ]dt = 14d[NO]dt = 1 d[H O]62dt Thus : \u2013 143 d[NH ]dt = 14d [N O ]dtor\u20133 d[NH ]dt = d[NO]dt= 3.6 \u00d7 10 mol \u20133L \u20131 s\u20131 Also14d[NO]dt = 1 d[H O]62dt32d[NO]dt=2d[H O]dt32 \u00d7 3.6 \u00d7 10 = \u201332d[H O]dt2d[H O]dt= 5.4 \u00d7 10 mol L s\u20133\u20131 \u20131Ex .7The following reaction was studied in a closed vessel.2N O (g) 25\uf088\uf088\uf086\uf087\uf088\uf088 4NO (g) + O (g)22It was found that concentration of NO increases by 4.0 \u00d7 10 mol L in five seconds, calculate2\u20132\u20131(a)the rate of reaction(b)the rate of change of concentration N O .25Sol.(a)Rate =2 d[NO ] 14dtBut 212 d[NO ]4.0 10 mol Ldt5 sec\uf02d\uf02d\uf0b4\uf03d= 8 \u00d7 10 mol L s\u20133\u20131\u20131\uf05cRate of reaction = 14\u00d7 8 \u00d7 10 mol L s = 2 \u00d7 10 mol L s\u20133\u20131\u20131\u20133\u20131\u20131(b)Rate of change of conc. of N O25= \uf02d\uf03d \uf02d25 d[N O ]1dt2\u00d7 Rate of formation of NO2= \u2013 1 [d[N O ]]225dt = \u201312 \u00d7 8 \u00d7 10 = 4 \u00d7 10 mol L s\u20133\u20133\u20131 \u20131Ex .8The rate of change in concentration of R in the reaction, 2P + Q \uf0be\uf0be\uf0ae 2R + 3S, was reported as1.0 mol L sec . Calculate the reaction rate as well as rate of change of concentration of P, Q and S.\u20131\u20131Sol.\uf02d 12d[P]dtd[Q]1 d[R]1 d[S]dt2 dt3 dt\uf03d \uf02d\uf03d\uf03d= Rate of reaction\uf051d[R]dt = 1.0 molL s\u20131\u20131\uf05cd[P]d[R]dtdt\uf02d\uf03d= 1.0 mol L s\u20131\u20131d[Q] 1 d[R] 1dt2 dt2\uf02d\uf03d\uf03d = 0.5 molL s\u20131 \u20131d[S]3 d[R]3dt2 dt2\uf02d\uf03d\uf03d \u00d7 1 = 1.5 molL s\u20131\u20131Rate of reaction =1 d[C]2dt=12\u00d7 1 = 0.5 mol L s\u20131 \u20131","6 .SPECIFIC REACTION RATE :Applying law of mass action to the reaction :m A + m B 12\uf0ae n C + n D12Rate \uf0b51 m[A]2 m[B]orv = k1 m[A]2 m[B]This equation is known as rate law. Where k is the proportionality constant and is called(i) Velocity constantor(ii) Velocity coefficientor(iii) Specific reaction rate.On putting [A] = [B] = 1, where have : v = kHence specific reaction rate is the rate of the reaction when the concentration of each reactant is taken asunit.\uf075Unit of Specific Reaction Ratev = k 1 m[A]. 2 m[B]conc.time= k12 m +m[conc.]12 [1\u2013(m +m )][conc.] \u00d7 [time] = k\u20131ork =[1\u2013(m + m )]12molelitre\uf0e9\uf0f9\uf0ea\uf0fa\uf0eb\uf0fb. [second ]\u201317 .DISTINCTION BETWEEN UNIT OF RATE AND RATE CONSTANT :\uf075Rate of a reaction : Its units are always mole litre time .\u20131\u20131\uf075Rate constant : Its unit depends upon the order of reaction.8 .RATE LAW :( a )It may also not depend upon the concentration of each reactant or product of the reaction.Suppose,mA + nB \uf0ae ProductmnR[A] [B]\uf0b5(b)Rate of a chemical reaction is directly proportional to the product of the concentration of reactantsraise to the power of their stoichiometric coffecient.( c )The rate law represents the experimentally observed rate of reaction which depends upon the sloweststep of the reaction.( d )Rate law cannot be deduce from the equation for a given reaction. It can be find by experimentsonly.(e)The rate law may not bear a simple relationship of the stoichiometric equation.Ex.9In the reaction, A + 2B \uf0ae 6C + 2D, if the initial rate \u2013d [A]dt at t = 0 is 2.6 \u00d7 10 M sec , what will be the\u20132\u20131value of \u2013d [B]dt at t = 0 ?[A] 8.5 \u00d7 10 M sec\u20132\u20131[B] 2.5 \u00d7 10 M sec\u20132\u20131[C] 5.2 \u00d7 10 M sec\u20132\u20131[D] 7.5 \u00d7 10 M sec\u20132\u20131Ans. [C]Sol.From the reaction it is evident that when a mole of A is reacting, 2 moles of B must react. Hence thedecrease in the concentration of B must be twice that of A\uf05c\u2013 d [B]dt= 2 d [A]\u2013dt\uf0e9\uf0f9\uf0ea\uf0fa\uf0eb\uf0fb= 2 \u00d7 2.6 \u00d7 10 = 5.2 \u00d7 10 M sec\u20132 \u20132\u20131","Ex.10The dimensions of rate constant of a second order reaction involves :[A] time and concentration[B] neither time nor concentration[C] time only[D] concentration only Ans. [A]Sol.k = 2Rate[A] = \u20131\u20131\u20131 2mol L s(mol L ) = \u20131\u20131smol L = (mol L ) s\u20131 \u20131\u20131Ex.11The rate constant of a reaction has same units as the rate of reaction. The reaction is of[A] zero order[B] first order[C] second order[D] none of theseAns. [A]Sol.For a zero order reaction, r = k[A]\u00b0. Thus the units of k are the same as that of rate of reaction.Ex.12The rate constant of n order reaction has units :th[A] litre mol sec1\u2013n1\u2013n\u20131[B] mol litre sec1\u2013n1\u2013n[C] 21\u2013nmol2nlitresec\u20131[D] mol litre sec1\u2013nn\u20131\u20131Ans. [D]Sol.For an n order reaction : rate = k [conc.]thnk = nrate[conc.]Units of k = \u20131\u20131\u20131 nmol Ls(mol L ) = mol1\u2013n L sn\u20131\u20131Ex. 13On which of the following factors, the rate constant does not depend ?[A] Temperature[B] Concentration[C] Presence of catalyst[D] Nature of reactantsAns. [B]Sol.Rate constant is independent of the conc. of the reactants.9 .ORDER OF REACTION :The sum of the power of the concentration terms on which the rate of reaction actually depends as observedexperimentally is called the order of the reaction. For example,Order of reaction = x + yThus, the order of reaction may also be defined as the sum of the exponents (powers) to which theconcentration terms in the rate law equation are raised in order to express the observed rate of the reaction.Thus, reaction is said to be of the first order if its rate is given by the expression of the type,r = k C1ASecond order if the rate is given by the expression of the type,r = k C22Aorr = k C C2ABthird order if the rate is given by the expression of the typer = k C or r = k C C or r = k C C or k C C C and so on3A33A 2B3AB23ABCFor zero order reaction, the rate equation is written as R = k . It is to be noted that the order of reaction0is essentially an experimental quantity.Note:Order may be zero, fractional, integer or negative.Example :ReactionExperimental rate equationor de rH + Cl 22\uf0ae 2HClv = kzeroH + Br 22\uf0ae 2HBrv = k [H ] [Br ]221\/2one and halfH + I 22\uf0ae 2HIv = k [H ] [I ]22two","\uf075Examples of fractional order reactionReaction :CO(g) + Cl (g) 2\uf0ae COCl (g)2v = k [CO] [Cl ] , order = 2.5221\/2Reaction :COCl (g) 2\uf0ae CO(g) + Cl (g)2v = k [COCl ] , order = 1.523\/2Ex.14For the chemical reaction,4HBr + O 2\uf0be\uf0be\uf0ae 2H O + 2Br22Rate = k [HBr] [O ]2What is the probable mechanism of the reaction ?Sol.HBr + O 2\uf0be\uf0be\uf0ae HOOBr(slow)HOOBr + HBr \uf0be\uf0be\uf0ae 2HOBr(fast)HOBr + HBr \uf0be\uf0be\uf0ae H O + [Br ] \u00d7 222(fast)Ex. 15Nitric oxide (NO) reacts with oxygen to produce nitrogen dioxide :2NO (g) + O (g) 2 \uf0be\uf0be\uf0ae 2NO (g)2What is the predicted rate law, if the mechanism is :-NO (g) + O (g) 2\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 k NO (g)3 (fast)NO (g) + NO (g) 3 \uf0be\uf0be\uf0ae 1 k 2NO (g)2(slow)Sol.From the slow step,Rate = k [NO ] [NO]1 3..........(i)From fast step :Equilibrium constant (K) = 32[NO ][NO][O ]..........(ii)Substituting the value of [NO ] from equation (ii) into equation (i), we get3Rate= k'[NO] [O ]221 0 .MOLECULARITY OF A REACTION :\u2018\u2018Molecularity is defined as the number of molecules, atoms, or radicals that must collide simultaneously inorder for the reaction to take place.\u2019\u2019 It is always a whole number and cannot be negative.In the elementary processes :Participating speciesMolecularityOne species participates.....unimolecular,1Two species participates.....bimolecular, 2Three species participates.....trimolecular, 3\uf075Example :N O 24\uf0ae 2NO2......unimolecularH + I 22\uf0ae 2HI......bimolecular2FeCl + SnCl32 \uf0ae 2FeCl + SnCl ......24trimolecularNote :If the reaction takes place in two or more steps then the overall molecularity of the reaction is monitored bythe slow or rate determining step.","1 1 .DIFFERENCE BETWEEN MOLECULARITY AND ORDER OF REACTION :MolecularityOrder of Reaction1. Molecularity can neither be zero nor fractional1. Order of a reaction can be zero, fractionalor integer2. It is the number of molecules of reactants2. It is sum of power raised or the rate expression.concentration terms taking part in elementarystep of a reaction.3. It can not have negative value.3. Order of a reaction may have negative value.4. Molecularity is a theoretical property.4. Order is an experimental property.5. Molecularity concerns with mechanism.5. Order concerns with kinetic (rate law).Ex.ReactionRate lawO r d e rCH CHO 3\uf0be\uf0be\uf0ae CH + CO4Rate \uf0b5\uf020[CH CHO]33\/21.5NH 3\uf0be\uf0be\uf0ae12N + 232H 2Rate \uf0b5\uf020[NH ]30 02Hl \uf0be\uf0be\uf0ae H + I22Rate \uf0b5\uf020[HI] , i.e. Rate = k0 0Order may change with change in experimental conditions while molecularity can't.Ex.\uf0be\uf0be\uf0be\uf0be\uf0be\uf0be\uf0aeisomerizationHC2CH 3This reaction follows first order kinetics at high pressure and 2 order kinetics at low pressure ofndcyclopropane.1 2 .PSEUDO UNIMOLECULAR REACTION :Consider the reaction : CH COOC H + H O 3252+H\uf0be\uf0be\uf0be\uf0aeCH COOH + C H OH325Since water is present in large excess, its concentration hardly changes during the course of the reaction.And as such rate depends only on the concentration of ester. The order is one but the molecularity is two.Such reactions are called pseudo unimolecular reaction.Ex. 16For a chemical reaction, A \uf0ae products, the rate of reaction doubles when the concentration of A is in-creased by 4 times. The order of reaction is[A] 4[B] 0[C] 1\/2[D] 1Ans. [C]Ans.r = k[A] ........ (i)n;2r = k [4A] ........... (ii)nDividing (ii) by (i) 2rr = kkn4AA \uf0e9\uf0f9\uf0ea\uf0fa\uf0eb\uf0fbor 2 = 2 or 2n = 1 or n = 1\/22nEx. 17For a hypothetical reactionA + B \uf0ae products, the rate law is, r = k [B] [A]\u00b0, the order of reaction is :[A] 0[B] 1[C] 2[D] 3Ans. [B]Sol.1 + 0 = 1Ex. 18The slowest step of a particular reaction is found to be12X + Y 22\uf0ae XY2The order of the reaction is[A] 2[B] 3[C] 3.5[D] 1.5Ans. [D]Sol.r = k[X ] [Y ]21\/221\uf05c Order = 0.5 + 1 = 1.5","Ex.19The rate of certain hypothetical reaction A + B + C \uf0ae products, is given byr =dAdt = k [A] [B] [C]1\/21\/31\/4The order of a reaction is given by[A] 1[B] 1\/2[C] 2[D] 13\/12Ans. [D]Sol.Order of reaction = 1116 + 4 + 313++==23412121 3 .ZERO ORDER REACTION :Reaction whose rate is not affected by concentration said to be of zero order reaction.Example :(i)Reaction between Acetone and Bromine(ii)Dissociation of HI on gold surface(A)Unit of Rate Constant :\u20131\u20131k = mol L secUnit of rate of reaction = Unit of rate constant.(B )Rate Constant of Zero Order Reaction :x = ktThe rate of reaction is independent of the concentration of the reaction substance.(C)Determination of Half life Period of Zero Order Reaction :Att = t\u00bd;x = a21\/2at=2kor1 \/ 2 ta \uf0b5The half life period is directly proportional to the initial concentration of the reactants.Ex.20The rate equation of a reaction is k[A] [B] [C] . What should be the order of the reaction ?1\/21\/2\u20131Sol.n = 12 + 12 \u2013 1 = 0\uf05cOrder of the reaction is zero.\uf075Graphical representationxdxdtttSlope = k000Example :\uf075Photochemical reactions, like H + Cl22\uf0be\uf0be\uf0ae2HCl, are zero order reaction.\uf075Decomposition of NH on platinum surface is also zero order reaction.3","1 4 .FIRST ORDER REACTION :When the rate of reaction depends only on one concentration term of reactant.A first order reaction is one whose rate varies as first power of the concentration of the reactant, i.e. therate increases as number of times as the concentration of reactant is increased.Let us, consider a unimolecular first order reaction represented by the general equation,A \uf0be\uf0be\uf0be\uf0be\uf0ae ProductAt t =0a0At t = ta \u2013 xxThe initial concentration of A is a mole L and its concentration after time t is (a \u2013 x) mole L . This means\u20131\u20131during the time interval t, x mole L of A has reacted.\u20131The rate of reaction at any time t is given by the following first order kinetics.\uf02dd(a x)dt\uf0b5 (a \u2013 x)ord(x)dt\uf0b5 (a \u2013 x)ordxdt= k (a \u2013 x)where k is the rate constant of the reaction.\uf02d dxa x= kdtThis is differential rate equation and can be solved by integration.\uf03d \uf02d \uf0f2\uf0f2dxk dta xor\u2013 ln (a \u2013 x) = k.t + C...............(1)where C is integration constant.The constant C can be evaluated by applying the initial condition of the reaction i.e. when t = 0,x = 0. Putting these in equation (1), we getC = \u2013 ln aPutting the value of C in equation (1), we get\u2013 ln (a \u2013 x) = k.t \u2013 lnaork =1tln\uf02d aa x=2.303tlog\uf02d aa x................(2)Also, k =212.303(tt )\uf02dlog\uf02d\uf02d12(a x )(a x )and k =\uf02d212.303(tt )log12RRwhere (a \u2013 x ) is concentration at time t and (a \u2013 x ) is concentration after time t and R is rate at time11221t and R is rate at time t .122If [A] and [A] be the concentrations of reactant at zero time and time t respectively, then Eq. (2) may0be put ask =1tln0 [A][A]Also, [A] = [A] e0\u2013kt","This is the integrated rate expression for first order reaction.As, k = 1tln\uf02d aa x\uf0delog(a \u2013 x) = log (a) \u2013kt2.303tSlope = OA = log AOl\u2013k2.303Akt = lna \u2013 ln (a \u2013 x)Also, (a \u2013 x) = ae\u2013ktx = a(1 \u2013 e )\u2013kt\uf05c Degree of dissociationxa\uf0e6 \uf0f6\uf0e7 \uf0f7\uf0e8 \uf0f8= (1 \u2013 e )\u2013kt\uf075Unit of Rate constantThe differential rate expression for n order reaction is as follows :th\u2013dxdt = k(a \u2013 x)nork =nndx(concentration)(a x) dt(concentration) time \uf03d\uf02d= (conc.)1\u2013n time\u20131If concentration be expressed in mole L and time in minutes, then\u20131k = (mole L )\u20131 1\u2013n min\u20131For zero order reaction, n = 0 and hence, k = mole L min\u20131\u20131For first order reaction, n = 1 and hence,k = (mole L ) min = min\u20131 0\u20131\u20131For second order reaction, n = 2 and hence,k = (mole L ) min = mole L min\u20131 \u20131\u20131\u20131\u20131The rate constant of a first order reaction has only time in its unit. It has no concentration term in theunit. This means the numerical value of k for a first order reaction is independent of the unit in whichconcentration is expressed. If concentration unit is changed, the numerical value of k for a first order reactionwill not change. However, it would change with change in time. Say, k is 6.0 \u00d7 10 min then it may also\u20133be written as 1 \u00d7 10 s , i.e., numerical value of k will decrease 60 times if time unit is changed from\u20134 \u20131hour to minute or from minute to second.\uf075Half - time or half - life period of a first order reaction :The half - time of a reaction is defined as the time required to reduce the concentration of the reactantto half of its initial value. It is denoted by the symbol t1\/2. Thus,When x =a2, t = t1\/2Putting these values in Eq. (2), we getk =1 \/ 22.303t logaaa2 \uf02d=1 \/ 22.303tlog 2 =1 \/ 22.303t \u00d7 0.30103( log 2 = 0.30103)\uf051t1\/2 =0.693k..............(3)","Since k is a constant for a given reaction at a given 020406080255075100t (min) \uf0aeC(%) \uf0ae0temperature and the expression lacks any concentra-tion term so from Eq. (3) it is evident that half-timeof a first order reaction is a constant independent ofinitial concentration of reactant. This means if we startwith 4 moles L of a reactant reacting by first order\u20131 kinetics, then after 20 minutes it is reduced to 2 molesL . That is, after 20 minutes from the start of reaction\u20131the concentration of the reactant will be 2 moles L\u20131after 40 minutes from the start of reaction, the concentration is 1 mole L . After 60 minutes from the\u20131start of reaction, the concentration of the reactant will be reduced to 0.5 mol L . In other words, if during\u2013120 minutes 50% of the reaction completes, then in 40 minutes 75%, in 60 minutes 85.5% of the reactionand so on, will complete as shown in the figure above.Thus, fraction left after n half-lives =n12\uf0e6 \uf0f6\uf0e7 \uf0f7\uf0e8 \uf0f8Concentration left after n half-lives, [A] =n12\uf0e6 \uf0f6\uf0e7 \uf0f7\uf0e8 \uf0f8[A]0It is also to be noted that Eq. (3) helps to calculate t1\/2 or k.A general expression for t1\/2 is as follows.\uf075Half - life of a nth order reactionLet us find out t1\/2 for n order reaction where n 1.th \uf0b9\uf05cd[A]dt\uf02d=k [A]nn \uf0dend[A][A]\uf02d= kdt \uf0de\uf020\u2013[ A ] 01 \/ 220tn[A ]0d[A]kdt[A]\uf03d\uf0f2\uf0f20[ A ] 02[A ]n[A] d[A] = k t\uf02d\uf0f2n 1\/200 \/ 2[A ]1 n[A ][A]1 n\uf02d\uf0e9\uf0f9\uf0ea\uf0fa\uf0eb \uf02d\uf0fb= k tn 1\/2\uf0de11 n\uf02d\uf05b \uf05d1 n1 n00[A]A2\uf02d\uf02d\uf0e6\uf0f6\uf0e9\uf0f9\uf02d\uf0e7\uf0f7\uf0ea\uf0fa\uf0eb\uf0fb\uf0e8\uf0f8= k tn 1\/2 \uf020\uf0de\uf0201 n0 [A]1n\uf02d\uf02d1 n112\uf02d\uf0e9\uf0f9\uf0e6 \uf0f6\uf02d \uf0e7 \uf0f7 \uf0ea\uf0fa\uf0e8 \uf0f8\uf0eb\uf0fb= k tn 1\/2\uf0de\uf05b \uf05dn 101(1 n) A\uf02d\uf02d[1 \u2013 2n\u20131] = k tn 1\/2 \uf0de \uf028\uf029n 1n021k (n 1)[A]\uf02d\uf02d\uf02d= t1\/2 (order n 1)\uf0b9Therefore, for n order reaction, the half-life is inversely related to the initial concentration raised to the powerth of (n \u2013 1).t 1\/2\uf0b5n 11a\uf02dwhere n = order of reaction.Example :(i)All radioactive reactions(ii)A \uf0ae Product(iii)2NO \uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 2N + O22(iv)2Cl O 27\uf0ae 2Cl + 7O22(A)Unit of rate constant of first order reactionK = (sec)\u20131\uf044n = 1(B )Velocity constant for first order reaction1102.303ak =logt(a \u2013 x)\uf0de1012.303at =logk(a \u2013 x)where t = time, a = initial concentration at t = 0(a \u2013 x) = concentration after time tK = Rate constant","(C)Graphical RepresentationGraph between t v\/s log a(a \u2013 x) is a straight line2.303tlog a\/(a\u2013x)Slope =kEx.21A first order reaction gets 90% completed in 40 minute. Find out the half-life period of the reaction.Sol.Suppose that the initial concentration of reactant (a) = 100t = 40 minutes90% of the reaction get completed in 40 minutes.Therefore,x = 90k =12.303tlogaa \u2013 x = 2.30340 log 100100 \u2013 90\uf020\uf0de\uf020 k = 12.30340 log 10 = 2.30340 \u00d7 1 = 5.757 \u00d7 10 minutes\u20132\u20131t = \u00bd\u201320.6935.757 \u00d7 10 = 10.3 minutesEx.22Prove with the help of the following data that hydrolysis of H O is a first order reaction. Initial22concentration in the reaction 25.0.Sol. Time, t1 02 03 0(in minutes) V2 0 . 01 5 . 71 2 . 5For a first order reaction, k = 12.303tlogaa \u2013 x ; Here, a = 25k at t = 10 minutes = 12.30310 log 2520 = 2.30310 \u00d7 0.0969 = 2.23 \u00d7 10\u20132k at t = 20 minutes = 12.30320 log 2515.7 = 2.30320 \u00d7 0.2020 = 2.32 \u00d7 10\u20132k at t = 30 minutes = 12.30330 log 2512.5 = 2.30330 \u00d7 0.3010 = 2.31 \u00d7 10\u20132Constant value of k shows that hydrolysis of H O in aqueous medium is a first order reaction.22","Ex.23The rate constant for an isomerization reaction, A \uf0be\uf0be\uf0ae B, is 2.5 \u00d7 10 min . If the initial concentration\u20133\u20131of A is 1M, calculate the rate of reaction after 1 hr.Sol.As, k =2.303tlog\uf02d a(a x), for first order reaction\uf0de2.5 \u00d7 10 =\u201332.30360log\uf02d 1(a x)\uf05c(a \u2013 x) = 0.8607The rate after 60 minutes = k (a \u2013 x) = 2.5 \u00d7 10 \u00d7 0.8607 = 2.1518 \u00d7 10\u20133\u20133Ex. 24The rate of a first order reaction is 0.08 mol L at 20 min. and 0.06 mol L at 40 min. after start of\u20131\u20131reaction. Find the half-life of reaction.Sol.As, rate = k[A]0.08 = k[A]100.06 = k[A]20\uf05c1020[A]0.084[A]0.063\uf03d\uf03dFor first order reaction :t = 2.303 log 1020[A][A]when t = (40 \u2013 20) = 20 min.\uf05c20 =2.303klog43\uf05ck = 2.30320log43 = 0.0144 min\u20131\uf05ct1\/2 = 0.6930.693k0.0144 \uf03d = 48.13 minEx. 25Radioactive decay of an atomic nucleus is a first order reaction. Half-life period of radium [ Ra88226] is1590 years. Find out its decay constant.Sol.t =\u00bd10.693k=1\/20.693t=0.6931590= 4.358 \u00d7 10 Y\u20134\u20131\uf071SECOND ORDER REACTIONA+A\uf0be\uf0be\uf0aeProductA+B\uf0be\uf0be\uf0aeProductAt t = 0aa0At t = t(a \u2013 x)(a \u2013 x)xAt t = t1(a \u2013 x )1(a \u2013 x )1x 1At t = t2(a \u2013 x )2(a \u2013 x )2x 2As per rate law, dxdt= k [A] = k [A] = k [A][B]2n222\uf05c\uf0e6\uf0f6\uf0e7\uf0f7\uf0e8\uf0f8 dxdt= k (a \u2013 x)22(k = rate constant for second order reaction)2Also, k =2\uf0e9\uf0f9\uf02d\uf03d\uf0ea\uf0fa\uf02d\uf02d\uf0eb\uf0fb1111at (a x) at a(a x)or k =22121111(tt ) (a x ) (a x )\uf0e9\uf0f9\uf02d\uf0ea\uf0fa\uf02d\uf02d\uf02d\uf0eb\uf0fbWhere (a \u2013 x ) and (a - x ) are the concentration of the reactant A at time t and t respectively. If reactant1212A and B have different concentrations a and b, then k =2\uf02d2.303t(a b)log10\uf02d\uf02db(a x)a(b x)","when a >> b then (a \u2013 b) \uf0bb\uf020a(a \u2013 x) a\uf0bbEquation reduces tok =22.303t a\uf0b4log 10\uf02d ba x\uf0dek ' = k \u00d7 a = 222.303tlog10\uf0e6\uf0f6\uf0e7\uf0f7\uf0e8\uf0f8 \uf02d bb x(equation for first order kinetics)This is an example of pseudo first order reaction.Equation for second order reaction can be rewritten as\uf02d a(a x) = k t + 21a\uf075Graphical RepresentationAt(i) Slope = k2(ii) Intercept, OA =01a1a \u2013 xIn general for n order reaction,thk =nn 1n 1111(n 1)t (a x)a\uf02d\uf02d\uf0e9\uf0f9\uf02d\uf0ea\uf0fa\uf02d\uf02d \uf0eb\uf0fbIn general for n order reaction,tht1\/2(n) =\uf02d\uf02d\uf02d\uf02dn 1n 1n21(n 1)k (a)(n 2)\uf0b3\uf0det1\/2(n)\uf0b5\uf02d (n 1)1a\uf0det1\/2(n) a\uf0b5(1\u2013n)Ex.26Initial concentrations of both the reactants of a second order reaction are equal and 60% of the reaction getscompleted in 3000 seconds. How much time will be taken in 20% completion of the reaction ?Sol.k = 21txa(ax) \uf02dSuppose, a = 1k = 213000 \u00d7 0.61(1 0.6)\uf02d = 13000\u00d7 0.60.4Now, for 20% completionk = 21txa(1x) \uf02d13000\u00d7 0.60.4 = \u00d7 1t(0.2)1(1 0.2)\uf02d13000\u00d7 0.60.4 = 1t \u00d7 14t = 30000.6 \u00d7 0.44t = 500 second","Ex.27A second order reaction requires 70 minutes to change the concentration of reactants from 0.08 M to0.01 M. How much time will it require to become 0.04 M.Sol.For second order reactionwhen, (a \u2013 x) = 0.01k =2xt.a(a \u2013 x)k = 20.0770 \u00d7 0.08 \u00d7 (0.01)......(1)(a \u2013 x) = 0.04k = 20.04t \u00d7 0.08 \u00d7 (0.04) .......(2)From the equation (1) and (2)0.0770 \u00d7 0.08 \u00d7 (0.01)=0.04t \u00d7 0.08 \u00d7 (0.04)t = 10 minutes1 6 .THIRD ORDER REACTION :A reaction is said to be of third order if its rate is determined by the variation of three concentration terms.When the concentration of all the three reactants is same or three molecules of the same reactant areinvolved, the rate expression is given as( i )2NO + O 2\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 2NO2(ii)A + B + C \uf0ae ProductRate constant of third order reaction3221x(2ax)k.t 2a (ax)\uf02d\uf03d\uf02dHalf life period1 \/ 223t2ka \uf03dThus, half life is inversely proportional to the square of initial concentration.\uf075nth order reaction :A \uf0ae Productk t = n1n 1\uf02dn 1n 111(ax)a\uf02d\uf02d\uf0ec\uf0fc\uf02d\uf0ed\uf0fd\uf02d \uf0ee\uf0fe[n 1, n = order]\uf0b9t1\/2 = n 1n 1n121.k (n 1)a\uf02d\uf02d\uf0e9\uf0f9 \uf02d\uf0ea\uf0fa\uf02d\uf0eb\uf0fb\uf075Side or concurent reaction :ABCk 1k2;ln 0t[A][A]= (k + k ) t ;1 212[B]k[C]k \uf03d\uf075Consecutive reaction :A 1 k\uf0be\uf0be\uf0ae B 2 k\uf0be\uf0be\uf0be\uf0ae C:t = max11221k n: [B] = [A]kkk \uf0e6\uf0f6\uf0e7\uf0f7\uf02d\uf0e8\uf0f8\uf06cmax0 212kkk21kk\uf02d\uf0e6\uf0f6\uf0e7\uf0f7\uf0e8\uf0f81 7 .THRESHOLD ENERGY AND ACTIVATION ENERGY :For a reaction to take place the reacting molecules must colloid together, but only those collisions, in whichcolliding molecules possess certain minimum energy is called threshold energy (E ).T","\uf075Activation energy (E ) :aThe extra energy needed for the reactant molecules to be able to react chemically is known as Activationenergy.E = Threshold energyTE =a Activation energy of forward reactionEaE'aETP2P1ReactantsEReaction co-ordinatesProductE' = activation energy of backward reactiona P =1 Potential energy of reactantsP = Potential energy of products21 8 .EFFECT OF CATALYST :A catalyst is a substance, which increases the rate of reaction without itself being consumed at the endof the reaction, and the phenomenon is called catalysis. There are some catalysts which decrease the rateof reaction and such catalysts are called negative catalyst. Obviously, the catalyst accelerating the rate willbe positive catalyst. However, the term positive is seldom used and catalyst itself implies positive catalyst.Catalysts are generally foreign substances but sometimes one of the product formed may act as a catalystand such a catalyst is called \\\"auto catalyst\\\" and the phenomenon is called auto catalysis.Thermal decomposition of KClO is found to be accelerated by the presence of MnO . Here, MnO (foreign322substance) act as a catalyst 2KClO + [MnO ] 32\uf0be\uf0be\uf0ae2KCl + 3O2\uf0ad\uf020+ [MnO ]2MnO can be received in the same composition and mass at the end of the reaction.2In the permanganate titration of oxalic acid in the presence of bench H SO (acid medium), it is found24that there is slow discharge of the colour of permanganate solution in the beginning but after sometimethe discharge of the colour becomes faster. This is due to the formation of MnSO during the reaction4which acts as a catalyst for the same reaction. Thus, MnSO is an \\\"auto catalyst\\\" for this reaction. This4is an example of auto catalyst.2KMnO + 4H SO + 5H C O 424222\uf0be\uf0be\uf0ae K SO + 8H O + 10CO2422\uf071General Characteristic of Catalyst\uf075A catalyst does not initiate the reaction, it simply fastens it.\uf075Only a small amount of catalyst can catalyse the reaction.\uf075A catalyst does not alter the position of equilibrium i.e. magnitude of equilibrium constant and hence\uf044G\u00b0. It simply lowers the time needed to attain equilibrium. This mean if a reversible reaction inabsence of catalyst completes to go to the extent of 75% till attainment of equilibrium, and this stateof equilibrium is attained in 20 minutes then in presence of a catalyst also the reaction will go to75% of completion before the attainment of equilibrium but the time needed for this will be lessthan 20 minutes.\uf075A catalyst drives the reaction through a different route for which energy barrier is of shortest heightand Hence, E is of lower magnitude. That is, the function of the catalyst is to lower down the ac-ativation energy.E = Energy of activation in absence of catalyst.aP.E.E aE'aProductsReactantsH PH RReaction CoordinateE' = Energy of activation in presence of catalyst.aE \u2013 E' = Lowering of activation energy by catalyst.aa","If k and k be the rate constant of a reaction at a given temperature T, and E and E' are the activationcataaenergies of the reaction in absence and presence of catalyst, respectively, the\uf02d\uf02d\uf03daaE ' \/ RTcatE \/ RTkAekAe\uf02d\uf03daa(EE ' ) \/ RTcat kAekSince E , E' is +ve, so k > k. The ratio aacatcat kkgives the number of times the rate of reaction will increaseby the use of catalyst at a given temperature and this depends upon E \u2013 E' . Greater the value ofaaE \u2013 E' , more number of times k is greater than k.aacatThe rate of reaction in the presence of catalyst at any temperature T may be made equal to the rate of1reaction in absence of catalyst but we will have to raise the temperature. Let, this temperature be T , then2\uf02d\uf02d \uf03da1a2E ' \/ RTE \/ RTeeor\uf03d aa12E 'ETTEx. 28A hydrogenation reaction is carried out at 500 K. If the same reaction is carried out in presence of catalystat the same rate, the temperature required is 400K. Calculate the activation energy of the reaction if thecatalyst lowers the activation energy barrier by 40 kJ\/mol.Sol.Let, E and E' be the energy of activation in absence and presence of catalyst for hydrogenation reaction,aaas k AeaE \/ RT\uf02d\uf03d\uf02d\uf0b4\uf03daE \/ R 5001 kAe(In absence of catalyst)\uf02d\uf0b4\uf03daE ' \/ R 4002 kAe(In presence of catalyst)Given, r = r ; Hence k = k1212aaE \/ R 500E ' \/ R 400ee\uf02d\uf0b4\uf02d\uf0b4\uf03d\uf0deaaEE 'R 500R 400 \uf03d\uf0b4\uf0b4oraaEE40500400\uf02d\uf03d(As E \u2013 E' = 40)aa\uf05cE = 200 kJ\/mola\u201311 9 .DETERMINATION OF ORDER OF REACTION :\uf075Integration MethodIn this method, value of k is determined by putting values of initial concentration of reactants and change inconcentration with time in kinetic equation of first, second and third order reactions. The equation by whichconstant value of k is obtained is called order of that reaction.k = 12.303tlogaax \uf02d(For first order reaction)k = 2 1txa(a \u2013 x)\uf0e9\uf0f9\uf0ea\uf0fa\uf0eb\uf0fb(For second order reaction)k = 312t22x(2a \u2013 x)a (a \u2013 x)\uf0e9\uf0f9\uf0ea\uf0fa\uf0eb\uf0fb(For third order reaction)Ex. 29For a reaction, A \uf0ae B, it has been found that the order of the reaction is zero with respect to A. Which ofthe following expression correctly describes the reaction ?[A] k = 0 [A]2.303logt[A][B] [A] \u2013 [A] = kt0[C] t =\u00bd0.693k[D] t \u00bd\uf020\uf0b501[A]Ans. [B]Sol.d[A]\u2013dt = k[A] , \u2013 d[A] = kdt0Integrating from t = 0 to t = t[A] \u2013 [A] = kt0","\uf071Graphical MethodIf a straight line is obtained on drawing a graph between log (a \u2013 x) and time then it is first order reaction.If a straight line is obtained on drawing a graph between (a \u2013 x) and 2dxdt, then it is second order reaction.Ex. 30Which of the following graphs is for a second order reaction ?[A] Rate[A]2[B] Rate[A][C] Rate[A]2[D] Rate[A]2Ans. [C]Sol.For second order reaction rate vs [A] is a straight line with slope equal to k rate = k[A]22If a straight line is obtained on drawing a graph between (a \u2013 x) and 3dxdt, then it is third order reaction.\uf071Half-life MethodRelation between half-life period of a reaction and initial concentration is as follows : t 1\/2\uf0b5n 11a\uf02dFor first order reaction(Half life a )\uf0b50For second order reaction(Half life 1\/a)\uf0b5For third order reaction(Half life 1\/a )\uf0b52Ex. 31For a first order reaction, t0.75 is 1386 seconds. Therefore, the specific rate constant is[A] 10 s\u20131\u20131[B] 10 s\u20133\u20131[C] 10 s\u20132\u20131[D] 10 s\u20134\u20131Ans. [B]Sol.t0.75 = 1386 s = 2 \u00d7 t0.5; t0.5 = 13862= 693 s; k = 0.693693 s = 1 \u00d7 10 s\u20133\u20131Ex. 32t of first order reactions is given by \u00bd0.693k, t would be equal to3\/4[A] 0.693k[B] 0.346k[C] 1.386k[D] 0.924kAns. [C]Sol.t = 2(t3\/4\u00bd ) = 2 \u00d7 0.693k = 1.386kEx. 33The t of a first order reaction is found to be 2 minutes. The percentage of the reactant left after\u00bd360 seconds is :[A] 12.5[B] 25[C] 15[D] 7.5Ans. [A]Sol.360 seconds = 6 min = 3 half-lives100 1\/2 t\uf0be\uf0be\uf0be\uf0ae 50 1\/2 t\uf0be\uf0be\uf0be\uf0ae 25 1\/2 t\uf0be\uf0be\uf0be\uf0ae 12.5\uf071Ostwald Isolation MethodThis method is used to find out the order of complex reactions. If nA, nB and nC molecules of substance A,B and C, respectively, are present in a reaction, then nA + nB + nC will be the order of reaction.When B and C are in excess, the order of reaction will be nA.When A and B are in excess, the order of reaction will be nC.When A and C are in excess, the order of reaction will be nB.Ex. 34When the initial concentration of a reaction was doubled, its half life become half, What should be the orderof the reaction ?Sol.Rate law for product of a reaction is as follows :Rate = k[A]nand 0.50.52tt=n 12aa\uf02d\uf0e9\uf0f9\uf0ea\uf0fa\uf0eb\uf0fb2 = [2]n\u20131 ;n \u2013 1 = 2n = 2","2 0 .TEMPERATURE EFFECT :The rate of reaction is dependent on temperature. This is expressed in terms of temperature coefficientwhich is a ratio of two rate constants differing by a temperature of 10\u00b0. Generally the temperature selectedare 298K and 308K. It is mathematically expressed as,rate constant at 308KTemperature coefficient = rate constant at 298K = ttk10k\uf02bThe value of temperature coefficient for most of the reactions lies between 2 to 3.2 1 .ARRHENIUS EQUATION :Arrhenius derived a mathematical expression to give a quantitative relationship between rate constant andtemperature. The expression isk = A.e\u2014Ea \/RT(Here, A = frequency factor; Ea = activation energy ; R = gas constant and T = temperature).If k and k are rate constants at temperature T and T then! 212a221112EkTTlogk2.303 RT T\uf0e9\uf0f9\uf02d\uf03d\uf0ea\uf0fa\uf0eb\uf0fbEx. 35Ethylene oxide is decomposed into CH and CO. Rate constant for this reaction may be described by4the equation log k (s ) = 14.34 \u2013\u2013141.25 10T \uf0b4(i)What will be the energy of activation of this reaction ?(ii)What will be the value of k at 670 K ?(iii)At what temperature will its half-life period be 25.6 minutes ?Sol.(i) We know, log k = log A \u2013 1010a E2.303RT.......(i)Given, log k (s ) = 14.34 \u2013 \u2013141.25 10T \uf0b4.......(ii)Comparing Eqs. (i) and (ii), we geta E2.303R = 1.25 \u00d7 104(ii)E = 1.25 \u00d7 10 \u00d7 2.303 \u00d7 8.314 \u00d7 10a4\u20133\uf05cE = 239.339 kJ\/molaSubstituting the value of T (670 K) in Eq. (ii),log k (s ) = 14.34 \u2013\u2013141.25 10670\uf0b4 = 4.3167\uf05ck = 4.82 \u00d7 10 s\u20135\u20131(iii)k =1 \/ 20.693t=0.693256 60\uf0b4= 0.000451 sec\u20131\uf0de\uf020log 0.000451 sec\u20131= 14.34 \u201341.25 10T \uf0b4\uf0de T = 706.79 K","Ex.36The rate constant of forward reaction a reaction increases by 6% when the temperature of the reaction isincreased from 300 to 301 K, whereas equilibrium constant increases by 2%. Calculate the activation energyfor the forward as well as backward reaction.Sol.According to Arrhenius equation, loga(f)221112EkTTk2.303RT T\uf02d \uf0e9\uf0f9\uf03d\uf0ea\uf0fa\uf0eb\uf0fbIf k = k at 300 K then at 301 K, k = k + k \u00d7 126100= 1.06 k\uf05clog10( f )a E1.06 k301 300k2.303 8.314 300 301\uf02d\uf0e9\uf0f9\uf03d\uf0ea\uf0fa\uf0b4\uf0b4\uf0eb\uf0fb\uf05c(f ) a E= log (1.06) \u00d7 2.303 \u00d7 8.314 \u00d7 300 \u00d7 301 = 43753 J\/mol = 43.753 kJ\/molAccording to van't Hoff equation, log221112kHTTk2.303RT T\uf044 \uf0b0\uf02d \uf0e9\uf0f9\uf03d\uf0ea\uf0fa\uf0eb\uf0fbIf k = k at 300 k, k = k + 122100 = 1.02 klog101.02 kH301 300k2.303 8.314 300 301\uf044 \uf0b0\uf02d\uf0e6\uf0f6\uf03d\uf0e7\uf0f7\uf0b4\uf0b4\uf0e8\uf0f8\uf05c\uf044H\u00b0 = (log 1.02) \u00d7 2.303 \u00d7 8.314 \u00d7 300 \u00d7 301 J\/mol= 14869 J mol = 14.87 kJ\/mol\u20131Thus, reaction is endothermic.For such a reaction, H\u00b0 = E \u2013 E\uf044a(f)a(b)\uf05cE = E \u2013 H\u00b0 = 43.753 \u2013 14.87 kJ\/mol = 28.883 kJ mola(b)a(f)\uf044\u20131Ex. 37Value of rate constant for a first order reaction at 500 K is 1.60 \u00d7 10 second , whereas at 600 K, it is\u20135\u201316.36 \u00d7 10 second . Find out the activation energy of the reaction.\u20133\u20131Sol.log12kk \uf0e6\uf0f6\uf0e7\uf0f7\uf0e8\uf0f8= a \u2013E2.303 R1211TT\uf0e9\uf0f9\uf02d\uf0ea\uf0fa\uf0eb\uf0fblog 1.60 \u00d7 10 \u2013 log 6.36 \u00d7 10\u20135\u20133= a E2.303 8.314\uf02d\uf0b411500600\uf0e9\uf0f9\uf0ea\uf0fa\uf0eb\uf0fb\u20135.241 \u2013 3.8035 = a E19.15 \u00d7 13000E = 2.5999 \u00d7 19.15 \u00d7 3000aE = 1.49 \u00d7 10a5Ex. 38An exothermic reaction, X \uf0ae Y, has an activation energy 30 kJ mol . If energy change ( E) during the\u20131\uf044reaction is \u2013 20kJ, then the activation energy for the reverse reaction is[A] \u201330kJ[B] 20 kJ[C] 50 kJ[D] 10 kJAns. [C]Sol.\uf044E = (f) a E \u2013 (b) a E;\u2013 20 = 30 \u2013 (b) a E;(b) a E = 50 kJEx. 39An endothermic reaction, A \uf0ae B, has an activation energy as x kJ mol of A. If energy change of the\u20131reaction is y kJ, the activation energy of reverse reaction is[A] \u2013 x[B] x \u2013 y[C] x + y[D] y \u2013 xAns. [B]Sol.\uf044E = (f) a E \u2013 (b) a E;y = x \u2013 (b) a E;(b) a E = x \u2013 yEx. 40Which of the following relations is correct ?[A] k = A eEa\/RT[B] ln k \u2013 ln A = a ERT[C] ln A \u2013 ln k = a ERT[D] ln A \u2013 ln k = \u2013a ERTAns. [C]Sol.k = Ae\u2013Ea\/RTln k = ln A \u2013a ERT orln A \u2013 ln k =a ERT","Ex.41Which of the following expression give the effect of temperature on the rate constant ?[A] ln A = RT ln E \u2013 ln ka[B] ln k = ln A \u2013 E \/RTa[C] k = AE \/RTa[D] None of theseAns. [B]Sol.The effect of temperature on rate constant is quantitatively given by Arrhenius equationk = Ae\u2013Ea\/RTor ln k = ln A \u2013 E \/RTaEx.42The plot of log k vs 1T helps to calculate[A] Energy of activation[B] Rate constant of the reaction[C] Order of the reaction[D] Energy of activations as well as the frequency factorAns. [D]Sol.According to Arrhenius equation : log k = log A \u2013 a E2.303 T.1Plot of log k vs. 1T is straight lineSlope = \u2013 a E2.303 R1\/Tlog kIntercept = log AEx.43The progress of the reaction given below, consider the reaction given below,CH COOC H + H O 3252H \uf02b\uf0be\uf0be\uf0be\uf0ae CH COOH + C H OH325the reaction can be followed by measuring the concentration of acid (HCl acid used as catalyst plus aceticacid formed during the reaction) by means of alkali titration. Calculate the volume of alkali (NaOH) neededfor the end point that will increase with time.Sol.CH COOC H + H O 3252H \uf02b\uf0be\uf0be\uf0be\uf0aeCH COOH + C H OH325At t = 0aexcess00At t = ta \u2013 xxxAt t = \uf0a50aaIf V , V and V are the volumes of NaOH solution needed for the end point of titration of the reaction0t\uf0a5mixture at zero time, time t and at infinity, i.e. after completion of the reaction the condition being achievedby heating the reaction mixture for some time, thenV [acid catalyst]0\uf0b5V t\uf0b5\uf020[acid catalyst] + xV \uf0a5\uf0b5\uf020[acid catalyst] + a\uf05cV \u2013 V a \u2013 x\uf0a5\uf020t\uf0b5V \u2013 V a\uf0a5\uf0200\uf0b5(since concentration of HCl acid acting as catalyst will remain constant).The above reaction which is of first order ( actually pseudo unimolecular) will, therefore, obey followingequation.k = 2.303tlog0tVVVV \uf0a5\uf0a5\uf02d\uf02d","Ex.44H O (aq.) 22\uf0be\uf0be\uf0ae H O + 212O 2\uf0adAt t = 0a00At t = ta \u2013 xxxSince H O acts as a reducing agent towards KMnO , so concentrations of H O at various time intervals22422may be determined by the titration of the reaction mixture against standard KMnO solution. The titre4value will go on decreasing with time. If V and V be the titre values at zero time and any time t then0tV 0\uf0b5\uf020a and V t\uf0b5\uf020a \u2013 xThe above reaction being first order, its rate constant may be expressed ask= 2.303tlog0tVVEx. 45The reaction mentioned below is first order w.r.t. sucrose and zero order w.r.t. water, since water is inlarge excess as compared to sucrose. That is, it is an example of pseudo unimolecular reaction.Sucrose, glucose and fructose all are optically active substances. Therefore, the progress of the reactioncan be followed by measuring angle of rotations of the reaction mixture at various time intervals.During the reaction, angle of rotation goes on decreasing and after sometime there is reversal of the directionof rotation, i.e. from dextro to laevo and Hence, the reaction is called \\\"inversion of cane sugar\\\" orinversion of sucrose.C H O + H O 1222112H \uf02b\uf0be\uf0be\uf0be\uf0ae C H O + C H O61266126d-Sucrosed-Glucose -Fructose\uf06cInitiallyaExcess00After time ta \u2013 xConstantxxAt infinity0ConstantaaAngle of optical rotation is measured by means of an instrument called polarimeter. Optical rotation ismathematically expressed as,R = .C. [ ]obs\uf06c\uf061 D twhere\uf06c = length of the polarimeter tubeC = concentration of test solution[ ] = specific rotation\uf061 D tFor a given sample and polarimeter, and [ ] are constant.\uf06c\uf061 D t R obs\uf0b5\uf020C, or R = kC,obsIf r , r and r be the observed angle of rotations of the sample at zero time, time t and infinity respectively,0t\uf0a5and k , k and k be proportionate in terms of sucrose, glucose and fructose, respectively.123Then,r = k a01r = k (a \u2013 x) + k x + k xt123r = k a + k a\uf0a523From these equations it can be shown that0tarra xrr\uf0a5\uf0a5\uf02d \uf03d \uf02d\uf02dSo, the expression for the rate constant of this reaction in terms of the optical rotational data may beput as k =2.303tlog0trrrr\uf0a5\uf0a5\uf02d\uf02d","Ex. 462N O 25\uf0be\uf0be\uf0ae 4NO + O22At t = 0P 0 0 0At t = tP \u2013 2x 4x x0At t =\uf0a50 2P \u00bd P00The progress of the reaction can be followed by measuring the pressure of the gaseous mixture in a closedvessel, i.e. at constant volume. The expression for the rate constant in terms of pressure data will be asgiven below.k =2.303tlog0tPP, where P = P \u2013 2xt0If total pressure after any time t and at is given then it is possible to find P and x and hence, k may\uf0a5, 0be calculated.Ex. 47Consider a first order reaction,A \uf0ae\uf020B + CAssume that A, B and C are gases. The given data isTime0TPartial pressure of AP 1P 2And we have to find the rate constant of the reaction.Sol.Since A is a gas assuming it to be ideal, we can state that P = [A] RT [From PV = nRT]A\uf05c At t = 0, P = [A] RT and at t =t, P = [A] RT. Thus, the ratio of the concentration of A at two different102ttime intervals is equal to the ratio of its partial pressure at those same time intervals.\uf05c01t2[A]P[A]P \uf03d\uf05ck = 1tln12PPEx. 48A \uf0ae\uf020B + CTime0tTotal pressure of A + B + CP 1P 2Find k.Sol.In this case, we are given total pressure of the system at these time intervals. The total pressure obviouslyincludes the pressure of A, B and C. At t = 0, the system would only have A. Therefore, the total pressureat t = 0 would be the initial pressure of A.\uf05c P is the initial pressure of A. At time t, let us assume moles of A will decompose to give B and C1because of which its pressure is reduced by an amount x while that of B and C is increased by x each.That is :A \uf0be\uf0be\uf0ae B + CInitialP 1 00At time tP \u2013 x1 xx\uf05c Total pressure at time t = P + x = P12\uf0dex = P \u2013 P21Now the pressure of A at time t would be P \u2013 x = P \u2013 (P \u2013 P ) = 2P \u2013 P112112\uf05ck = ln 01t12[A]Pln[A](2PP )\uf03d\uf02d","Ex.49For the given following first order reaction,A \uf0be\uf0be\uf0ae B + CTimeT\uf0a5Total pressure of A + B + CP 2P 3Calculate k.Sol.Here means that the reaction is complete. Now, we have\uf0a5A \uf0be\uf0be\uf0aeB+CAt t = 0P 100At t = t(P \u2013 x)1xxAt t = \uf0a50P 1P 1\uf05c2P = P13\uf0deP =13 P2At time t,P + x = P12\uf0de3 P2 + x = P 2\uf0de x = P \u2013 23 P2\uf0deP \u2013 x = 13 P2 \u2013 (P \u201323 P2) = P \u2013 P32k = 1tln 03t32[A]1P \/ 2ln[A]t(PP )\uf03d\uf02d = 3321Plnt2(PP )\uf02d2 2 .PARALLEL REACTIONS :These are reactions in which reaction substances do not follow a particular path to give a particular set ofproducts. It follows one or more paths to give different products, e.g.k1k2BCAThe reactant A follows two different paths to form B and C as shown below :CBtConc.ARate = \u2013d[A]dt= k [A] + k [A]12= k[A] [As, (k + k ) = k]12% yield of B =112kkk\uf02b\u00d7 100","Ex.50A follows parallel path, first order reaction giving B and C ask1k2B2C5AIf initial concentration of A is 0.25 M, calculate the concentration of C after 5 hours of reaction.[Given : k = 1.5 \u00d7 10 s , k = 5 \u00d7 10 s ]1\u20135\u201312\u20136\u20131Sol.k =t2.303[A]logt[A](k = k + k = 1.5 \u00d7 10 + 5 \u00d7 10 = 2 \u00d7 10 s )12\u20135\u20136\u20135\u20131\uf0de2 \u00d7 10 =\u20135t2.3030.25logt[A]\uf05c[A] = 0.1744 Mt\uf05c[A]decomposed = [A] \u2013 [A] = 0.25 \u2013 0.1744 = 0.0756 M0tFraction of C formed =212k(kk )\uf02b\u00d7 [A]decomped \u00d7 25=655 102 10\uf02d\uf02d\uf0b4\uf0b4 \u00d7 0.0756 \u00d7 25= 7.56 \u00d7 10 M\u20133(5 moles of A are used to give 2 moles of C)2 3 .SEQUENTIAL REACTIONS :These are reactions which proceed from reactants to product through one or more intermediate stages,e.g.A1 k\uf0be\uf0be\uf0aeB 2 k\uf0be\uf0be\uf0aeC\uf075Graphical representationd[A]dt\uf02d= k [A]1.............(1)CBtConc.Ad[B]dt= k [A] \u2013 k [B]12.............(2)d[C]dt= k [B]2.............(3)Integrating Eq. (1), we get[A] =1 k t0[A] e\uf02dNow, we shall integrate Eq. (2) and find the concentration of B related to time t.d[B]dt= k [A] \u2013 k [B] 12\uf0ded[B]dt+ k [B] = k [A]21Substituting [A] as 1 k t0[A] e\uf02d\uf0ded[B]dt+ k [B] = k211 k t0[A] e\uf02d.............(4)Integration of the above equation is not possible as we are not able to separate the two variables, [B] andt. Therefore, we multiply Eq. (4) by integrating factor e2 k t, on both the sides of the equation.221k t(kk )t210d[B]k [B] ek [A] edt\uf02d\uf0e6\uf0f6\uf02b\uf03d\uf0e7\uf0f7\uf0e8\uf0f8We can see that the left hand side of the equation is a differential of [B] e2 k t.\uf05c221k t(kk )t10d([B]e ) k [A] edt\uf02d\uf03d","221k t(kk )t10d ([B]e ) k [A] edt\uf02d\uf03dIntegrating within the limits 0 to t.221tk t(kk )t100d ([B]e ) k [A]edt\uf02d\uf03d\uf0f2\uf0f2\uf0de212t(kk )tk t10210e[B]ek [A](kk )\uf02d\uf0e9\uf0f9\uf03d\uf0ea\uf0fa\uf02d\uf0eb\uf0fb\uf0de\uf05b\uf05d221k t(kk )t1021k [A][B]ee1(kk )\uf02d\uf03d\uf02d\uf02d\uf0de [B]\uf05b\uf05d221k t(kk )t1021k [A]ee1(kk )\uf02d\uf02d\uf03d\uf02d\uf02d\uf05b\uf05d12k tk t1021k [A][B]eekk\uf02d\uf02d\uf03d\uf02d\uf02d.............(5)Now, in order to find [C], substitute Eq. (5) in Eq. (3), we get12k tk t12021d[C]k k [A][ee]dtkk\uf02d\uf02d\uf03d\uf02d\uf02d\uf05cd[C] =\uf05b\uf05d12k tk t12021k k [A]eedtkk\uf02d\uf02d \uf02d\uf02dOn integrating, we get12tk tk t120210k k [A]d[C][ee]dt(kk )\uf02d\uf02d\uf03d\uf02d\uf02d\uf0f2\uf0f2\uf0de12ttk tk t120211200k k [A]ee[C](kk )kk\uf02d\uf02d\uf0e9\uf0f9\uf0e6\uf0f6\uf0e6\uf0f6\uf0ea\uf0fa\uf03d\uf02d\uf0e7\uf0f7\uf0e7\uf0f7\uf02d\uf02d\uf02d\uf0e8 \uf0ea\uf0f8\uf0e8\uf0f8 \uf0fa\uf0eb\uf0fb\uf0de12k tk t1202112k ke1e1[C][A](kk )kk\uf02d\uf02d\uf0e9\uf0e6\uf0f9 \uf0f6\uf0f6 \uf0e6 \uf02d\uf02d\uf03d\uf02d\uf0ea\uf0fa\uf0e7\uf0f7 \uf0e7\uf0f7\uf02d\uf02d\uf02d\uf0e8\uf0f8 \uf0e8\uf0f8\uf0eb\uf0fb\uf0de12k tk t1202112k k1 e1 e[C][A]kkkk\uf02d\uf02d\uf0e9\uf0e6 \uf02d\uf0f9 \uf0f6\uf0f6 \uf0e6 \uf02d\uf03d\uf02d\uf0ea\uf0fa\uf0e7\uf0f7 \uf0e7\uf0f7\uf02d\uf0e8\uf0f8 \uf0e8\uf0f8\uf0eb\uf0fb12k tk t0221[A][C][k (1 e)(1 e)]kk\uf02d\uf02d\uf03d\uf02d\uf02d\uf02dBmax and tmax : We can also attempt to find the time when [B] becomes maximum. For this, we differentiateEq. (5) and findd[B]dtand equate it to zero.\uf05cd[B]dt=1021k [A](kk )\uf02d12k tk t12e( k )e(k )0\uf02d\uf02d\uf0e9\uf0f9 \uf03d\uf02d\uf02b\uf0eb\uf0fb\uf0de12k tk t12k ek e\uf02d\uf02d\uf03d\uf028\uf02912kkt12kek\uf02d\uf03d, taking log of both the sides\uf05ctmax =11221klnkkk\uf02d...........(6)Substituting Eq. (6) in Eq. (5)Bmax= [A]0212k \/ kk21kk\uf02d\uf0e9\uf0f9\uf0ea\uf0fa\uf0eb\uf0fb","2 4 .HYPOTHESIS OF STEADY STATE :There are many reactions which involve multi steps and the intermediates (one or more) do not appearin the overall equation, e.g. 2NO + O 2\uf0be\uf0be\uf0ae 2NO2Steps involved in the above reaction areNO + O 2\uf0be\uf0be\uf0ae NO3NO + NO 3\uf0be\uf0be\uf0ae 2NO22NO + O 2\uf0be\uf0be\uf0ae 2NO2In the above steps, NO is an intermediate species and do not appear in the overall balanced equation.3Usually these intermediates are very reactive and do not accumulate to any significant extent during thereaction.For hypothetical reaction,A \uf0be\uf0be\uf0ae BThe reaction can proceeds in the following steps :A 1 k\uf0be\uf0be\uf0ae II 2 k\uf0be\uf0be\uf0ae BThe concentration of the intermediate [I] is much less than the reactant [A] as well as the product [B].Accordingly the formation of intermediate will start at zero, rises to maximum, and then fall back to zero.If the concentration of intermediate remains small during the reaction, then the curves of reactants, in-termediate and product versus t will be given as below :From the plot, it is clear that the slope of the curve for intermediate tConc.ABmuch less than those for reactants A and products B. It is, thereforegood approximation to take d[I]dt= 0, for each reaction intermediate.This is steady state (stationary state) approximation.2 5 .RADIOACTIVITY :All radioactive decay follows first order kinetics and this is where the similarity ends. This will be explainedlater in the chapter.We have measured the rate of reaction in chemical kinetics based on the rate of change of concentrationof reactants or products. But this procedure will not work for calculating the rate of a radioactive reaction.This is because most of the time the radioactive substance is a solid. Therefore, its concentration wouldbe a constant with time (assuming it to be pure and that the product does not remain with the reactants).Therefore, the rate of radioactive reactions is measured by calculating the rate of change of number ofnuclei of the radioactive substance.For a radioactive decay A \uf0ae B, the rate of reaction is calculated asAAdNNdt\uf02d\uf03d \uf06cwhere = decay constant of reaction.\uf06cN = number of nuclei of the radioactive substance at the time when rate is calculated.AAs you can see, the above rate law is very much similar to the rate law of a first order chemical reaction,but all other similarities ceases here. For example unlike a chemical reaction the decay constant ( ) does\uf06cnot depend on temperature. Arrhenius equation is not valid for radioactive decay.AAdNNdt\uf02d\uf03d \uf06c","Integrating the differential rate law, we get\u2013t0NtAAN0dNdtN\uf03d \uf06c\uf0f2\uf0f2log 0tNN= t\uf06cwhereN0 = number of nuclei of A, at t = 0N = number of nuclei of A, at t = tt\uf06c = decay constantThe expression can be rearranged to giveN = N et0\u2013 t \uf06c..............(1)This suggest that the number of nuclei of radioactive substance A at any instant of time can be calculated,if we know the number of nuclei at t = 0, its decay constant and the time.\uf075Half-LifeJust like a first order reaction, the half-life of radioactive decay is given byt1\/2 = 0.693\uf06c[Note : Let us start with 10 nuclei. If the half-life is 5 minutes, then at the end of first 5 minutes, numberof nuclei would be 5. Now, what would be the number of nuclei after next 5 minutes ? Will it be 2.5 or2 or 3 ? We can clearly see that it cannot be 2.5 and if it is 2 or 3 then it cannot be called as half-life.This dilemma can be overcome by understanding that all formula relating to kinetics are only valid whenthe sample size is very large and in such a large sample size, a small difference of 0.5 will be insignificant.The fact that radioactive decay follows the exponential law implies that this phenomenon is statistical innature. Every nucleus in a sample of a radionuclide has a certain probability of decaying, but there is noway to know in advance which nuclei will actually decay in a particular time span. If the sample is largeenough, i.e. if many nuclei are present - the actual fraction of it that decays in a certain time span willbe very close to the probability for any individual nucleus to decay. To say that, a certain radioisotope hasa half-life of 5 hr. signifies that every nucleus of this isotope has a 50 percent chance of decaying in every5 hr. period. This does not mean probability of 100 percent decaying is 10 hr. A nucleus does not havea memory, and its decay probability per unit time is constant until it actually does decay. A half-life of 5hr. implies a 75 probability of decay in 10 hr., which increases to 87.5% in 15 hr., to 93.75% in 20 hr.,and so on, because in every 5 hr. The probability of decay is 50 percent.\uf075Average\u2013Life TimeAverage life time is defined as the life time of a single isolated nucleus. Let us imagine, a single nucleuswhich decays in 1 second. Assuming 1 second time interval to be very small, the rate of change of nucleiwould be 1\/1 (because \u2013dN = 1 and dt = 1). We can also see that since dNN , for a single isolateddt\uf02d\uf03d \uf06cnucleus N = 1,dNdt\uf02d\uf03d \uf06c. Therefore, in this present case, = 1.\uf06cNow, let us assume, the same nucleus decays in 2 seconds, we can see that dNdt\uf02d, i.e. is equal to \u00bd .\uf06cYou will also notice that in the 1 case the nucleus survived for 1 second and in the second case it survivedstfor 2 seconds. Therefore, the life time of a single isolated nucleus is 1\uf06c .\uf05ct =av1\uf06c","\uf075Activity :Activity is the rate of decay of a radioactive element. It is represented as 'A' and is equal to N. By no\uf06cmeans should activity be confused with rate of change of radioactive nuclei represented by dNdt\uf02d. Thisis because dNdt\uf02d talks about the overall change in the number of nuclei in a given instant of time whileactivity only talks about that change which is decay. For example, if you go to a market with Rs. 50 inyour pocket and you spend Rs. 20 in 5 minutes then your rate of change of money in the wallet is Rs.4\/min and in fact the rate of spending the money is also Rs. 4\/min. Here, you can see both are same.But if while spending Rs. 20 in 5 minutes, somebody keeps Rs. 10 in your wallet, then the rate of changeof money in your wallet would become Rs. 2.5 \/min while the rate of spending the money is Rs. 4\/min.This implies that as long as the radioactive substance is only decaying the rate of change of nuclei andactivity are same and Eq.(1) in terms activity of radioactive substance can be written as At = A e . But0\u2013 t \uf06cif the radioactive substance is also being produced, then dNdt = rate of production \u2013 activity (of courseit's a different matter that rate of production may or may not be a constant).\uf075Specific ActivityIt is defined as per unit mass of the sample. Let, radioactive sample weighing w g have a decay constant\uf06c. The number of nuclei in the w g would be 0wN , where M = molecular weight of the radioactiveM \uf0b4substance and N = Avogadro's number.0\uf05cSpecific activity =0wN Mw\uf0e6\uf0f6\uf06c \uf0b4\uf0b4\uf0e7\uf0f7\uf0e8\uf0f8=0 NM \uf06c \uf0b4It should be remembered that if a radioactive sample is pure and the product does not remain with reactant,then specific activity is a constant.\uf075Units of ActivityThe unit of radioactivity of a substance is measured as the rate at which it changes into daughter nucleus.It has been derived on the scale of disintegration of radium.Let us consider, 1g of radium (atomic mass = 226 and t1\/2 = 1600 yrs) undergoes decay, thenRate of decay of radium = \u00d7 Number of nuclei of Ra in 1g\uf06c= 0.6931600365246060\uf0b4\uf0b4\uf0b4\uf0b4 \u00d7 231 6.023 10226\uf0b4\uf0b4 = 3.7 \u00d7 10 dps = 3.7 \u00d7 10 becquerel1010= 1 curie ( 1 Ci = 3.7 \u00d7 10 dps)\uf05110= 3.7 \u00d7 10 Rutherford ( 1Rd = 10 dps)4\uf0516The SI unit of activity is dps or Becquerel.Ex. 5184210Po decays with emission of -particle to \uf06120682Pb with a half-life period of 138.4 days. If 1g of 21084Po isplaced in a sealed tube, how much helium will be accumulated in 69.2 days ? Express the answer in cm3at STP.Sol.21084Po \uf0ae20682Po + He2 4Amount of 21084Po left after 69.2 days can be calculated by applyingN = N (1\/2)0nn = t\/t1\/2 = 69.21138.22 \uf03d\uf05cN = 1 \u00d7 1 \/ 212\uf0e6 \uf0f6\uf0e7 \uf0f7\uf0e8 \uf0f8= 0.702 g","Amount of polonium disintegrated = 1 \u2013 0.7072 = 0.2928 gMoles of polonium in 0.2928 g =0.2928210Moles of helium atoms formed =0.2928210\uf05cVolume of helium collected =0.2928210 \u00d7 22400 = 31.23 cm32 6 .CARBON DATING :The cosmic ray generates neutrons in the atmosphere which bombards the nucleus of atmospheric nitrogento form radioactive C hence C in the atmosphere has been remaining constant over thousands of years.1414In living materials, the ratio of C to C remains relatively constant. When the tissue in an animal or1412plant dies, assimilation of radioactive C ceased to continue. Therefore, in the dead tissue the ratio of1414C to C would decrease depending on the age of the tissue.127 14N + n 0 1\uf0be\uf0ae C + p6141 1614C + N + e714\u201310A sample of dead tissue is burnt to give carbon dioxide and the carbon dioxide is analysed for the ratioof C to C. From this data, age of dead tissue (plant or animal) can be determined.1412Age (t) =\uf06c0 N2.303logN\uf0de Age =141\/2 2.303 \u00d7 t( C)0.693log\uf0e6\uf0f6\uf0e7\uf0f7\uf0e8\uf0f8 0 NNN = ratio of C\/ C in living plant0 1412N = ratio of C\/ C in the wood1412Age =1\/2 2.303 \u00d7 t0.693log\uf0e6\uf0f6\uf0e7\uf0f7\uf0e8\uf0f8 0 AAA = Original activity0 A = Final activityAlso, N =n01N 2\uf0e6 \uf0f6\uf0e7 \uf0f7\uf0e8 \uf0f8\uf0de where n =1 \/ 2tt2 7 .ROCK DATING :It is based on the kinetics of radioactive decay. It is assumed that no lead was originally present in thesample and whole of it came from uranium.Initial no. of mole (N ) = [U] + [Pb]0Final no. of mole (N) = [U]\uf02b\uf03d 0 N[U] [Pb]N[U]= 1 +[Pb][U]t =\uf06c2.303log\uf0e9\uf0f9\uf0ea\uf0fa\uf0eb\uf0fb0 NN\uf0de\uf020t =\uf06c2.303log 1\uf0e9\uf0f9\uf02b \uf0ea\uf0fa\uf0eb\uf0fb[Pb][U]Also, 1\uf0e9\uf0f9\uf02b \uf0ea\uf0fa\uf0eb\uf0fb[Pb][U] = (2)n\uf0e6\uf0f6\uf03d \uf0e7\uf0f7\uf0e8\uf0f81 \/ 2tnt","Ex. 52A sample of uranium mineral was found to contain 206Pb and 238U in the ratio of 0.008 : 1. Estimate theage of the mineral (half-life of 238U is 4.51 \u00d7 10 years).9Sol.t =\uf06c2.303log\uf0e9\uf0f9\uf02b \uf0ea\uf0fa\uf0eb\uf0fb206238Pb1Ut =\uf0b41 \/ 22.303t0.693\u00d7 log 1\uf0e9\uf0f9\uf02b \uf0ea\uf0fa\uf0eb\uf0fb206238PbU\uf0deRatio by mass 206Pb : 238U = 0.008 : 1Ratio by moles 206Pb : 238U =0.0081 :206238= 0.0092\uf05c t =\uf0b4\uf0b492.303 4.51 100.693log (1 + 0.0092)= \uf0b4\uf0b492.3034.51 100.693\u00d7 0.00397 =\uf0b490.041210 = 5.945 \u00d7 10 years0.6937 2 8 .STABILITY OF NUCLEI WITH RESPECT TO NEUTRON - PROTON RATIO :If number of neutrons is plotted against the number of protons, the 0 20406080 10012014020 40 60 80 100 120 140Number of protons (p)Number of neutrons (n)Zone ofstabilityUnstable regionUnstable regionn\/p=1stable nuclei lie within well-defined region called zone of stability.All the nuclei falling outside this zone are invariably radioactiveand unstable in nature. Nuclei that fall above the stability zonehas an excess of neutrons while those lying below have moreprotons. These nuclei attain stability by making adjustment inn\/p ratio.\uf075When (n\/p) ratio is higher than that required for stability :Such nuclei have tendency to emit -rays (transforming a neutron into proton).\uf06210n \uf0be\uf0ae p + e ( -particle) 110\u20131\uf062146U \uf0be\uf0ae N + e 1470\u20131np1.33 18736Kr \uf0be\uf0ae Rb + e 87370\u20131np :51365037\uf075When (n\/p) ratio is lower than that required for stability :Such nuclei have tendency to increase n\/p ratio by adopting any of the following three ways.\uf075By emission of an -particle (natural radioactivity).\uf06129892U \uf0be\uf0ae\uf02023490Th + He ( -particle) 42\uf061nP\uf0e6 \uf0f6\uf0e7 \uf0f7\uf0e8 \uf0f814692 \uf0e6\uf0f6\uf0e7\uf0f7\uf0e8\uf0f814490 \uf0e6\uf0f6\uf0e7\uf0f7\uf0e8\uf0f8=1.50 =1.60\uf075By emission of positron137N \uf0be\uf0ae 136C + e 0+1np6776\uf075By K-electron capture19479Au + e 0\u20131\uf020\uf0be\uf0ae\uf02019478PtnP\uf0e6 \uf0f6\uf0e7 \uf0f7\uf0e8 \uf0f81157911678\uf061-emission is usually observed in natural radioactive isotopes while emission of positron or K-electron captureis observed in artificial radioactive isotopes. The unstable nuclei continue to emit or -particle until stable\uf061\uf062nucleus comes into existence.","2 9 .NUCLEAR FISSION :It is a nuclear reaction in which heavy nucleus splits into lighter nuclei of comparable masses with releaseof large amount of energy by bombardment with suitable sub-atomic particles, i.e.10n +23592U56360Ba + Kr + 3 n140 93 154380Xe + Sr + 2 n144 90 155370Cs + Rb + 2 n144 90 1If the neutrons from each nuclear fission are absorbed by other U92235 nuclei, these nuclei split and releaseeven more neutrons. Thus, a chain reaction can occur. A nuclear chain reaction is a self sustaining seriesof nuclear fissions caused by the previous neutrons released from the previous nuclear reactions.1235n + U09256Ba + n140 1092235U+ n 1092U 23536Kr + n931092U 2350n 192235U0n 192235U0n 192235UThere should be critical amount of the fissionable material to maintain fission chain. This in turn requires,minimum critical mass of the fissionable material. It is the small mass of the fissionable material in whichchain reaction can be sustained. If mass is larger than critical mass (supercritical mass), then the numberof nuclei that split, multiplies rapidly. An atomic bomb is detonated with small amount of chemical explosivethat push together two or more masses of fissionable material to get a supercritical mass.A nuclear fission reactor is a device that permits a controlled chain nuclear fissions. Control rods made ofelements such as boron and cadmium, absorb additional neutrons and can therefore, slow the chain reactions.3 0 .NUCLEAR FUSION :It is a nuclear reaction in which two lighter nuclei are fused together to form a heavier nuclei. To achievethis, colliding nuclei must posses enough kinetic energy to overcome the initial force of repulsion betweenthe positively charged core. At very high temperature of the order of 10 to 10 K, the nuclei may have67the sufficient energy to overcome the repulsive forces and fuse. Such reactions are therefore also knownas thermonuclear reactions.23411120HHHen 17.8MeV\uf02b\uf0be\uf0be\uf0ae\uf02b\uf02b224112HHHe 24.9 MeV\uf02b\uf0be\uf0be\uf0ae\uf02b134112HHHe 20.0 MeV\uf02b\uf0be\uf0be\uf0ae\uf02b714312LiHHe 17.7 MeV\uf02b\uf0be\uf0be\uf0ae\uf02bThe energy of fusion process is due to mass defect (converted into binding energy). The high temperaturerequired to initiate such reaction may be attained initially through fission process.Hydrogen bomb is based on the principle of fusion reactions. Energy released is so enormous that it is about1000 times that of atomic bomb. In hydrogen bomb, a mixture of deuterium oxide (D O) and tritium oxide2(T O) is enclosed in space surrounding an ordinary atomic bomb. The temperature produced by the explosion2of the atomic bomb initiates the fusion reaction between 21H and 31H releasing huge amount of energy..It is believed that the high temperature of stars including the sun is due to fusion reactions. E. Salt Peterin 1953, proposed a proton-proton chain reaction.11201111HHHe\uf02b\uf02b\uf0be\uf0be\uf0ae\uf02b\uf02b \uf067213112HHHe\uf02b\uf0be\uf0be\uf0ae\uf02b \uf06731402121HeHHee\uf02b\uf02b\uf0be\uf0be\uf0ae\uf02b\uf02b \uf0671401214 HHe 2 e 24.7MeV\uf02b\uf0be\uf0be\uf0ae\uf02b\uf02b","MEMORY TIPS1 .Expression for rate constants for reaction of different orders.Type of reactionIntegrated rate Unit of rate Half-lifet3\/4 life equation constant periodperiodZero order reaction00d[A]k [A]dt \uf02d\uf03d Concentration10a t =2k2 -- Differentiation formdxdt=k time\u20131First order reactionk =1102.303alogt(a \u2013 x) time\u2013110.693t =K1231140.693 1.382t = 2\u00d7=kkSecond order22.303b(ax)klogt(ab)a(bx)\uf02d\uf03d\uf02d\uf02d Mole litre time \u20131\u201311221tk a \uf03d3243t =k areaction Differential form2dxk(ax)dt\uf03d\uf02dThird order reaction322x (2ax)kt2a (ax)\uf0b4\uf02d\uf03d\uf02d Litre mole time2\u20132\u2013112323t =2k a --- Differential form3dxk(ax)dt\uf03d\uf02d2 .Some typical linear plots for reactions of different orders :( i )(a )0Zero order r(a )01 Firstorderr(a )02Second orderr(a )03r Third order(ii)tZeroorder(a\u2013x)t First orderlog (a\u2013x)tSecond order(a\u2013x)1tThirdorder(a\u2013x)2 1(iii)t1\/2Zeroorder(a )0t1\/2Firstorder(a )0t1\/2Second order(a )01t1\/2Third ordera0 2 13 .Amount left after n half-lives =n01[A]2\uf0e6 \uf0f6\uf0e7 \uf0f7\uf0e8 \uf0f8No. of half-lives =12Total timet","4 .Exponential form of expression for rate constant for reaction of 1 order : [A] = [A] e or C = C est 0 \u2013ktt0\u2013kt5 .12n 101t[A]\uf02d\uf0b56 .Arrhenius equation for effect of temperature on rate constant,k = Ae\u2013E \/RTaon log k = log R \u2013a E2.303RTAlso,d ln kdT=a2ERTIf k and k are rate constants at temperature T and T , then1212log21kk=a E2.303R1212TTT T\uf02d \uf0e9\uf0f9\uf0ea\uf0fa\uf0eb\uf0fb7 .Examples of reactions of 1 order and their formula for rate constantsst(i)2N O 25\uf0ae 4NO + O ; k =222.303tlogtVVV \uf0a5\uf0a5\uf02dwhere V = volume of O gas collected at infinite time\uf0a52 V = volume of O gas collected at time tt2(ii)NH NO 42\uf0ae 2H O + N ; k =222.303tlogtVVV \uf0a5\uf0a5\uf02dwhere V and V are volumes of N gas collected after infinity time and after time t respectively.\uf0a5t 2(iii)H O 22\uf0ae H O + 212O ; k =22.303tlogDtVVwhere V and V are the volumes of KMnO solution used for titrating a definite volume of the reactionDt4mixture at t = D and at time t respectively.(iv)CH COOC H + H O 3252\uf0ae CH COOH + C H OH325k =2.303tlogDtVVVV \uf0a5\uf0a5\uf02d\uf02dwhere V , V and V are the volume of NaOH solution used for titration mixture at zero time, afterDt\uf0a5time t and after infinity respectively.(v)C H O + H O 1222112H \uf02b\uf0be\uf0be\uf0be\uf0ae C H O + C H O61266126 Glucose Fructosek =2.303tlog0t(rr )(rr )\uf0a5\uf0a5\uf02d\uf02dwhere r , r and r are the polarimetric reading at zero time, after time t and after infinity respectively.0t\uf0a5","8 .Rate law equation for reactions involving parallel reactionA k1k2B (90%)C (10%)Rate = \u2013d[A][dt]= k [A] + k [A] = [k + k ] [A]1 2 129 .Degree of dissociation at any time t = (1 \u2013 e )\u2013kt","SOLVED OBJECTIVE PROBLEMSEx .1Find the activation energy [kJ\/mol] for the reaction, A(g) + B(g) \uf0ae C(g) + D(g).From the plot given below :100806040200Reaction courseC+DA + B[A] 20[B] 60[C] 40[D] 80Sol.E = 100 \u2013 40 = 60 kJ mola\u20131Hence the answer is [B].Ex .2In an endothermic reaction, H represents the enthalpy of reaction in kJ\/mol, the minimum value for\uf044the energy of activation will be.(A) less than H\uf044(B) zero(C) more than H\uf044(D) equal to H\uf044Sol.Progress of reaction\uf044HE > Ha\uf044Hence, (C) is the correct answer.Ex .3Consider the following first order competing reactionsA \uf0be\uf0ae\uf020B, C \uf0be\uf0ae\uf020D,the ratio of 12kk, if only 25% of A have been reacted whereas 50% of C has been reacted, calculate theratio of 12kk(A) 0.415(B) 0.246(C) 2.06(D) 0.06Sol.k =112.303tlog10075 for 25% (A) reactedk =222.303tlog10050for 50% (C) reacted\uf05c1221kt0.1249kt0.3010\uf03d\uf0b4Since t = t21\uf05c12k0.1249k0.3010 \uf03d = 0.415Hence, (A) is the correct answer.","Ex .4For a reaction A k 1\uf0be\uf0be\uf0ae B k 2\uf0be\uf0be\uf0aeC. If the reactions are of 1st order then d [B]dt is equal to[A] \u2013 k [B]2[B] +k [A][C] k [A] \u2013 k [B]12[D] k [A] + k [B]12Sol.Rate of increase in [B] = k [A]1Similarly rate of decrease in [B] = k [B]2Thus,d [B]dt= k [A] \u2013 k [B]12Hence the answer is [C]Ex .5The half life period t is independent of initial concentration of reactant when the order of reaction is1\/2[A] Negative[B] 0[C] 1[D] FractionalSol.t\u00bd of a reaction of an order n is related to initial concentration by the expressiont \u00bd \uf0b5n\u2013101C(Here, n = order of reaction)for n = 1, t\u00bd is independent of concentration term. Hence the answer is [C].Ex .6For a first order reaction, A \uf0be\uf0ae\uf020B, the rate of reaction at [A] = 0.1 M is 1.0 \u00d7 10 mol L min . The\u20131 \u20131 \u20131half-life period for the reaction is(A) 42 sec(B) 21 sec(C) 20 sec(D) 28 secSol.r = k[A]k = 1r10[A]0.1\uf02d\uf03d= 1120.693tk\uf03d=0.6931= 0.693 min = 0.693 \u00d7 60 \uf0bb 42 secHence, (A) is the correct answer.Ex .7A catalyst lowers the activation energy of a reaction from 30 kJ mol to 15 kJ mol . The temperature\u20131 \u20131at which the uncatalysed reaction will have the same rate as that of the catalysed at 27\u00b0C is(A) \u2013123\u00b0C(B) 327\u00b0C(C) \u2013327\u00b0C(D) +23\u00b0CSol.aa12E 'ETT \uf03d\uf0de21520300T \uf03d\uf05c T = 600 K = 327\u00b0C2Hence, (B) is the correct answer.Ex .8SO Cl 22\uf088\uf088\uf088\uf086\uf087\uf088\uf088\uf088 SO + Cl , is the first order gas reaction with k = 2.2 \u00d7 10 sec at 270\u00b0C. The percentage22\u20135 \u20131 of SO Cl decomposed on heating for 50 minutes is22(A) 1.118(B) 0.1118(C) 18.11(D) 6.39Sol.k = 2.303alogt(ax) \uf02d\uf0de logakt(ax)2.303 \uf03d\uf02d\uf0de log aax \uf02d=52.2 1050 602.303\uf02d\uf0b4\uf0b4\uf0b4= 0.0286Hence, a(ax) \uf02d= 1.068 \uf0de\uf020ax0.936a\uf02d\uf03d\uf0de\uf020 \uf02dx1a= 0.936 \uf0de\uf020xa = 0.068 = 6.39 %Hence, (D) is the correct answer.","Ex .9In the decomposition of N O , the plot between the reciprocal of concentration of the reactant and the time25was found to be linear as shown in figure. Determine the order of reaction.Time 1Conc.Sol.The reaction is of second order, because for II order,k = 1t.xa(a \u2013 x) ort = 1k.xa(a \u2013 x)ortime (t) vs. 1conc. graph is linearEx. 10The inversion of cane sugar proceeds with half-life of 250 minutes at pH = 4 for any concentration ofsugar. However, if pH = 5, the half-life changes to 25 minutes. The rate law expression for the sugar inversioncan be written as(A) r = k [sugar] [H]2 6(B) r = k [sugar] [H ]1 + 0(C) r = k [sugar] [H ]1 + 1(D) r = k [sugar] [H ]0 + 1Sol.At pH = 4, the half-life is 250 minutes for all concentrations of sugar that is t1\/2 \uf0b5\uf020[sugar] .0The reaction is first order with respect to sugar.Let, rate = k[sugar] [H ]1+ xFor [H ] t+1\/2 \uf0b5[H ]+ 1\u2013x\uf0de 250 (10 )\uf0b5\u20134 (1\u2013x)......(1)At pH = 5, the half life is 25 minutes so 50 (10 )\uf0b5\u20135 1\u2013x ......(2)\uf05c\uf020\uf02010 = 10(1 \u2013 x)\uf0de (1 \u2013 x) = 1 \uf05c x = 0Therefore, rate = k [sugar] [H ]1+ 0Hence, (B) is the correct answer.Ex. 11In a hypothetical reaction x \uf0ae y, the activation energies for the forward and backward reactions are13 and 8 kJ\/mol respectively. The potential energy of x is 10 kJ\/mol, then(A) the threshold energy of the reaction is 23 kJ\/mol(B) potential energy of y is 15 kJ(C) heat of reaction is 5 kJ(D) the reaction is endothermicSol.13kJ8kJ(y)(z)(x)Energy10kJEfHence, (A), (B), (C) and (D) are correct answer.","Ex. 12The reactions, CH COOC H + NaOH 325\uf0be\uf0ae CH COONa + C H OH, is325(A) biomolecular reaction(B) second order reaction(C) third order reaction(D) none of the aboveSol.(A) and (B)Ex. 13The activation energies of two reactions are E and E ' with E > E '. If the temperature of the reactingaa aasystems is increased from T and T predict which alternative is correct. k ' and k ' are rate constants at1212higher temperature. Assume A being the same for both the reactions.(A) 1212k 'k 'kk \uf03d(B) k < k and k' < k '1212(C) k > k and k' < k '1212(D) 1212''k2kkk \uf03cSol.More is energy of activation lesser is rate constant.k = Ae\u2013E \/RTak < k and k ' < k '1212Hence, (B) is the correct answer.Ex. 14Fill in the blank :2351921920360Un?Kr3 n\uf02b\uf0be\uf0be\uf0ae \uf02b\uf02b(A) 14156Ba(B) 13956Ba(C) 13954Ba(D) 14154BaSol.92 + 0 = Z + 36 + 0 \uf0de Z = 56235 + 1 = A + 92 + 3\uf05c A = 141Missing nucleide is 14156BaHence, (A) is the correct answer.Ex. 15In the nuclear reaction, 2342229286URn \uf0be\uf0be\uf0ae, the number of and -particles lost would be :\uf061\uf062(A) 1, 3(B) 2, 3(C) 3, 0(D) 0, 3Sol.23422240928621URnxy \uf02d\uf0be\uf0be\uf0ae\uf02b\uf061 \uf02b\uf062234 = 222 + 4x \uf0de x = 392 = 86 + 2x \u2013 y or y = 0Hence, (C) is the correct answer.Ex. 16The rate constant of a reaction is given by k = 3.2 \u00d7 101027002.303R e \uf02dIt means that(A) log k vs 1T will be straight line with slope =27002.303R \uf02d(B) log k vs 1T will be a straight line with intercept on log k axis = log 3.2 \u00d7 1010(C) the number of effective collisions are 3.2 \u00d7 1010 cm sec\u20133 \u20131(D) half-life of the reaction increases with increase in temperatureSol.(A) and (B) are correct, (C) is wrong because frequency factor gives total number of collisions and notthe effective collision cm sec , (D) is wrong because half-life of the reaction decreases with increase\u20133\u20131in temperature (as reaction becomes faster).Hence, (A) and (B) are correct answer.","Ex. 17Two substances x and y are present such that [x ] = 2[y ] and half-life of x is 6 minutes and that of y is0018 minutes. If they start decaying at the same time following first order kinetics how much time later willthe concentration of both of them would be same?(A) 15 minutes(B) 9 minutes(C) 5 minutes(D) 12 minutesSol.Amount of x left in n half-lives =11 n01[x ]2\uf0e6 \uf0f6\uf0e7 \uf0f7\uf0e8 \uf0f8Amount of y left in n half-lives =22 n01[y ]2\uf0e6 \uf0f6\uf0e7 \uf0f7\uf0e8 \uf0f8At the end, 1200nn[x ][x ]22 \uf03d\uf0de21 nn2122 \uf03d, {[x ] = 2[y ]}00\uf05c\uf020nn22= 2 \uf0de\uf02021 nn2\uf02d= (2)1\uf05c n \u2013 n = 112 n = (n \u2013 1)21...(1)Also, t = n \u00d7 t11\/2(x) ; t = n \u00d7 t21\/2(y)(Let, concentration of both become equal after time t)\uf05c11 \/ 2(x)21 \/ 2(y )ntnt\uf0b4\uf0b4= 1 \uf0de12n6n18 \uf0b4\uf0b4 = 1 \uf0de12n3n\uf03d...(2)From Eqs. (1) and (2), we getn = 0.5, n' = x21.5t = 0.5 \u00d7 18 = 9 minutesHence (B) is the correct answer.Ex. 18Consider a gaseous reaction, the rate of which is given by k[x] [y], the volume of the reaction vessel containingthese gases is suddenly increased to 3 of the initial volume. The rate of reaction relative to the originalrdrate would be(A) 9\/1(B) 1\/9(C) 6\/1(D) 1\/6Sol.By increasing volume to 3 the concentration will become rd13 times, hence rate 19 times.Hence, (B) is the correct answer.Ex. 19The rate constant for the reaction, 2N O 25\uf0be\uf0ae\uf0204NO + O , is 4.0 \u00d7 10 sec . If the rate of reaction22 \u20135 \u20131is 4.80 \u00d7 10 molL sec , the concentration of N O (molL ) is :\u20135 \u20131 \u20131 25\u20131(A) 1.4(B) 1.2(C) 0.04(D) 0.8Sol.r = k [N O ]25\uf05c [N O ] = 2555r4.80 10k4.0 10\uf02d\uf02d\uf0b4\uf03d\uf0b4= 1.2 molL\u20131Hence, (B) is the correct answer.","Ex. 20The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at 25\u00b0C are2 \u00d7 10 , s , 114.4 kJ mol\u20134\u20131\u20131 and 6.0 \u00d7 1014 s\u20131 respectively, the value of the rate constant atT \uf0ae \uf0a5 is.(A) 2.0 \u00d7 1018 \u20131s(B) 3.6 \u00d7 1030 \u20131s(C) \uf0a5(D) 6.0 \u00d7 1014 \u20131sSol.k = Ae\u2013E \/RTaWhen T \uf0ae\uf020\uf0a5k \uf0ae AA = 6 \u00d7 10 s14\u20131Hence, (D) is the correct answer.Ex. 21If a reaction A + B \uf0ae C, is exothermic to the extent of 40 kJ\/mol and the forward reaction has an activationenergy 60 kJ\/mol, the activation energy for the reverse reaction is(A) 30 kJ\/mol(B) 40 kJ\/mol(C) 70 kJ\/mol(D) 100 kJ\/molSol.Progress of reactionEnergy60kJEnergy100kJ40kJActivation energy for backward reaction = 100 kJHence, (D) is the correct answer.Ex. 22The rate of reaction is doubled for every 10\u00b0 rise in temperature. The increase in reaction rate as a resultof temperature rise from 10\u00b0 to 100\u00b0 is.(A) 112(B) 512(C) 400(D) 614Sol.Increase in steps of 10\u00b0 has been made 9 times. Hence, rate of reaction should increase 2 times i.e.,9 512 times.Hence, (B) is the correct answer.Ex2 3.Van't Hoff equation is(A) (d\/dT) lnK = (\u2013 E\/RT )\uf0442(B) (d\/dT) lnK = + (E\/RT )2(C) (d\/dT) lnK = \u2013( E\/RT)\uf044(D) K = Ae\u2013E \/RTaSol.(B) and (D).Ex. 24The rate of chemical reaction (except zero order)(A) decreases from moment to moment(B) remains constant throughout(C) depends upon the order of reaction(D) none of the aboveSol.(A) and (C)Ex. 25 The accompanying figure depicts the change in concentration of species A and B for the reaction A \uf0ae B, asa function of time the point of inter section of the two curves represents.Timeconc.[B][A][A] t1\/2[B] t3\/4[C] t2\/3[D] data insufficient to predictSol.The intersection point indicates that half life of the reactant A is converted into B.Hence the answer is [A]."]