6 . Compressibility factor for H behaving as real gas is : 2 (A) 1 (B) 1 a (C) 1 Pb RTV RTV RT (D) (1 a) 7 . Equal masses of methane and oxygen are mixed in an empty container at 25°C. The fraction of the total pressure exerted by oxygen is : (A) 1/3 (B) 1/2 (C) 2/3 (D) (1/3) (273/298) 8 . The temperature at which a real gas obeys the ideal gas laws over a wide range of pressure is : (A) critical temperature (B) Boyle temperature (C) boiling temperature (D) reduced temperature 9 . Helium atom is two times heavier than a hydrogen molecule. At 298 K, the average kinetic energy of a helium atom is : (A) two times that of a hydrogen molecules (B) same as that of a hydrogen molecules (C) four times that of a hydrogen molecules (D) half that of a hydrogen molecules 1 0 . In vander Waal's equation of state for a non-ideal gas the term that accounts for intermolecular forces is : (A) (V – b) (B) p a (C) RT (D) (RT)–1 v2 1 1 . The compressibility of a gas is less than unity at STP. Therefore : (A) V > 22.4 L (B) V < 22.4 L (C) V = 22.4 L (D) V 44.8 L m m m m 1 2 . If two moles of an ideal gas at 546 K occupies a volume of 44.8 litres, the pressure must be : (A) 2 atm (B) 3 atm (C) 4 atm (D) 1 atm 1 3 . At STP the order of root mean square velocity of molecules of H , N , O and HBr is : 22 2 (A) H > N > O > HBr(B) HBr > O > N > H (C) HBr > H > O > N (D) N > O > H > HBr 22 2 22 2 222 222 1 4 . The density of a gas at 27°C and 1 atm is d. Pressure remaining constant at which of the following temperatures will its density become 0.75 d? (A) 20 °C (B) 30 °C (C) 400 K (D) 300 K 1 5 . At 27°C the ratio of rms velocities of ozone to oxygen is : (A) 3 / 5 (B) 4 / 3 (C) 2 / 3 (D) 0.25 1 6 . A real gas most closely approaches the behaviour of an ideal gas at : (A) 15 atm and 200 K (B) 1 atm and 273 K (C) 0.5 atm and 500 K (D) 15 atm and 500 K 1 7 . The temperature at which the second virial coefficient of a real gas is zero is called : (A) Critical temperature (B) Boyle's temperature (C) Boiling point (D) none of these 1 8 . V vs T curves at constant P and P for an ideal gas are shown in figure. Which is correct : 12 (A) P > P V 12 P2 (B) P < P P1 12 (C) P = P 12 (D) all T
1 9 . At STP, 2.8 litres of hydrogen sulphide were mixed with 1.6 litres of sulphur dioxide and the reaction occurred according to the equation : 2H2S (g) + SO2 (g) 2H2O (l ) + 3S (s) Which of the following shows the volume of the gas remaining after the reaction? (A) 0.2 litres of SO (g) (B) 0.4 litres of H (g) 2 2 (C) 1.2 litres of H S (g) (D) 1.2 litres of SO (g) 2 2 2 0 . The rates of diffusion of SO , CO , PCl and SO are in the following order : 32 3 2 (A) PCl > SO > SO > CO (B) CO > SO > PCl > SO 3 32 2 2 2 33 (C) SO > SO > PCl > CO (D) CO > SO > SO > PCl 2 3 32 2 23 3 2 1 . Which of the following curves does not represent Boyle's law : (A) P (B) log P (C) P P (D) V log V 1/V V 2 2 . A closed vessel contains equal number of nitrogen and oxygen molecules at pressure of P mm. If nitrogen is removed from the system, then the pressure will be : (A) P (B) 2P (C) P/2 (D) P2 2 3 . The ratio of average molecular kinetic energy of PCl to that of O both at 300 K is : 52 (A) 1 : 1 (B) 7 : 2 (C) 176 : 1 (D) 2 : 7 2 4 . The correct order of normal boiling points of O , N , NH and CH for whom the values of vander waals 22 3 4 constant 'a' are 1.360, 1.390, 4.170 and 2.253 L2.atm.mol–2 respectively, is : (A) O < N < NH < CH (B) O < N < CH < NH 22 3 4 22 4 (C) NH < CH < N < O (D) NH < CH < O < N 3 42 2 3 422 2 5 . A molecule in a cube of length has a velocity v in the x direction. The number of collisions with one x face per second for this molecule with be given by which one of the following : (A) v / (B) v /2 (C) ½ (D) N mv2/3 x x 2 6 . The rate of diffusion of methane at a given temperature is twice that of gas X. The molecule weight of X is : (A) 64.0 (B) 32.0 (C) 4.0 (D) 8.0 2 7 . For a fixed volume of a gas the pressure is increases when the temperature increases. This is due to : (A) increase in the average molecular speed (B) increase in the rate of collisions (C) increase in the molecular attraction (D) decrease in the mean free path 2 8 . The ratio, rms velocity of SO2 , of sulphur dioxide and helium gases at 30°C is equal to : rms velocity of He (A) 4 (B) 0.25 (C) 0.10 (D) 8
2 9 . A certain volume of argon gas (Mol Wt. 40) requires 45 s to effuse through a hole at a certain pressure and temperature. The same volume of another gas of unknown molecular weight requires 60 s to pass through the same hole under the same conditions of temperature and pressure. The molecular weight of the gas is : (A) 53 (B) 35 (C) 71 (D) 120 3 0 . On the surface of the earth at 1 atm pressure, a balloon filled with H gas occupies 500 mL. This volume 2 5/6 of its maximum capacity. The balloon is left in air. It starts rising . The height above which the balloon will burst if temperature of the atmosphere remains constant and the pressure decreases 1 mm for every 100 cm rise of height is : (A) 120 m (B) 136.67 m (C) 126.67 m (D) 100 m 3 1 . A chemist has synthesized a greenish yellow gaseous compound of chlorine and oxygen and finds that its density is 7.71 g/L at 36°C and 2.88 atm. Then the molecular formula of the compound will be : (A) ClO (B) ClO (C) ClO (D) Cl O 3 2 22 3 2 . In the following arrangement the pressure of the confined gas will be : (A) 15 cm of Hg Patm=76cm of Hg Pgas=? (B) 25 cm of Hg =27.2gm/cc (C) 35 cm of Hg 10cm (D) 45 cm of Hg =6.8gm/cc 30cm 3 3 . Which of the following graphs represent Boyle's law : =1gm/cc 13.6cm 76cm 15cm Hg P PV PV PV (B) (C) (D) (A) V PV P 3 4 . What conclusions would you draw from the following graphs : PV 0 K(–273.15°C) T 0 K(–273.15°C) T (A) As the temperature is reduced, the volume as well as pressure increase. (B) As the temperature is reduced, the volume becomes zero and the pressure reaches infinity. (C) As the temperature is reduced, the volume as well as the pressure decrease. (D) A point is reached where theoretically, the volume as well as the pressure become zero. 3 5 . Which of the following statements are correct : (A) Helium diffuses at a rate 8.65 times as much as CO does (B) Helium escapes at a rate 2.65 times as fast as CO does. (C) Helium escapes at a rate 4 times as fast as CO does. 2 (D) Helium escapes at a rate 4 times as fast as SO does. 2
3 6 . When a gas is expanded at constant temperature : (A) the pressure decreases (B) the kinetic energy of the molecules remains the same (C) the kinetic energy of the molecules decreases (D) the number of molecules of the gas decreases 3 7 . Which of the following are correct statements : (A) vander waals constant 'a' is a measure of attractive force (B) vander waals constant 'b' is also called co-volume or excluded volume (C) 'b' is expressed in L mol–1 (D) 'a' is expressed in atm L2 mol–2 3 8 . Which of the following statements are incorrect : (A) Molar volume of every gas at STP is 22.4 L (B) Under critical states compressibility factor is 1 (C) All gases will have equal value of average KE at a given temperature 3 (D) At absolute zero, KE is R. 2 3 9 . At what temperature will the total KE of 0.3 mol of He be the same as the total KE of 0.40 mol of Ar at 400 K : (A) 533 K (B) 400 K (C) 346 K (D) 300 K 4 0 . If molecules of the gas are spherical of radius 1 Å, the volume occupied by the molecules in 1 mol of a gas is : (A) 22400 mL (B) 22.4 L (C) 2.52 mL (D) 4.22 mL 4 1 . See the figure : NH3 HCl XB A CY The valves of X and Y are opened simultaneously. The white fumes of NH Cl will first form at : 4 (A) A (B) B (C) C (D) A,B and C simultaneously 42. Calculate (ratio of C and C) for triatomic linear gas at high temperature. Assume that the contribution p v of vibrational degree of freedom is 75% : (A) 1.222 (B) 1.121 (C) 1.18 (D) 1.33 4 3 . 11 moles N and 12 moles of H mixture reacted in 20 litre vessel at 800 K. After equilibrium was reached, 22 6 mole of H was present. 3.58 litre of liquid water is injected in equilibrium mixture and resultant gaseous 2 mixture suddenly cooled to 300 K. What is the final pressure of gaseous mixture ? Negative vapour pressure of liquid solution. Assume (i) all NH dissolved in water (ii) no change in volume of liquid (iii) no reaction 3 of N and H at 300 K : 22 N2 :11 moles T=800K N2,H2 T = 300 K ; P = ? H2 :12 moles V=20L NH3(aq) solution Initial condition (A) 18.47 atm (B) 60 atm (C) 22.5 atm (D) 45 atm
4 4 . Two closed vessel A and B of equal volume containing air at pressure P and temperature T are connected 11 to each other through a narrow open tube. If the temperature of one is now maintained at T and other 1 at T (where T > T ) then that what will be the final pressure ? 2 12 (A) T1 (B) 2P1T2 (C) 2P1T2 (D) 2 P1 2P1 T2 T1 T2 T1 T2 T1 T2 4 5 . A balloon containing 1 mole air at 1 atm initially is filled further with air till pressure increases to 4 atm. The initial diameter of the balloon is 1 m and the pressure at each stage is proportion to diameter of the balloon. How many no. of moles of air added to change the pressure from 1 atm to 4 atm : (A) 80 (B) 257 (C) 255 (D) 256 4 6 . What is the density of wet air with 75% relative humidity at 1 atm and 300 K? Given : vapour pressure of H O is 30 torr and average molar mass of air is 29 g mol–1 : 2 (A) 1.614 g/L (B) 0.96 g/L (C) 1.06 g/L (D) 1.164 g/L 4 7 . Calculate minimum number of balloons each of volume 82.1 L required to lift a mass of 1 kg to a height of 831 m. Given : molar mass of air = 29 g/mol, temperature is constant at 290 K and mass of each balloons is 40 g. [Use e–0.1 = 0.9, pressure at sea level = 1 atm, acceleration due to gravity (g) = 10 m/s2] : (A) 10 (B) 20 (C) 25 (D) 50 4 8 . If a real gas following equation P(V – nb) = nRT, at low pressure then find the intercept and slope of graph d between p v/s P are respectively : (A) MR , M(RT)2 M Mb Mb M RT b Tb (B) RT , (RT)2 (C) RT , (RT)2 (D) M , M(RT)2 BRAIN TEASERS ANSWER KEY EXERCISE -2 Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. C D A C A C A B B B B A A C C Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Ans. C B A A D D C A B B A A B C C Que. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 An s . B B B,C C,D B,D A,B All B,D A C C C C B C Que. 46 47 48 Ans. D B B
EXERCISE–03 MISCELLANEOUS TYPE QUESTIONS TRUE / FALSE 1 . The volume occupied by 32 g of oxygen is greater than that occupied by 16 g of methane, both being at the same T and P. 2 . A real gas can be liquefied if its temperature is greater than its critical temperature. 3 . Kinetic energy of gaseous molecules is zero at 0°C. 4 . The term V – b in vander waals equation represents the available volume where molecules of the gas can m move. 5 . The average speed of a gas varies linearly with increases of temperature. 6 . A mixture of ideal gaseous is cooled upto liquid helium temperature (4.22 K) to form an ideal solution. 7 . For a vander waals gas VC = 3b, PC = a/27 b2, TC = 8a/27b R. Numerically the compressibility factor of a vander waals gas at the critical points is Z = 0.375. FILL IN THE BLANKS 1 . The value of PV for 5.6 L of an ideal gas is ......................... RT at STP. 2 . The density of an ideal gas ......................... with increase in temperature of the gas, provided the pressure remains constant. 3 . The unit of vander waals constant 'a' in SI units is ......................... . 4 . 0.5 L of a certain gas at STP weighs 0.58 g. Its molar mass is .........................g mol–1. 5 . The root means square speed of methane gas will be ......................... that of helium gas both having the same temperature and pressure. 6 . (i) The rate of diffusion of a gas is ......................... proportional to ......................... or square root of molecular mass. (ii) The total energy of one mole of gas (ideal monoatomic) at 27°C is ......................... calorie. (iii) Equal masses of SO2 and O2 are kept in a vessel at 27°C. The total pressure is 2 atm. The partial pressure of SO is ......................... . 2 7 . (i) There is no effect of ......................... on the motion of gas molecules. (ii) If the density of a gas at 27°C and 1 atmospheric pressure is 1.8 g lit–1, its molecular mass is ......................... . (iii) If the speed of a molecule at 27°C is 0.25 metre sec–1, its speed at 927°C will be ......................... . (iv) A steel vessel of capacity 22.4 litre contains 2 g of hydrogen, 8g of oxygen and 22 g of carbon dioxide at a temperature of 0°C. The total pressure of the gas is ......................... . 8 . A gas has a volume of 580 cm3 at a certain pressure. If its pressure is increased by 0.96 atm, its volume becomes 100 cm3. The new pressure of the gas is ......................... .
MATCH THE COLUMN 1. Column-I Column-II (p) Dalton's law of partial pressures at constant (A) P V = P V = P V = .......... 11 22 33 temperature. (q) Kinetic equation of ideal gases. (B) V1 V2 V3 =......... at constant T1 T2 T3 (r) 22.4 litre pressure. (s) Isotherm (C) r 1 (t) Isobar d (u) Charles' law (D) P = P + P + P + ............ (v) Graham's law 123 (w) Boyle's law (x) Equation for real gases. (E) (V – b) P a RT V2 (y) Boltzmann's constant (F) R/N (G) Molar volume (H) PV 1 mnc2 = 3 (I) Graph between P and V at constant temperature. (J) Graph between V and T at constant pressure. ASSERTION & REASON These questions contains, Statement I (assertion) and Statement II (reason). (A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I. (B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I (C) Statement-I is true, Statement-II is false (D) Statement-I is false, Statement-II is true 1 . Statement-I : Plot of P Vs. 1/V (volume) is a straight line. Because Statement-II : Pressure is directly proportional to volume. 2 . Statement-I : Absolute zero is a theoretically possible temperature at which the volume of the gas becomes zero. Because Statement-II : The total kinetic energy of the molecule is zero at this temperature. 3 . Statement-I : Pressure exerted by a mixture of reacting gases is equal to the sum of their partial pressures. Because Statement-II : Reacting gases react to form a new gas having pressure equal to the sum of both. 4 . Statement-I : Critical temperature of the gas is the temperature at which it occupies 22.4 L of volume. Because Statement-II : Molar volume of every gas at NTP is 22.4 L. 5 . Statement-I : Excluded volume or co-volume equals to (v–nb) for n moles. Because Statement-II : Co-volume depends on the effective size of gas molecules. 6 . Statement-I : Carbondioxide has greater value of root mean square velocity u than carbon monoxide. rms Because Statement-II : u is directly proportional to molar mass. rms
7 . Statement-I : Most probable velocity is the velocity possessed by maximum fraction of molecules at the same temperature. Because Statement-II : On collision, more and more molecules acquire higher speed at the same temperature. 8 . Statement-I : The effusion rate of oxygen is smaller than that of nitrogen. Because Statement-II : Molecular size of nitrogen is smaller than that of oxygen. COMPREHENSION BASED QUESTIONS Comprehension # 1 The rate of change of pressure (p) of a gas at constant temperature and constant external pressure due to effusion of gas from a vessel of constant volume is related to rate of change of number of molecules present by dp kT dN dt V dt where k = Boltzmann constant, T = temperature, V = volume of vessel & N = No. of molecules and dN pA 0 , where A = area of orifice and m = mass of molecule. dt (2mkT )1 / 2 0 1 . Time required for pressure inside vessel to reduce to 1/e of its initial value in (ln e = 1) 2m 1 / 2 V kT 1 / 2 V 2 m kT 1 / 2 2m V (A) kT A 0 (B) 2m A 0 (D) kT A0 (C) A0 2 . If the gas inside the vessel had molecular weight 9 times the gas in previous example and area of orifice was doubled and temperature maintained at 4T, time required for pressure to fall to 1/e times of its initial value would be (t = answer of previous option) (A) 1.33 t (B) 4.24 t (C) 0.75 t (D) 1.125 t 3 . The incorrect statement (s) is/are. [I] Pressure will not fall to zero in infinite time. [II] Time required for pressure to decrease to half its initial value is independent of initial pressure. [III] The relations given above are true for real gases also. (A) I (B) II (C) III (D) I and III Comprehension # 2 Sketch shows the plot of Z v/s P for a hypothetical gas for one mole at three distinct temperature. 200K 2 500K 0°C H2 1000K Z1 1 Z CO2 250 500 750 P P(atm.) Boyle's temperature is the temperature at which a gas shows ideal behaviour over a pressure range in the low a pressure region. Boyle's temperature (Tb) = Rb . If a plot is obtained at temperatures well below Boyle's temperature then the curve will show negative deviation, in low pressure region and positive deviation in the high pressure region. Near critical temperature the curve is more likely as CO2 and the temperature well above critical temperature curve is more like H2 at 0°C as shown above. At high pressure suppose all the constant temperature curve varies linearly with pressure according to the following equation Z = 1 Pb (R = 2 cal mol–1 K–1) + RT
1 . Which of the following is correct : (A) a < 0.4 k cal mol–1 (B) 0.4 k cal mol–1 a < 2 k cal mol–1 b < b (C) a > 0.4 k cal mol–1 (D) a = 1 K cal mol–1 b b 2 . For 500 K plot value of Z changes from 2 to 2.2 if pressure is varied from 1000 atm to 1200 atm (high pressure) then the value of b will be :- RT (A) 10–3 atm–1 (B) 2 × 10–3 atm–1 (C) 5 × 10–4 atm–1 (D) 10–4 atm–1 3 . As shown in the figure at 200 K and 500 atm value of compressibility factor is 2 (approx). Then volume of the gas at this point will be :- (A) 0.01 L (B) 0.09 L (C) 0.065 L (D) 0.657 L 4 . Plot at Boyle's temperature for the gas will be : ZZ ZZ 1 1 1 1 (A) (B) (C) (D) P P P P 5 . In very high pressure region if Z v/s P is plotted at 1200 K for the above gas then it will have greatest slope. (A) true (B) false (C) can't say (D) not related to the paragraph MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE -3 True / False 2. F 3. F 4. T 1. F 6. F 7. T 5. F Fill in the Blanks 1. 0.25 2. decreases 3. Pa m6 mol–2 4. 26 5. One half 6. (i) inversely, square root of density, (ii) 900, (iii) 2/3 7. w (i) gravity (ii) 44.33 (iii) 0.5 m/s (iv) 1.75 atm 8. 1.16 atm Match the Column 1. A - (w), B - (u), C - (v), D - (p), E - (x), F - (y), G - (r), H - (q), I - (s), J - (t) Assertion - Reason Questions 1. C 2. B 3. E 4. D 5. D 6. E 7. C 8. C Comprehension Based Questions Comprehension #1 : 1. (A) 2. (C) 3. (C) 4. (C) 5. (B) Comprehension #2 : 1. (B) 2. (A) 3. (C)
EXERCISE–04 [A] CONCEPTUAL SUBJECTIVE EXERCISE 1 . 3.6 g of an ideal gas was injected into a bulb of internal volume of 8 L at pressure P atm and temp T K. The bulb was then placed in a thermostat maintained at (T + 15)K 0.6 g of the gas was let off to keep the original pressure. Find P and T if mol weight of is 44. 2 . A toy balloon originally held 1.0 g of He gas and had a radius 10 cm. During the night, 0.25 g of the gas effused from the balloon. Assuming ideal gas behaviour, under these constant P and T conditions, what was the radius of the balloon the next morning. 3 . If a scuba diver is to remain submerged for 1 hr, what pressure must be applied to force sufficient air into the tank to be used. Assume 0.5 dm3 of air per breath at standard atmospheric pressure, a respiration rate of 38 breaths per minute, and a tank capacity of 30 dm3. 4 . While resting, the average human male use 0.2 dm3 of O per hour at STP for each kg of body mass. Assume 2 that all this O is used to produce energy by oxidising glucose in the body. What is the mass of glucose required 2 per hour by a resting male having mass 60 kg. What volume, at STP of CO would be produced. 2 5 . 12 g N , 4 g H and 9 g O are put into a one litre container at 27°C. What is the total pressure. 22 2 6 . 1.0 × 10–2 kg of hydrogen and 6.4 × 10–2 kg of oxygen are contained in a 10 × 10–3 m3 flask at 473 K. Calculate the total pressure of the mixture. If a spark ignities the mixture, What will be the final pressure. 7 . At room temp, NH gas at one atm & HCl gas at \"P\" atm are allowed to effuse through identical pin holes 3 to the opposite ends of a glass tube 1 m long & uniform cross-section. A white deposit is observed at a distance of 60 cm from the HCl end. What is \"P\". 8 . A gas mixture contains equal number of molecules of N and SF , some of it is passed through a gaseous 26 effusion apparatus. Calculate how many molecules of N are present in the product gas for every 100 2 molecules of SF . 6 9 . Two gases NO and O2 were introduced at the two ends of a one metre long tube simultaneously (tube of uniform cross-section). At what distance from NO gas end, Brown fumes will be seen. 1 0 . At 20°C two balloons of equal volume and porosity are filled to a pressure of 2 atm, one with 14 kg N 2 1 & other with 1 kg H . The N balloon leaks to a pressure of atm in one hour. How long will it take 22 2 1 for H balloon to leaks to a pressure of atm. 22 1 1 . Pure O diffuses through an aperture in 224 sec, where as mixture of O and another gas containing 80% 22 O diffuses from the same in 234 sec. What is molecular weight of the gas. 2 1 2 . A space capsule is filled with neon gas 1.00 atm and 290 K. The gas effuses through a pin - hole into outer space at such a rate that the pressure drops by 0.3 torr/sec. (a) If the capsule were filled with ammonia at the same temperature and pressure, what would be the rate of pressure drop. (b) If the capsule were filled with 30.0 mol % helium, 20.0 mol% oxygen & 50.0 mol% nitrogen at a total pressure of 1.00 atm & a temp of 290 K, what would be the corresponding rate of pressure drop. 1 3 . Show that the height at which the atmospheric pressure is reduced to half its value is given by h = 0.693RT . Mg 1 4 . Calculate the pressure of a barometer on an aeroplane which is at an altitude of 10 Km. Assume the pressure to be 101.325 kPa at sea level & the mean temperature 243 K. Use the average molar mass of air (80% N , 20% O ) 22
1 5 . An iron cylinder contains helium at a pressure of 250 kPa and 27°C. The cylinder can withstand a pressure of 1 × 106 Pa. The room in which cylinder is placed catches fire. Predict whether the cylinder will blow up before it metls or not. [melting point of cylinder = 1800 K] 1 6 . Determine the molar mass of a gas if its pressure is to fall to one-half of its value in a vertical distance of one meter at 298 K. 1 7 . The time taken for a given volume of gas E to effuse through a hole is 75 sec. Under identical conditions the same volume of a mix of CO & N (containing 40% of N by volume) effused in 70 sec. Calculate 22 (i) the relative mol mass of E, and (ii) the RMS velocity (in ms–1 units) of E at 0°C. 1 8 . At what temperature in °C, the U of SO is equal to the average velocity of O at 27°C. rms 2 2 1 9 . The density of CO at 273 K and 1 atm is 1.2504 kg m–3. Calculate (a) root mean square speed (b) the average speed and (c) most probable speed. 2 0 . Calculate the temperature values at which the molecules of the first two members of the homologous series C H will have the same rms speed as CO gas at 770 K. The normal b.p. of n-butane is 273 K. Assuming n 2n+2 2 ideal gas behaviour of n-butane upto this temperature, calculate the mean velocity and the most probable velocity of its molecules at this temperature. 2 1 . Calculate the temperature at which the root mean square velocity, average velocity and most probable velocity of oxygen gas are all equal to 1500 ms–1. 2 2 . Calculate the fraction of N molecules at 101.325 kPa and 300 K whose speeds are in the range of 2 u – 0.005 u to u + 0.005 u . mp mp mp mp 2 3 . What is the ratio of the number of molecules having speeds in the range of 2ump and 2ump + du to the number of molecules having speeds in the range of u and u + du ? mp mp 2 4 . The density of mercury is 13.6 g/cm3. Estimate the b value. 2 5 . Calculate the pressure exerted by 22 g of carbon dioxide in 0.5 dm3 at 298. 15 K using : (a) the ideal gas law and and (b) vander waals equation. Given : [a = 363.76 kPa dm6 mol–2 b = 42.67 cm3 mol–1] 2 6 . The compressibility factor for N at –50°C and 800 atm pressure is 1.95 and at 100°C and 200 atm, 2 it is 1.10. A certain mass of nitrogen occupied one litre at –50°C and 800 atm. Calculate the volume occupied by the same quantity of N at 100°C and 200 atm. 2 2 7 . At 273.15 K and under a pressure of 10.1325 MPa, the compressibility factor of O is 0.927. Calculate 2 the mass of O necessary to fill a gas cylinder of 100 dm3 capacity under the given conditions. 2 2 8 . The vander waals constant for O are a = 1.36 atm L2 mol–2 and b = 0.0318 L mol–1. Calculate the temperature 2 at which O gas behaves, ideally for longer range of pressure. 2 2 9 . The vander waals constants for gases A,B and C are as follows : Gas a/dm6 kPa mol–2 b/dm3 mol–1 A 405.3 0.027 B 1215.9 0.030 C 607.95 0.032 Which gas has (i) the highest critical temperature , (ii) the largest molecular volume, and (iii) most ideal behaviour around STP? 3 0 . A commercial cylinhder contains 6.91 m3 of O at 15.18 M Pa and 21°C. The critical constants for O 22 are T = –118.4°C, P = 50.1 atm. Determine the reduced pressure and reduced temperature for O under CC 2 these conditions.
3 1 . Show that at low densities, the vander waals equation a p (V – b) = RT Vm2 m and the Dieterici's equation P (Vm – b) = RT exp (–a/RTVm) given essential the same value of p. 3 2 . Calculate from the vander waal's equation, the temperature at which 192 g of SO would occupy a vol. 2 of 10 dm3 at 15 atm pressure. [a = 6.7 atm L2 mol2, b = 0.0564 L mol–1] 3 3 . Calculate the pressure of 15 mol neon at 30°C in a 12 litre container using (i) the ideal gas equation (ii) the vander waals equation [a = 0.2107 atm L2 mol2, b = 0.0171 L mol–1] 3 4 . What will be the temperature difference needed in a hot air balloon to lift 1.0 kg of mass? Assume that the volume of balloons is 100 m3, the temperature of ambient air is 25°C, the pressure is 1 bar, and air is an ideal gas with an average molar mass of 29 g mol–1(hot and cold both). 3 5 . One mole of a non linear triatomic gas is heatted in a closed rigid container from 500°C to 1500°C. Calculate the amount of energy required if vibrational degree of freedom become effective only above 1000°C. CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) 1 . P = 0.062 atm, T = 75 k 2 . 9.08 cm 3 . 3.8 × 103 kPa 4 . 16.07 g, 12 dm3 5 . 66.74 atm 6 . P = 27.54 × 105 N/m2 , P = 19.66 × 105 N/m2 7 . 2.19 atm total final 1 0 . 16 min 8 . 228 9 . 50.8 cm 11 . 46.6 1 2 . (a) 0.32 Torr/sec (b) 0.29 Torr/sec 13. 1 4 . 25.027 kPa 1 5 . Yes 1 6 . 175.133 kg mol–1 1 7 . 32.14 g/mol, 460.28 m/s 18 . 236°C 1 9 . U = 493 m/s, U = 403 m/s, U = 454.4 m/s RMS mp av 2 0 . 280 K, 525 K, 3.157 × 102 m/s, 2.798 × 102 m/s 2 1 . R 2886 K, T = 3399 K, T = 4330 K 2 2 . 8.303 × 10–3 RMS av mp 2 4 . 58.997 cm3 2 3 . 0.199 2 5 . (a) 2.479 × 103 kPa (b) 2225.55 kPa 2 6 . 3.77 L 2 7 . 15.40 kg 2 8 . 521 K 2 9 . (i) B (ii) C (iii) A 3 0 . Pr = 2.99, Tr = 1.90 3 2 . 350.5°C 3 3 . (i) 31.1 atm (ii) 31.4 atm 3 4 . 2.53°C 3 5 . 4500 RJ
EXERCISE–04 [B] BRAIN STORMING SUBJECTIVE EXERCISE 1 . The respiration of a suspension of yeast cells was measured by determining the decrease in pressure of the gas above the cell suspension. The apparatus was arranged so that the gas was confined to a constant volume, 16 cm3 and the entire pressure change was caused by uptake of oxygen by the cells. The pressure was measured in a monometer, the fluid of which had density of 1.034 g/cm3. The entire apparatus was immersed in a thermostat at 37°C. In a 30 minute observation period the fluid in the open side of the manometer dropped 37 mm. Neglecting the solubility of oxygen in the yeast suspension, compute the rate oxygen consumption by the cells in mm3 of O2 (STP) per hour. 2 . In a basal metabolism measurement timed at 6 minutes, a patient exhaled 52.5 L of air measured over water at 20°C. The vapour pressure of water at 20°C is 17.5 torr. The barometric pressure was 750 torr. The exhaled air analyzed 16.75 volume % oxygen and the inhald air 20.32 volume % oxygen. Both on a dry basis neglecting any solubility of the gases in water and any difference in the total volumes of inhaled and exhaled air, calculate the rate of oxygen consumption by the patient in mL (STP) per minute. 3 . The temperature and the relative humidity of air are 20°C and 80% on a certain day. Find the fraction of the mass of water vapour that will condense if the temperature falls to 5°C. Saturation vapour pressures at 20°C and 5°C are 17.5 mm and 6.5 mm of Hg respectively. 4 . 6.0 g of He having average velocity 4 × 102 ms–1 is mixed with 12.0 g of Ne20 having the same average velocity. What is the average kinetic energy per mole in the mixture? 5 . Molar volume of He at 10.1325 MPa and 273 K is 0.011075 times its molar volume at 101.325 kPa. Calculate radius of He atom assuming negligible 'a'. 6 . The Viral equation for ethane gas is given by PV = RT + BP. At 0°C, B = –0.1814 L/mol. Calculate volume of one mole of ethane at 10 atm, and 'a'. 7 . Pressure of He gas confined in a steel chamber drops from 4.0 to 1.0 atmosphere in 4.0 hours due to diffusion through a pin-hole in the steel chamber. If an equimolar mixture of He and methane gas at 20 atmosphere and the same temperature are confined in the same chamber, what will be the parital pressure of He and methane after 1.0 hour. Assume rate of diffusion to be linear function of gas pressure and inverse function of square root of molar masses. 8 . A one litre flask containing NH (g) at 2.0 atmoshpere and 300 K is connected to another 800 mL flask 3 containing HCl (g) at 8.0 atmosphere and 300 K by means of a narrow tube of negligible volume and gases were allowed to react quantitatively as : NH (g) + HCl (g) NH Cl (s) ; H = – 43 kJ/mol 3 4 If heat capacity of HCl (g) C is 20 JK–1 mol–1, determine final pressure inside the flask assuming negligible V heat capacity of flask and negligible volume of solid NH Cl. 4 9 . Calculate the value of , Z1 and Z11 for nitrogen molecules at 25°C and at pressure of 10–3 mm Hg. Given that b for nitrogen is 39.1 cm3 mol–1. 1 0 . The mean free path of the molecule of a certain gas at 300 K is 2.6 × 10–5 m. The collision diameter of the molecule is 0.26 nm. Calculate (a) pressure of the gas, and (b) number of molecules per unit volume of the gas. 1 1 . There are two vessels of same consisting same no of moles of two different gases at same temperature. One of the gas is CH4 & the other is unknown X. Assuming that all the molecules of X are under random motion whereas in CH4 except one all are stationary. Calculate Z1 for X in terms of Z1 of CH4. Given that the 1 collision diameter for both gases are same & (Urms)x = 6 (Uav )CH4 .
1 2 . A mixture of CH4 & O2 is used as an optimal fuel if O2 is present in thrice the amount required theoretically for combustion of CH4. Calculate number of effusions steps required to convert a mixture containing 1 part of CH4 in 193 parts mixture (parts by volume). If calorific value (heat evolved when 1 mole is burnt) of CH4 is 100 cal/mole & if after each effusion 90% of CH4 is collected. find out what initial mole of each gas in initial mixture required for producing 1000 cal of energy after processing. [Given (0.9)5 = 0.6] 1 3 . A closed vessel of known volume containing known amount of ideal gaseous substance 'A' was observed for variation of pressure with temperature. The expected graph was to be like as in (i) However actual observations revealed the graph to be like. (ii) The deviation was attributed to polymerisation of gas molecules an nA (g) An(g). If it is known that the above reaction given only 50% yield. (a) Calculate the ratio of n experiment (where nexp. = Total no. of gaseous mole actually present n theoritical ntheoritical = Total no. of mole original taken) (b) Find the value of n to which the gas A is being polymerised into 1.54atm 3.2atm 1.25atm 2.2875atm (T) (T+10) (T) (T+10) P(atm) P(atm) T(°C) T(°C) (i) (ii) Expected Actual 1 4 . During one of his adventure, Chacha chaudhary got trapped in an underground cave which was sealed two hundred year back. The air inside the cave was poisonous, having some amount of carbon monoxide in addition to O2 and N2. Sabu, being huge could not enter into the cave, so in order to save chacha choudhary be started sucking the poisonous air out of the cave by mouth. Each time, he filled his lunge with cave air and exhaled it out in the surroundings. In the mean time fresh air from surrounding effused into the cave till the pressure was again one atmosphere. Each time sabu sucked out some air, the pressure in the cave dropped to half of its initial value of one atmosphere. If the initial sample of air from the cave contain 5% by volume CO. If the safe level of CO in the atmosphere is less than 0.001% by volume how many times does Sabu need to such out air in order to save Chacha choudhary. 1 5 . A closed vertical cylinder is divided into two parts by a frictionless piston, each part contains 1 mole of air. At 27°C the volume of the upper part is 4 times than that of the lower part. Calculate the temperature when volume of the upper part will be three times that of the lower part. 1 6 . A water gas mixture has the composition by volume of 50% H2, 40% CO and 5% CO2. (i) Calculate the volume in litres at STP of the mixture which on treatment with excess steam will contain 5 litres of H2. The stoichiometry for the water gas shift reaction is CO + H2O CO2 + H2 (ii) Find the density of the water gas mixture in kg/m3. (iii) Calculate the moles of the absorbants KOH, Ca(OH)2 and ethanolamine. HO – CH2 – CH2 – NH2 required respectively to collect the CO2 gas obtained. 1 7 . A gas present in a container connected to frictionless, weightless piston operating always at one atmosphere pressure such that it permits flow of gas outside (with no adding of gas). The graph of n vs T (Kelvin) was plotted & was found to be a straight line with co-ordinates of extremen points as (300, 2) & (200, 3). Calculate : (i) relationship between n & T (ii) relationship between V & T (iii) Maxima or minima value of ' V '
1 8 . Find the critical constant (Pc , Vc and Tc) in terms of A and B, also find compressibility factor (z) for the following equation of state. A 2B PV = RT – V V 2 where A and B are constant, P = pressure and V = molar volume. 19. Calculate the volume occupied by 14.0 g N2 at 200 K and 8.21 atm prressure if PC VC 3 and Pr Vr =2.2. RTC 8 Tt BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) 1 . 100 mm3/hr 2 . 280 mL/min 3 . 0.51 4 . 807.84 J 5 . 134 pm 6 . 0.918, a = 3.77 bar L2 mol–27 . PHe = 7.07 atm, PCH4 8.4 atm 8 . 10.3 atmosphere 9 . 314 pm, 7.015 cm, 6742 s–1, 1.09 × 1017 cm–3 s–1 1 0 . (a) 1.281 × 1023 m–3 (b) 5.306 × 102 Pa 22 1 1 . 3 Z1 1 2 . 10 Steps, 27.78 mol CH4, 5333.3 mol O2 1 3 . (a) 0.625, (b) 4 1 4 . 1 3 1 5 . 421.9 K 16. (i) 5.263 L, (ii) 0.7kg/m3, (iii) KOH = 0.2348 moles, Ca(OH)2 = 0.1174 moles, ethanolamine = 0.2348 moles 17. n= T 5, V R T 2 5RT, 51.3125 = 100 100 18. VC 6B , A2 , PC A3 , compressibility factor PC VC 1 1 9 . 0.825 L = T= = 108B2 = RTC 3 A C 6RB
EXERCISE–05 (A) PREVIOUS YEARS QUESTIONS 1 . The no. of moles per litre in the equation PV = nRT is expressed by : [AIEEE-02] P PV RT (4) None (1) (2) (3) PV RT RT 2 . The correct value of R is - [AIEEE-02] (1) R = 0.082 litre atm (2) R = 8.314 × 107 erg –K–1 mol–1 (3) R = 2K–1 mol–1 (4) None 3 . According to the kinetic theory of gases, in an ideal gas, between two successive collisions a gas molecule travels - [AIEEE-03] (1) In a straight line path (2) with an accelerated velocity (3) In a circular path (4) In a wavy path 4 . What volume of hydrogen gas, at 273K and 1 atm, pressure will be consumed in obtaining 21.6g of elemental boron (atomic mass = 10.8) from the reduction of boron trichloride by hydrogen ? [AIEEE-03] (1) 44.8 L (2) 22.4 L (3) 89.6 L (4) 67.2 L 5 . As the temperature is raised from 20°C to 40°C, the average kinetic energy of neon atoms changes by factor of which of the following ? [AIEEE-04] (1) 1/2 (2) (313 / 293) (3) 313/293 (4) 2 6 . In vanderwaals equation of state of the gas law, the constant 'b' is a measure of : [AIEEE-04] (1) intermolecular repulsions (2) intermolecular attraction (3) volume occupied by the molecules (4) intermolecular collisions per unit volume 7 . An ideal gas expands in volume from 1 × 10–3 m3 to 1 × 10–2 m3 at 300 K against a constant pressure of 1 × 105 Nm–2. The work done is - [AIEEE-04] (1) – 900 J (2) – 900 K (3) 2710 kJ (4) 900 kJ 8 . For which of the following parameters the structural isomers C2H5OH and CH3OCH3 would be expected to have the same values ? (Assume ideal behavior) (1) Heat of vaporization [AIEEE-05] (2) vapour pressure at the same temperature (3) Boiling points (4) Gaseous densities at the same temperature and pressure 9 . Equal masses of methane and oxygen are mixed in an empty container at 25°C. The fraction of the total pressure exerted by oxygen is - [AIEEE-06] 1 273 1 1 (1) 2/3 (2) (3) 3 (4) 3 298 2 PREVIOUS YEAR QUESTIONS ANSWER KEY EXERCISE-05(A) Que. 1 2 3 4 5 6 7 8 9 Ans 1 2 1 4 3 3 1 4 3
EXERCISE–05 (B) PREVIOUS YEARS QUESTIONS 1 . A mixture of ideal gases is cooled upto liquid He temperature (4.22 K) to form an ideal solution. Is this statement true or false. Justify your answer in not more than two lines. [JEE 1996] 2 . The ratio between the r.m.s. velocity of H at 50 K and that of O at 800 K is. [JEE 1996] 22 (A) 4 (B) 2 (C) 1 (D) 1/4 3 . X ml H gas effuses through a hole in a container in 5 sec. The time taken for the effusion of the same 2 volume of the gas specified below under identical conditions is [JEE 1996] (A) 10 sec, He (B) 20 sec, O (C) 25 sec, CO (D) 55 sec, CO 2 2 4 . One mole of N O (g) at 300 K is kept in a closed container under one atm. It is heated to 600 K when 24 20% by mass of N O (g) decomposes to NO (g). The resultant pressure is. [JEE 1996] 24 2 (A) 1.2 atm (B) 2.4 atm (C) 2.0 atm (D) 1.0 atm 5 . The absolute temperature of an ideal gas is ........ to/than the average kinetic energy of the gas molecules. [JEE 1997] 6 . One way of writing the equation for state for real gases is, PV RT 1 B ......... where B is constant. V Derive an approximate expression for 'B' in terms of Vander Waals constant 'a' & 'b'. [JEE 1997] 7 . Calculate the total pressure in a 10 litre cylinder which contains 0.4 g He, 1.6 g oxygen and 1.4 g nitrogen at 27°C. Also calculate the partial pressure of He gas in the cylinder. Assume ideal behavious for gases.[JEE 1997] 8 . According to Graham's law, at a given temperature the ratio of the rates of diffusion rA of gases A and rB B is given by. [JEE 1997] (A) PA MA 1 / 2 (B) M A PA 1 / 2 (C) PA MB 1 / 2 (D) MA PB 1 / 2 PB MB M B PB PB MA MB PA 9 . An evacuated glass vessel weighs 50.0 g when empty, 148.0 g when filled with a liquid of density 0.98 g /mL and 50.5 g when filled with an ideal gas at 760 mm Hg at 300 K. Determine the molecular weight of the gas. [JEE 1998] 1 0 . Using Vander Waals equation, calculate the constant \"a\" when 2 moles of a gas confined in a 4 litre flask exerts a pressure of 11.0 atm at a temperature of 300 K. The value of \"b\" is 0.05 litre mol–1.[JEE 1998] 1 1 . The pressure exerted by 12 g of an ideal gas at temperature t°C in a vessel of volume V is one atm. When the temperature is increased by 10 degrees at the same volume, the pressure increases by 10%. Calculate the temperature 't' and volume 'V'. [molecular weight of gas = 120] [JEE 1999] 1 2 . One mole of N gas at 0.8 atm takes 38 sec to diffuse through a pin hole, whereas one mole of an unknown 2 compound of Xenon with F at 1.6 atm takes 57 sec to diffuse through the same hole. Calculate the molecular formula of the compound. [JEE 1999] 1 3 . A gas will approach ideal behaviour at : [JEE 1999] (A) low temperature and low pressure (B) low temperature and high pressure (C) low pressure and high temperature (D) high temperature and high pressure 1 4 . The compressibility of a gas is less than unity at STP. Therefore : [JEE 2000] (A) V > 22.4 L (B) V < 22.4 L (C) V = 22.4 L (D) V = 44.8 L m m m m
1 5 . The r.m.s. velocity of hydrogen is 7 times the r.m.s. velocity of nitrogen. If T is the temperature of the gas : [JEE 2000] (A) T(H ) = T(N ) (B) T(H ) > T(N ) (C) T(H ) < T(N ) (D) T(H ) = 7 T(N ) 22 22 22 2 2 1 6 . The pressure of a fixed amount of an ideal gas is proportional to its temperature. Frequency of collision and their impact both increase in proportion to the square root of temperature. True/False.[JEE 2000] 1 7 . Calculate the pressure exerted by one mole of CO gas at 273 K, if the Vander Waals constant a = 3.592 2 dm6 atm mol–2. Assume that the volume occupied by CO molecules is negligible. [JEE 2000] 2 1 8 . The root mean square velocity of an ideal gas at constant pressure varies with density as : [JEE 2001] (A) d2 (B) d (C) d1/2 (D) 1/d1/2 1 9 . The compression factor (compressibility factor) for one mole of a vander Waals gas at 0°C and 100 at- mosphere pressure is found to be 0.5. Assuming that the volume of a gas molecule is negligible, calculate the vander waals constant 'a'. [JEE 2001] 2 0 . Which one of the following V, T plots represents the behaviour of one mole of an ideal gas at one atm? [JEE 2002] 38.8 L 28.8 L 30.6 L 14.2 L V(L) 373 K V(L) 373 K V(L) 373 K V(L) 373 K 22.4 L 20.4 L 22.4 L 22.4 L (A) 273 K (B) 273 K (C) 273 K (D) 273 K T(K) T(K) T(K) T(K) 2 1 . The density of the vapour of a substance at 1 atm pressure and 500 K is 0.36 Kg m–3. The vapour effuse through a small hole at a rate of 1.33 times faster than oxygen under the same condition.[JEE 2002] (a) Determine (i) mol. wt. ; (ii) molar volume ; (iii) compressibility factor (z) of the vapour (iv) which forces among the gas molecules are dominating the attractive or the repulsive. (b) If the vapour behaves ideally at 1000 K, determine the average translational K.E. of a molecule. 2 2 . The average velocity of gas molecules is 400 m/sec. Calculate its (rms) velocity at the same temperature.[JEE 2003] 2 3 . C value of He is always 3R/2 but C value of H is 3R/2 at low temperature and 5R/2 at moderate v v2 temperature and more than 5R/2 at higher temperature explain in two to three lines. [JEE 2003] 2 4 . Positive deviation from ideal behaviour takes place because of. [JEE 2003] (A) molecular interaction between atoms and PV >1 nRT (B) molecular interation between atoms and PV <1 nRT (C) finite size of atoms and PV >1 nRT (D) finite size of atoms and PV <1 nRT 2 5 . For a real gas obeying vander Waal's equation a graph is plotted between PV (y-axis) and P(x-axis) where m V is molar volume. Find y-intercept of the graph. [JEE 2005] m 2 6 . The ratio of the rate of diffusion of helium and methane under identical condition of pressure and tem- perature will be : [JEE 2005] (A) 4 (B) 2 (C) 1 (D) 0.5
CA Z Ideal gas where Z = PV , 27. nRT B P a = Vander Waal's constant for pressure correction b = Vander Waal's constant for volume correction Pick the only incorrect statement (A) for gas A, if a = 0, the compressibility factor is directly proportional to pressure (B) for gas B, if b = 0, the compressibility factor is directly proportional to pressure (C) for gas C, a 0, b 0, it can be used to calculate a and b by giving lowest P value and its intercept with Z = 1. (D) slope for all three gases at high pressure (not shown in graph) is positive. [JEE 2006] PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE -5(B) 1 . Yest it is false statement 2. C 3. B 4. B 5 . directly proportional 6. B = b – a 7 . 0.492 atm, 0.246 atm 8. C RT 1 0 . 6.46 atm L2 mol–2 1 1 . –173°C, 0.82 L 9 . 123 1 2 . XeF 13. C 14. B 15. C 6 1 7 . 0.99 atm 18. D 1 6 . both statements are correct 1 9 . 1.2544 atm L2 mol–2 20. C 2 1 . (a) (i) 18.1 g/mol (ii) 50.25 L mol–1 (iii) 1.224 (iv) repulsive (b) 2.07 × 10–20 J 2 2 . 434.17 m/sec 23. Since H is diatomic and He is monoatomic degree of freedom for mono is 3 and only translational but for 2 diatomic, vibrational and rotational are also to be considered. 24. C 25. RT 26. B 27. B
EXERCISE-01 CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1 . Which of the following factors affects the adsorption of a gas on solid ? (A) T (critical temp.) (B) Temperature of gas (C) Pressure of gas (D) All of them c 2 . The volume of gases NH , CO and CH adsorbed by one gram of charcoal at 298 K are in 32 4 (A) CH > CO > NH (B) NH > CH > CO (C) NH > CO > CH (D) CO > NH > CH 423 342 324 234 3 . The heat of physisorption lie in the range of (A) 1 to 10 kJ mol–1 (B) 20 to 40 kJ mol–1 (D) 40 to 200 kJ mol–1 (D) 200 to 400 kJ mol–1 4 . Which of the following is not a gel ? (A) Cheese (B) Jellies (C) Curd (D) Milk 5 . Which of the following is used to adsorb water (A) Silica gel (B) Calcium acetate (C) Hair gel (D) Cheese 6 . An emulsion is a colloidal system of (A) two solids (B) two liquids (C) one gas and one solid (D) one gas and one liquid 7 . Which of the following is a lyophobic colloid ? (A) Gelatin (B) Sulphur (C) Starch (D) Gum arabic 8 . The nature of bonding forces in adsorption (A) purely physical such as Vander Waal's forces (B) purely chemical (C) both chemical and physical always (D) none of these 9 . The Tyndall effect associated with colloidal particles is due to (A) presence of electrical charges (B) scattering of light (C) adsorption of light (D) reflection of light 1 0 . Which one of the following is not applicable to chemisorption ? (A) Its heat of adsorption is high (B) It takes place at high temperature (C) It is reversible (D) It forms mono-molecular layers 1 1 . In the colloidal state the particle size ranges (A) below 1 nm (B) between 1 nm to 100 nm (C) more than 100 nm (D) none of the above 1 2 . All colloids (A) are suspensions of one phase in another (B) are two phase systems (C) contain only water soluble particles (D) are true solutions 1 3 . Colloids can be purified by (A) condensation (B) peptization (C) coagulation (D) dialysis 1 4 . Milk is an example of (A) emulsion (B) suspension (C) foam (D) sol 1 5 . Fog is a colloidal system of (A) gas in liquid (B) liquid in gas (C) gas in gas (D) gas in solid
1 6 . When a colloidal solution is observed under ultramicroscope, we can see (A) light scattered by colloidal particle (B) size of the colloidal particle (C) shape of the colloidal particle (D) relative size of the colloidal particle 1 7 . Colloidal solutions are classified on the basis of (A) molecular size (B) organic or inorganic (C) surface tension value (D) pH value 1 8 . The electrical charge on a colloidal particle is indicated by (A) Brownian movement (B) electrophoresis (C) ultra microscope (D) molecular sieves 1 9 . Crystalloids differ from colloids mainly in respect of (A) electrical behaviour (B) particle nature (C) particle size (D) solubility 2 0 . Which one of the following is lyophillic colloid? (A) Milk (B) Gum (C) Fog (D) Blood 2 1 . Small liquid droplets dispersed in another liquid is called (A) suspension (B) emulsion (C) gel (D) true solution 2 2 . The process which is catalysed by one of the product is called (A) acid-base catalysis (B) autocatalysis (C) negative catalysis (D) homogeneous catalysis 2 3 . Tyndall effect would be observed in a (A) solution (B) solvent (C) precipitate (D) colloidal sol 2 4 . Adsorption is multilayer in case of (A) physical adsorption (B) chemisorption (C) in both (D) none of these 2 5 . A liquid is found to scatter a beam of light but leaves no residue when passed through the filter paper. The liquid can be described as (A) a suspension (B) oil (C) a colloidal sol (D) a true solution 2 6 . The ability of an ion to bring about coagulation of a given colloid depends upon (A) its charge (B) the sign of the charge alone (C) the magnitude of the charge (D) both magnitude and sign of charge 2 7 . An emulsifier is a substance (A) which stabilises an emulsion (B) which breaks the emulsion into its constituent liquids (C) which can convert liquid into an emulsion (D) which bring about coagulation of an emulsion CHECK YOUR GRASP ANSWER KEY EXERCISE -1 Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. D C B D A B B D B C B B D A B Que. 16 17 18 19 20 21 22 23 24 25 26 27 Ans. A A B C B B B D A C D A
EXERCISE–02 BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THEN ONE CORRECT ANSWERS) 1 . Which gas will be adsorbed on a solid to greater extent. (A) A gas having non polar molecule (B) A gas having highest critical temperature (T ) (C) A gas having lowest critical temperature. c (D) A gas having highest critical pressure. 2 . Colloidal particles in a sol. can be coagulated by (A) heating (B) adding an electrolyte (C) adding oppositely charged sol (D) any of the above methods 3 . Emulsifier is an agent which (A) accelerates the dispersion (B) homogenizes an emulsion (C) stabilizes an emulsion (D) aids the flocculation of an emulsion 4 . Given below are a few electrolytes, indicate which one among them will bring about the coagulation of a gold sol. quickest and in the least of concentration? (A) NaCl (B) MgSO (C) Al (SO ) (D) K [Fe(CN) ] 4 2 43 46 5 . The minimum concentration of an electrolyte required to cause coagulation of a sol is called (A) flocculation value (B) gold number (C) protective value (D) none of these 6 . Smoke precipitator works on the principle of (A) distribution law (B) neutralization of charge on colloids (C) Le-Chaterlier's principle (D) addition of electrolytes 7 . Which one of following statements is not correct in respect of lyophilic sols ? (A) There is a considerable interaction between the dispersed phase and dispersion medium (B) These are quite stable and are not easily coagulated (C) They carry charge (D) The particle are hydrated 8 . As S sol is (B) negative colloid (C) neutral colloid (D) none of the above 23 (A) positive colloid 9 . Which of the following electrolyte will be most effective in coagulation of gold sol ? (A) NaNO (B) K [Fe(CN) ] (C) Na PO (D) MgCl 3 46 34 2 1 0 . At the critical micelle concentration (CMC) the surfactant molecules (A) decompose (B) dissociate (C) associate (D) become completely soluble 1 1 . Alums purify muddy water by (A) dialysis (B) absorption (C) coagulation (D) forming true solution 1 2 . Solute dispersed in ethanol is called (A) emulsion (B) micelle (C) hydrophilic sol. (D) alcosols 1 3 . An arsenious sulphide sol. carries a negative charge. The maximum precipitating power of this sol. is possessed by (A) K SO (B) CaCl (C) Na PO (D) AlCl 24 2 34 3 1 4 . Reversible adsorption is (A) chemical adsorption (B) physical adsorption (C) both (D) none
1 5 . The function of gum arabic in the preparation of Indian ink is (A) coagulation (B) peptisation (C) protective action (D) adsorption 1 6 . Which of the following is an example of associated colloid? (A) Protein + water (B) Soap + water (C) Rubber + benzene (D) As O + Fe(OH) 3 23 1 7 . Adsorption of gases on solid surface is generally exothermic because (A) enthalpy is positive (B) entropy decreases (C) entropy increases (D) free energy increases 1 8 . Which of the following is/are correct statements (A) Hardy Schulz rule is related to coagulation (B) Brownian movement and Tyndall effect are shown by colloids (C) When liquid is dispersed in liquid, it is called gel. (D) Gold number is a measure of protective power of lyophillic colloid. 1 9 . Which statement is/are correct ? (A) Physical adsorption is multilayer non-directional and non-specific (B) Chemical adsorption is generally monolayer and specific in nature (C) Physical adsorption is due to free valence of atoms (D) Chemical adsorption is stronger than physical adsorption 2 0 . Which is the following is/are correct for lyophillic sols ? (A) Its surface tension is lower than that of H O 2 (B) Its viscosity is higher than that of water (C) Its surface tension is higher than that of water (D) Its viscosity is equal to that of water 2 1 . Which statement(s) is/are correct (A) A solution is prepared by addition of excess of AgNO solution in KI solution. The charge likely to develop 3 on colloidal particle is positive. (B) The effects of pressure on physical adsorption is high if temperature is low. (C) Ultracentrifugation process is used for preparation of lyophobic colloids. (D) Gold number is the index for extent of gold plating done. 2 2 . Colloidal solution can be purified by (A) Dialysis (B) Electrodialysis (C) Electrophoresis (D) ultrafiltration 2 3 . Coagulation of colloids can be achieved by (A) Centrifugation (B) Adding electrolyte (C) Change in pH (D) Adding water 2 4 . When –vely charged colloid like As S sol is added to +vely charged Fe(OH) sol in suitable amounts 23 3 (A) Both the sols are precipitated simultaneously (B) This process is called mutual coagulation (C) They become +vely charged colloid (D) They become –vely charged colloid 2 5 . Which of the following is not lyophillic (A) Gelatin sol (B) Silver sol (C) Sulphur sol (D) As S sol 23 2 6 . Colloidal Gold can be prepared by (A) Bredig's are method (B) Reduction of AuCl (C) Hydrolysis (D) Peptization 3
2 7 . The coagulation of sol particles may be brought about by (A) heating (B) addition oppositely charged sol (C) addition electrolyte (D) persistent dialysis 2 8 . Which one is not lyophobic in nature ? (A) Gelatin (B) Sulphur (C) Starch (D) Protein 2 9 . Which of the following are colloids? (A) Milk (B) Ice cream (C) Urea solution (D) Blood 3 0 . Which are the properties of sols. (A) Adsorption (B) Tyndall effect (C) Flocculation (D) Paramagnetism 3 1 . The migration of colloidal particles under the influence of an electrical field is known as (A) electro osmosis (B) electrophoresis (C) electrodialysis (D) None BRAIN TEASERS ANSWER KEY EXERCISE -2 12 13 14 15 Que. 1 2 3 4 5 6789 10 11 DD BC C C A C C 27 28 29 30 Ans . B D 18 19 20 BC BD 25 26 B,D A,C,D A,B,D A,B,C A,B,D A,B,D A,B B,C,D A,B Que. 16 17 21 22 23 24 Ans . B B A,B A,B A,B,C A,B Que. 31 Ans . B
EXERCISE–03 MISCELLANEOUS TYPE QUESTIONS TRUE / FALSE 1 . Physisorption is non-specific. 2 . Chemisorption needs activation energy. 3 . A graph of x/m vs temperature at constant pressure is called adsorption isotherm. 4 . Suspensions have solute particles with size less than 1 nm. 5 . Fe(OH) sol contains positively charged colloidal particles. 3 6 . Chemisorption is irreversible. 7 . Adsorption isobars of chemisorption and physisorption are of the same type. 8 . Milk is an example of water in oil emulsions. 9 . Gold sol can be prepared by Bredig's arc method. 1 0 . Gel is a system in which liquid is the dispersed phase and solid is the dispersion medium. FILL IN THE BLANKS 1 . The substance on whose surface adsorption takes place is called an ...................... . 2 . Removal of adsorbate from the surface of adsorbent is called ...................... . 3 . Migration of colloidal particles under the effect of electric field is called ...................... . 4 . The heat of adsorption in case of physisorption is approximately ...................... . 5 . The phenomenon of zig-zag motion of colloidal particles is known as ...................... . 6 . Lyophilic sols are ...................... stable than lyophobic sols. 7 . Electrical properties of a colloidal solution are demonstrated by ...................... . 8 . Tyndall effect takes place due to ...................... of light by colloidal particles. 9 . The liquid-liquid colloidal dispersions are called ...................... . 1 0 . The movement of dispersion medium under the influence of an electric field is called ...................... . 1 1 . Smoke is a colloidal solution of ...................... in ...................... . 1 2 . The adhering of the molecules of a gas on the surface of a solid is called ...................... . 1 3 . The protective action of different colloids is compared in term of ...................... . 1 4 . The colloidal dispersion of a liquid in a liquid is called ...................... . 1 5 . The colloidal dispersions of liquids in solid media are called ...................... . ASSERTION & REASON These questions contains, Statement I (assertion) and Statement II (reason). (A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I. (B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I (C) Statement-I is true, Statement-II is false (D) Statement-I is false, Statement-II is true 1 . Statement-I : Isoelectric point is pH at which colloidal can move towards either of electrode Because Statement-II : At isoelectric point, colloidal particles become electrically neutral. 2 . Statement-I : When AgNO is treated with excess of potassium iodide, colloidal particles gets attracted 3 towards anode. Because Statement-II : Precipitate adsorb common ions (excess) and thus become charged.
3 . Statement-I : For adsorption G, H, S all have –ve values Because Statement-II : Adsorption is a spontaneous exothermic process in which randomness decreases due to force of attraction between adsorbent and adsorbate. 4 . Statement-I : A gas with higher critical temperature gets adsorbed to more extent than a gas with lower critical temperature. Because Statement-II : The easily liquifiable gases get adsorbed to more extent. 5 . Statement-I : Micelles are formed by surfactant molecules above the critical micellar concentration (CMC). Because Statement-II : The conductivity of a solution having surfactant molecules decreases sharply at the CMC. COMPREHENSION BASED QUESTIONS Comprehension # 1 Question No. 46 to 50 (5 questions) Whenever a mixture of gases is allowed to come in contact with a particular adsorbent under the same conditions, the more strong adsorbate is adsorbed to greater extent irrespective of its amount present, e.g., H O is adsorbed to more extent on silica gel than N and O . This shows that some adsorbates are preferentially 2 22 adsorbed. It is also observed that preferentially adsorbable adsorbents can displace a weakly adsorbed substance from the surface of an adsorbent. 1 . Which of the following gases is adsorbed to maximum extent : (A) He (B) Ne (C) Ar (D) Xe 2 . Which of the gas can displace remaining all the gases (A) O (B) N (C) CO (D) H 2 2 2 3 . When temperature is increased (A) extent of adsorption increases (B) extent of adsorption decreases (C) no effect on adsorption (D) extent of adsorption first decreases, then increases 4 . Chromatogarphic separations are based on (A) differential solubility (B) differential adsorption (C) differential absorption (D) None of these 5 . Activated charcoal is prepared by (A) heated charcoal with steam so that it becomes more porous (B) addition Ca (PO ) to charcoal 3 42 (C) addition impurity to charcoal (D) reacted with conc. HNO 3 Comprehension # 2 Question No. 51 to 54 (4 questions) The clouds consist of charged particles of water dispersed in air. Some of them are +vely charged, others are negatively charged. When +vely charged clouds come closer they have cause lightening and thundering whereas when +ve and –ve charged colloids come closer they cause heavy rain by aggregation of minute particles. It is possible to cause artificial rain by throwing electrified sand or silver iodide from an aeroplane and thus coagulation the mist hanging in air.
1 . When excess of AgNO is treated with KI solution, AgI forms 3 (A) +ve charged sol (B) –vely charged sol (C) neutral sol (D) true solution 2 . Clouds are colloidol solution of (A) liquid in gas (B) gas in liquid (C) liquid in liquid (D) solid in liquid 3 . AgI helps in artificial rain because (A) It helps in coagulation (B) It helps in dispersion process (C) Both (D) None 4 . Electrical chimneys are made on the principle of (A) Electroosmosis (B) Electrophoresis (C) Coagulation (D) All of these Comprehension # 3 Question No. 55 to 58 (4 questions) In macromolecular type of colloids, the dispersed particles are themselves large molecules (usually polymers). Since these molecules have dimensions comparable to those of colloidal particles, their dispersions are called macromolecular colloids. Most lyophilic sols belong to this category. There are certain colloids which behave as normal strong electrolytes at low concentrations, but exhibit colloidal properties at higher concentrations due to the formation of aggregate particles. These are known as micelles or associated colloids. Surface active agents like soaps and synthetic detergent belong to this class. Critical micelle concentration (CMC) is the lowest concentration at which micelle formation appears. CMC increases with the total surfactant concentration. At concentration higher than CMC, they form extended parallel sheets known as lamellar micelles which resemble biological membranes. With two molecules thick, the individual molecule is perpendicular to the sheets such that hydrophilic groups are on the outside in aqueous solution and on the inside in a non-polar medium In concentrated solution, micelles take the form of long cylinders packed in hexagonal arrays and are called lytotropic mesomorphs. In an aqueous solution (polar medium), the polar group points towards the periphery and the hydrophobic hydrocarbon chains point towards the centre forming the core of the micelle. Micelles from the ionic surfactants can be formed only above a certain temperature called the Kraft temperature. They are capable of forming ions. Molecules of soaps and detergents consist of lyophilic as well as lyophobic parts which associate together to form micelles. Micelles may contains as many as 100 molecules or more. 1 . Select incorrect statement(s) : (A) Surface active agent like soaps and synthetic detergents are micelles (B) Soaps are emulsifying agents (C) C H (hydrocarbon part) and –COO– (carboxylate part) of stearate ion (C H COO–) both are hydrophobic 17 35 17 35 (D) All are incorrect statements 2 . Which part of the soap (RCOO–) dissolved grease and forms micelle ? (A) R part (called tail of the anion) (B) –COO– part (called head of the anion) (C) both (A) and (B) (D) none of these 3 . In multimolecular colloidal sols, atoms or molecules are held together by: (A) H-bonding (B) vander Waals forces (C) ionic bonding (D) polar covalent bonding
4 . Cleansing action of soap occurs because : (A) oil and grease can be absorbed into the hydrophobic centres of soap micelles and washed away (B) oil and grease can be absorbed into hydrophilic centres of soap micelles acid washed away (C) oil and grease can be absorbed into both hydrophilic and hydrophobic centres but not washed away (D) cleansing action is not related to micelles Comprehension # 4 Question No. 59 to 61 (3 questions) The protective power of the lyophilic colloids is expressed in terms of gold number a term introduced by Zsigmondy. Gold number is the number of milligram of the protective colloid which prevent the coagulation of 10 mL of red gold sol. when 1 mL of a 10 percent solution of sodium chloride is added to it. Thus, smaller the gold number of lyophilic colloid, the greater is the protective power. 1 . On addition of one mL of solution of 10% NaCl to 10 mL of red gold sol in presence of 0.025 g of starch, the coagulation is just prevented. The gold number of starch is (A) 0.025 (B) 0.25 (C) 2.5 (D) 25 2 . Which of the following statement(s) is/are correct (A) Higher the gold number, more protective power of colloid (B) Lower the gold number, more the protective power (C) Higher the coagulation value, more the coagulation power (D) Lower the coagulation value, higher the coagulation power 3 . Gold number gives an indication of (A) protective nature of colloids (B) purity of gold in suspension (C) the charge on a colloidal solution of gold (D) g-mole of gold per litre MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE -3 True / False 1. T 2. T 3. F 4. F 5. T 6. T 7. F 8. F 9. T 10. T Fill in the Blanks 1. absorbent 2. desorption 3. electrophoresis 4. 20–40 kJ-mol–1 6. more 7. electrophoresis 5. Brownian movement 10. electro-osmosis 11. solid in gas 14. emulsion 15. gel 8. scattering 9. emulsions 12.adsorption 13. gold number Assertion - Reason Questions 1. B 2. C 3. C 4. C 5. D Comprehension Based Questions C o mp re he n s i o n # 1 : 1. (D) 2. (C) 3. (B) 4. (B) 5. (A) C o mp re he n s i o n # 2 : 1. (A) 2. (A) 3. (C) 4. (B) C o mp re he n s i o n # 3 : 1. (C,D) 2. (C) 3. (B) 4. (A) C o mp re he n s i o n # 4 : 1. (D) 2. (B,D) 3. (A)
EXERCISE - 04 [A] JEE-[MAIN] : PREVIOUS YEAR QUESTIONS 1 . The volume of a colloidal particle, V as compared to the volume of a solute particle in a true solution C V , could be :- [AIEEE-2005] S (1) VC 1 (2) VC 1023 (3) VC 10 3 (4) VC 103 VS VS VS VS 2 . The disperse phase in colloidal iron (III) hydroxide and colloidal gold is positively and negatively charged, respectively. Which of the following statement is NOT correct [AIEEE-2005] (1) Magnesium chloride solution coagulates, the gold sol more readily than the iron (III) hydroxide sol (2) Sodium sulphate solution cause coagulation in both sols (3) Mixing the sols has no effect (4) Coagulation in both sols can be brought about by electrophoresis 3 . Which of the following statements is incorrect regarding physissorptions ? [AIEEE-2009] (1) Under high pressure it results into multi molecular layer on adsorbent surface (2) Enthalpy of adsorption (Hadsorption) is low and positive (3) It occurs because of Van der Waal's forces (4) More easily liquefiable gases are adsorbed readily 4 . According to Freundlich adsorption isotherm, which of the following is correct ? [AIEEE-2012] (1) x p0 (2) x p1 (3) x p1/ n m m m (4) All the above are correct for different ranges of pressure MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE -4(A) Que. 1 2 3 4 Ans 4 3 2 4
EXERCISE - 04 [B] JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS Q.1 Among the following, the surfactant that will from micelles in aqueous solution at the lowest molar concentration at ambient conditions is [JEE 2008] (A) CH3(CH2)15N+(CH3)3Br¯ (B) CH3(CH2)11OSO3¯Na+ (C) CH3(CH2)6COO¯Na+ (D) CH3(CH2)11N+(CH3)3Br¯ Q.2 Among the electrolytes Na2SO4, CaCl2, Al2(SO4)3 and NH4Cl, the most effective coagulation agent for Sb2S3 sol is [JEE 2009] (A) Na2SO4 (B) CaCl2 (C) Al2(SO4)3 (D) NH4Cl Q.3 The correct statement(s) pertaining to the adsorption of a gas on a solid surface is (are) - (A) Adsorption is always exothermic [JEE 2011] (B) Physisorption may transform into chemisorption at high temperature (C) Physisorption increases with increasing temperature but chemisorption decreases with increasing temperature (D) Chemisorption is more exothermic than physisorption, however it is very slow due to higher energy of activation Q.4 The given graphs / data I, II, III and IV represent general trends observed for different physisorption and chemisorption processes under mild conditions of temperature and pressure. Which of the following choice(s) about I, II, III and IV is (are) correct ? [JEE 20112] Amount of gas adsorbed (I) P constant Amount of gas absorbed (II) P constant TT Amount of gas adsorbed (III) 200K Potential energy (IV) 250K Distance of molecule from the surface 0 Huds=150kJmol–1 P (A) I is physisorption and II is chemisorption (B) I is physisorption and III is chemisorption (C) IV is chemisorption and II is chemisorption (D) IV is chemisorption and III is chemisorption
Q.5 Choose the correct reason(s) for the stability of the lyophobic colloidal particle. [JEE 2012] (A) Preferential adsorption of ions on their surface from the solution (B) Preferential adsorption of solvent on their surface from the solution (C) Attraction between different particles having opposite charges on their surface (D) Potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE -4(B) Q.3 A,B,D Q.1 A Q.2 C Q.4 A, C Q.5 A, D
EXERCISE-01 CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1 . Which of the following equations represents a reaction that provides the heat of formation of ethane (CH3CH3)? (A) 2 C(s) + 6 H(g) CH3CH3(g) (B) 2 C(s) + 3 H2(g) CH3CH3(g) (C) CH2 = CH2(g) + 2 H2(g) CH3CH3(g) (D) CH–CH(g) + 2 H2O(g) CH3CH3(g) + O2(g) 2 . Which of the following equations represents a reaction that provides the heat of formation of CH3Cl? (A) C(s) + HCl(g) + H2(g) CH3Cl(g) (B) C(s) + 3/2 H2(g) + 1/2 Cl2(g) CH3Cl(g) (C) C(s) + 3 H(g) + Cl(g) CH3Cl(g) (D) CH4(g) + Cl2(g) CH3Cl(g) + HCl(g) 3 . Use the given standard enthalpies of formation to determine the heat of reaction of the following reaction: TiCl4(g) + 2 H2O(g) TiO2(g) + 4 HCl(g) Hof TiCl4(g) = –763.2 kJ/mole Hof TiO2(g) = –944.7 kJ/mole Hof H2O(g) = –241.8 kJ/mole Hof HCl(g) = –92.3 kJ/mole (A) – 278.1 (B) + 369.2 (C) + 67.1 (D) – 67.1 4 . The heats of formation of CO2(g) and H2O(l) are –394 kJ/mole and –285.8 kJ/mole respectively. Using the data for the following combustion reaction, calculate the heat of formation of C H (g). 38 C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) Ho = –2221.6 kJ (A) 212.2 (B) – 143.3 (C) 185.4 (D) – 103.6 5 . The heats of formation of CO2(g) and H2O(l) are –394 kJ/mole and –285.8 kJ/mole respectively. Using the data for the following combustion reaction, calculate the heat of formation of C2H2(g). 2 C H (g) + 5 O (g) 4 CO (g) + 2 H O(l) 22 2 2 2 Ho = – 2601 kJ (A) – 238.6 (B) 253.2 (C) 238.7 (D) 226.7 6 . Using the following information calculate the heat of formation of NaOH. 2 Na(s) + 2 H2O(l) 2 NaOH(s) + H2(g) Ho = – 281.9 kJ Hof H2O(l) = –285.8 kJ/mole (A) – 141.6 (B) – 712.6 (C) – 426.8 (D) – 650.4 7 . Using the following information calculate the heat of formation of CH4. CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) Ho = – 890.4 kJ Hof CO2(g) = – 393.5 kJ/mole Hof H2O(l) = – 285.9 kJ/mole (A) – 98.6 (B) – 65.5 (C) – 74.9 (D) – 43.5
8 . The heat of formation of CO (g) is –394 kJ/mole and that of H O(l) is –286 kJ/mole. The heat of combustion 22 of C2H4 is –1412 kJ/mole. What is the heat of formation of C2H4? C H (g) + 3 O (g) 2 CO (g) + 2 H O(l) 24 2 2 2 (A) +1360 (B) – 108 (C) + 52 (D) +108 9 . What amount of heat energy (kJ) is released in the combustion of 12.0 g of C3H4? (Atomic weights: C = 12.01, H = 1.008, O = 16.00). C3H4(g) + 4 O2(g) 3 CO2(g) + 2 H2O(l) Ho = –1939.1 kJ (A) 725 (B) 504 (C) 783 (D) 581 1 0 . The standard heat of combustion of solid boron is equal to :- (A) H° (B O ) (B) 1 H ° (B O ) (C) 2 H ° (B O ) (D) 1 H° ( B O) f2 3 2 f 23 f 2 3 2 f 23 11. In the reaction, CO (g) + H (g) CO(g) + H O(g); H = 2.8 kJ H represents :- 2 2 2 (A) heat of reaction (B) heat of combustion (C) heat of formation (D) heat of solution 1 2 . A gas mixture 3.67 L in volume contain C H and CH is proportion of 2 : 1 by moles and is at 25°C 24 4 and 1 atm. If the HC (C H ) and HC (CH ) are –1400 and –900 kJ/mol find heat evolved on burning 24 4 this mixture :- (A) 20.91 kJ (B) 50.88 kJ (C) 185 kJ (D) 160 kJ 1 3 . The bond dissociation energy of gaseous H , Cl and HCl are 104, 58 and 103 kcal mol–1 respectively. 22 The enthalpy of formation for HCl gas will be :- (A) –44.0 kcal (B) –22.0 kcal (C) 22.0 kcal (D) 44.0 kcal 14. The average, S – F bond energy in SF if the H° value are –1100, +275 and +80 kJ/mol respectively 6 f for SF (g), S(g) and F(g) is :- 6 (A) 390.1 kJ/mol (B) 103.9 kJ/mol (C) 903.1 kJ/mol (D) 309.1 kJ/mol 1 5 . The average O–H bond energy in H O with the help of following data :- 2 (1) H O ( ) HO ; H = + 40.6 kJ mol–1 2 (g) 2 (2) 2H H ; H = – 435.0 kJ mol–1 (g) 2(g) (3) O2(g) 2O(g) ; H = + 489.6 kJ mol–1 (4) 2H + O 2H O() ; H = – 571.6 kJ mol–1 2(g) 2(g) 2 (A) 584. 9 kJ mol–1 (B) 279.8 kJ mol–1 (C) 462.5 kJ mol–1 (D) 925 kJ mol–1 1 6 . The enthalpy change for the following reaction is 514 kJ. Calculate the average Cl – F bond energy. ClF3(g) Cl(g) + 3 F(g) (A) 1542 (B) 88 (C) 171 (D) 514 1 7 . Given the following equations and Ho values, determine the heat of reaction (kJ) at 298 K for the reaction: 4 H2O(g) + 3 Fe(s) Fe3O4(s) + 4 H2(g) H2(g) + 1/2 O2(g) H2O(g) Ho/kJ = –285.8 Ho/kJ = –302.4 3 FeO(s) + 1/2 O2(g) Fe3O4(s) Ho/kJ = –13.8 (C) – 602.0 FeO(s) + H2(g) Fe(s) + H2O(g) (A) +391.7 (B) – 24.8 (D) +24.8
1 8 . For the following reaction : C + O CO (g) ; H = –94.3 kcal Diamond 2 2 C + O CO (g); H = –97.6 kcal Graphite 2 2 The heat required to change 1 g of C C is :- diamond graphite (A) 1.59 kcal (B) 0.1375 kcal (C) 0.55 kcal (D) 0.275 kcal 1 9 . The standard heat of combustion of Al is –837.8 kJ mol–1 at 25°C which of the following releases 250 kcal of heat :- (A) The reaction of 0.624 mol of Al (B) The formation of 0.624 mol of Al O 23 (C) The reaction of 0.312 mol of Al (D) The formation of 0.150 mol of Al O 23 20. If heat of dissociation of CHCl COOH is 0.7 kcal/mole then H for the reaction :- 2 CHCl COOH + KOH CHCl COOK + HO 2 2 2 (A) –13 kcal (B) + 13 kcal (C) –14.4 kcal (D) –13.7 kcal 2 1 . A solution is 500 mL of 2 M KOH is added to 500 mL of 2 M HCl and the mixture is well shaken. The rise in temperature T1 is noted. The experiment is then repeated using 250 mL of each solution and rise in temperature T is again noted. Assume all heat is taken by the solution :- 2 (A) T = T (B) T is 2 times as large as T 12 12 (C) T is twice of T (D) T is 4 times as large as T 21 12 2 2 . Anhydrous AlCl is a covalent compound. From the data given below, predict whether it would remain 3 covalent or become ionic in an aqueous solution . (Ionisation energy of Al = 5137 kJ mol–1), H hydration for Al3+ = –4665 kJ mol–1. H hydration for Cl– = –381 kJ mol–1 (A) Ionic (B) Covalent (C) Partially ionic (D) Partially covalent 2 3 . Given, H (g) + Br (g) 2HBr(g), H° and standard enthalpy of condensation of bromine is H° , standard 22 1 2 enthalpy of formation of HBr at 25°C is :- (A) H1 (B) H 1 H 1 (D) (H1 H2 ) 2 2 H2 (C) 2 H2 2 2 4 . From the following data of H, of the following reaction, 1 C(s) +2 O (g) CO(g) H = –110 kJ 2 C(s) + H O(g) CO(g) + H (g) H = 132 kJ 2 2 What is the mole composition of the mixture of steam and oxygen on being passed over coke at 1273K, keeping temperature constant :- (A) 0.5 : 1 (B) 0.6 : 1 (C) 0.8 : 1 (D) 1 : 1 CHECK YOUR GRASP ANSWER KEY EXERCISE -1 Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. B B D D D C C C D B A C B D C Que. 16 17 18 19 20 21 22 23 24 Ans. C D D B A A A D B
EXERCISE–02 BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THEN ONE CORRECT ANSWERS) 1 . The standard enthalpy of formation of propene, C H , is +20.6 kJ/mole. Calculate the heat of combustion 36 of one mole of C3H6. The heats of formation of CO2(g) and H2O(l) are –394 kJ/mole and –285.8 kJ/mole respectively. (A) 1721.2 (B) –1939.1 (C) 2060.0 (D) 2221.6 2 . The fat, glyceryl trioleate, is metabolized via the following reaction. Given the enthalpies of formation, calculate the energy (kJ) liberated when 1.00 g of this fat reacts. (Atomic weights: C = 12.01, H = 1.008, O = 16.00). C57H104O6(s) + 80 O2(g) 57 CO2(g) + 52 H2O(l) Ho C57H107O6 = –70870 kJ/mole Ho H O(l) = –285.8 kJ/mole 2 Ho CO2(g) = –393.5 kJ/mole (A) 40.4 (B) 33.4 (C) 37.8 (D) 42.6 3 . Using the enthalpies of formation, calculate the energy (kJ) released when 3.00 g of NH reacts according 3(g) to the following equation. (Atomic weights: B = 10.81, O = 16.00, H = 1.008). 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) Ho NH3(g) = –46.1 kJ/mole Ho NO(g) = +90.2 kJ/mole Ho H2O(g) = –241.8 kJ/mole (A) 34.3 (B) 30.8 (C) 39.9 (D) 42.6 4 . A sheet of 15.0 g of gold at 25.0oC is placed on a 30.0 g sheet of copper at 45.0oC. What is the final temperature of the two metals assuming that no heat is lost to the surroundings. The specific heats of gold and copper are 0.129 J/goC and 0.385 J/goC respectively. (A) 42.1 (B) 40.1 (C) 41.1 (D) 43.4 5 . Calculate the heat of combustion (kJ) of propane, C H using the listed standard enthapy of reaction data: 38 C H (g) + 5 O (g) 3 CO (g) + 4 H O(g) 38 2 22 3 C(s) + 4 H2(g) C3H8(g) Ho/kJ = – 103.8 C(s) + O2(g) CO2(g) Ho/kJ = – 393.5 H (g) + 1/2 O (g) H O(g) Ho/kJ = – 241.8 22 2 (A) – 2043.9 (B) – 1532.9 (C) – 1021.9 (D) –739.1 6 . Calculate the value of Ho/kJ for the following reaction using the listed thermochemical equations: 2 C(s) + H2(g) C2H2(g) 2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l) Ho/kJ = –2600 kJ Ho/kJ = –390 kJ C(s) + O2(g) CO2(g) Ho/kJ = –572 kJ (C) +202 2 H2(g) + O2(g) 2 H2O(l) (A) +184 (B) +214 (D) +234
7 . Use the given standard enthalpies of formation to determine the heat of reaction of the following reaction: C H OH(l) + 3 O (g) 2 CO (g) + 3 H O(g) 25 2 2 2 Hof C H OH(l) = – 277.7 kJ/mole 25 Hof CO (g) = – 393.5 kJ/mole 2 Hof H O(g) = – 241.8 kJ/mole 2 (A) – 1456.3 (B) – 1234.7 (C) – 1034.0 (D) – 1119.4 8 . Calculate Ho/kJ for the following reaction using the listed standard enthapy of reaction data: 2 N2(g) + 5 O2(g) 2 N2O5(s) N2(g) + 3 O2(g) + H2(g) 2 HNO3(aq) Ho/kJ = –414.0 Ho/kJ = –86.0 N2O5(s) + H2O(l) 2 HNO3(aq) Ho/kJ = –571.6 2 H2(g) + O2(g) 2 H2O(l) (C) – 71.2 (A) – 84.4 (B) – 243.6 (D) – 121.8 9 . Determine Ho/kJ for the following reaction using the listed enthalpies of reaction: 4 CO(g) + 8 H2(g) 3 CH4(g) + CO2(g) + 2 H2O(l) C(graphite) + 1/2 O2(g) CO(g) Ho/kJ = –110.5 kJ CO(g) + 1/2 O2(g) CO2(g) Ho/kJ = –282.9 kJ H2(g) + 1/2 O2(g) H2O(l) Ho/kJ = –285.8 kJ C(graphite) + 2 H2(g) CH4(g) Ho/kJ = –74.8 kJ (A) – 622.4 (B) – 686.2 (C) – 747.5 (D) – 653.5 1 0 . Use the given standard enthalpies of formation to determine the heat of reaction of the following reaction : 2 LiOH(s) + CO (g) Li CO (s) + H O(l) 2 23 2 Hof LiOH(s) = – 487.23 kJ/mole Hof Li2CO3(s) = – 1215.6 kJ/mole Hof H O(l) = – 285.85 kJ/mole 2 Hof CO2(g) = – 393.5 kJ/mole (A) +303.4 (B) – 133.5 (C) – 198.6 (D) +198.6 1 1 . NH (g) + 3Cl (g) NCl (g) + 3HCl (g) ; –H 32 31 N (g) + 3H (g) 2NH (g) ; H 22 32 H (g) + Cl (g) 2HCl (g) ; H 22 3 The heat of formation of NCl (g) in the terms of H , H and H is ? 3 12 3 (A) H = –H + H 2 3 H (B) H = H + H2 – 3 H – f 1 223 f 12 23 (C) H = H – H2 – 3 H (D) None f 12 23 1 2 . Determine Ho of the following reaction using the listed heats of formation : 4 HNO3(l) + P4O10(s) 2 N2O5(s) + 4 HPO3(s) Hof HNO3(l) = –174.1 kJ/mole Hof N2O5(s) = –43.1 kJ/mole Hof P4O10(s) = –2984.0 kJ/mole Hof HPO3(s) = –948.5 kJ/mole (A) –176.3 (B) – 199.8 (C) +276.2 (D) – 242.4 1 3 . If x , x and x are enthalpies of H–H, O=O and O–H bonds respectively, and x is the enthalpy of vaporisation 12 3 4 of water, estimate the standard enthalpy of combustion of hydrogen. (A) x + x2 –2x + x (B) x + x2 –2x – x (C) x + x2 –x + x (D) 2x – x – x2 – x 12 34 12 34 12 3 4 1 12 4
1 4 . Use the given bond enthalpy data to estimate the Ho (kJ) for the following reaction. (C – H = 414 kJ, H – Cl = 431 kJ, Cl – Cl = 243 kJ, C – Cl = 331 kJ). CH (g) + 4 Cl (g) CCl (g) + 4 HCl(g) 4 2 4 (A) 620 (B) 330 (C) 420 (D) 105 1 5 . For which of the following change H E ? (A) H (g) + I (g) 2HI (g) (B) HCI (aq) + NaOH (aq) NaCl(aq) + H O (l) 22 2 (C) C(s) + O (g) CO (g) (D) N (g) + 3H (g) 2 NH (g) 22 22 3 1 6 . H of which of the following reactions is zero ? r (A) H (g) 2H+ (g) + 2e– (B) 2H (g) + aq 2H+ (aq) + 2e– 2 (C) 2H (g) 2H+ (g) + 2e– (D) H (g) + aq 2H+ (aq) + 2e– 2 1 7 . H0f of water is –285.8 kJ mol–1. If enthalpy of neutralisation of monoacid strong base is –57.3 kJ mol–1, H0f of OH– ion will be ? (A) –228.5 kJ mol–1 (B) 228.5 kJ mol–1 (C) 114.25 kJ mol–1 (D) –114.25 kJ mol–1 1 8 . 4 grams of sodium hydroxide pellets were dissolved in 100 cm3 of water. The temperature before adding the sodium hydroxide pellets was 25 degrees C, and after adding the pellets it was 35 degrees C. Calculate the enthalpy change in kJ/mole of the reaction (Specific heat capacity of water = 4.2 J/K/g) (A) 42 kJ/mole (B) 4.2 kJ/mole (C) 4200 kJ/mole (D) none 1 9 . 50.0 mL of 0.10 M HCl is mixed with 50.0 mL of 0.10 M NaOH. The solution temperature rises by 3.0°C. Calculate the enthalpy of neutralization per mol of HCl. (A) –2.5 × 102 kJ (B) –1.3 × 102 kJ (C) –8.4 × 101 kJ (D) –6.3 × 101 kJ 2 0 . Which of the following statements is (are) correct ? (A) the reaction between the strong acid and strong base takes place with the evolution of heat (B) Hneut. of weak acid/strong base is less than the Hneut. of strong acid/strong base (C) Hneut. of strong acid/strong base is equal to the H of formation of H2O(l) from its ions in the aqueous medium (D) H – H = Hneut.(weak acid/strong base) ioni(weak acid) neut.(strong acid/strong base) 2 1 . The standard enthalpies of formation of CO , and HCOOH(l) are –393.7 kJ/mol and –409.2 kJ/mol 2(g) respectively :- (A) –393.7 kJ/mol is the enthalpy change for the reaction C(s) + O2(g) CO2(g) (B) the enthalpy change for the reaction CO2(g) + H2(g) HCOOH(l) would be –15.5 kJ/mol (C) the enthalpy change for the reaction H2O + CO HCOOH is –409.2 kJ/mol (D) the enthalpy change for the reaction H2(g) + CO2(g) H2O(l) + CO(g) is –409.2 kJ/mol 2 2 . Ethanol can undergoes decomposition to form two sets of products ? C H OH (g) 1 C2H4(g) + H2O (g) H° = 45.54 kJ 25 2 CH3CHO(g) + H2 (g) H° = 68.91 kJ if the molar ratio of C H to CH CHO is 8 : 1 in a set of product gases, then the energy involved in the 24 3 decomposition of 1 mole of ethanol is ? (A) 65.98 kJ (B) 48.137 kJ (C) 48.46 kJ (D) 57.22 kJ
2 3 . Reactions involving gold have been of particular interest to a chemist. Consider the following reactions, Au(OH) + 4 HCl HAuCl + 3 H O, H = –28 kcal 3 42 Au(OH) + 4 HBr HAuBr + 3 H O, H = –36.8 kcal 3 42 In an experiment there was an absorption of 0.44 kcal when one mole of HAuBr was mixed with 4 moles 4 of HCl. What is the percentage conversion of HAuBr into HAuCl ? 44 (A) 0.5 % (B) 0.6 % (C) 5 % (D) 50% 2 4 . Which of the following statement is (are) correct ? (A) for an exothermic reactions, H (p ro duct s) < H (rea ct a nt s ) f f (B) H of CO (g) is same as the H° of carbon graphite f 2 comb. (C) all exothermic reactions have a free energy change negative (D) for a reaction N + O 2NO , the heat at constant pressure and the heat at constant volume 2(g) 2(g) (g) at a given temperature are same BRAIN TEASERS ANSWER KEY EXERCISE -2 Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. C C C A A D B A C B A B B C D Que. 16 17 18 19 20 21 22 23 24 Ans. D A A A A ,B,C ,D A ,B B C A ,B,D
EXERCISE–03 MISCELLANEOUS TYPE QUESTIONS TRUE / FALSE 1 . The enthalpies of elements are always taken to be zero. 2 . Heat of neutralisation of weak acid-strong base is not constant. 3 . Resonance energy is always negative. 4 . For reaction 2A(g) + B(g) 3C (g) H = -x kJ 3 Bx then for reaction 2 C(g) A(g) + 2 (g) H = 2 kJ. 5 . H°f (C, diamond) 0 6 . For a particular reaction E = H + P.V 7 . If BE (bond energy) of N N bond is x1 that of H – H bond is x2 and N – H bond is x3 then enthalpy change of the reaction is N2 + 3H2 2NH3 Hr = x1 + 3x2 – 2x3 8 . Enthalpy change is dependent on temperature and on the path adopted. 9 . Heat of hydrogenation of ethene is x1 and that of benzene is x2, hence resonance energy of benzene is (x1 – x2). FILL IN THE BLANKS 1 . The combustion of a substance is always ................... . 2 . If heat content of X is smaller than that of Y then the process X Y is ................... . 3 . C (Diamond) C (Graphite) + x kJ. The heat change in this process is called ................... . 4 . As per reaction, N2(g) + 2O2(g) 2NO2(g) –66 kJ the value of Hf of NO2 is ................... . 5 . Heats of combustion of methane, carbon and hydrogen are –212, –94, –68 kcal mol–1. The heat of formation of CH4 is ................... . 6 . The heat of neutralisation of 1 mole of HClO4 with 1 gm-equivalent of NaOH in aqueous solutions is ................... kJ mol–1. 7 . The heat of combustion of graphite and carbon monoxide respectively are –393.5 kJ mol–1 and –283 kJ mol–1. Thus heat of formation of carbon monoxide in kJ mol–1 is ................... . 8 . For the reaction, C H (g) + 5O (g) 3CO (g) + 4H O() 38 2 2 2 at constant temperature, H – U is ................... . MATCH THE COLUMN 1 . Match the reaction (In Column I) with relation between H and E (in Column II) : Column-I Column-II (A) C(s) + O (g) CO (g) (p) H = E + RT 22 (q) H = E (r) H = E – 2RT (B) N (g) + 3H (g) 2N H ( g ) (s) H = E + 2RT 2 2 (t) H = E – RT 3 (C) NH HS(s) NH (g ) + H S(g) 4 3 2 (D) PCl (g) PC l (g) + Cl 5 2 3 (E) 2SO (g) + O (g) 2 S O (g ) 2 2 3
2 . List equation/law (in Column I) with statement (in Column II) : Column-I Column-II (A) Arrhenius equation (p) Variation of enthalpy of a reaction with (B) Kirchhoff equation temperature (C) Second law of thermodynamics (q) Variation of rate constant with temperature (r) Entropy of an isolated system tends to (D) Hess's law of constant heat summation increase and reach a maximum value (s) Enthalpy change in a reaction is always constant and independent of the manner in which the reaction occurs. 3 . Column-I Column-II (p) Heat of solution (A) S(g) + O (g) SO2(g) ; H (q) Heat of neutralisation 2 (r) Heat of formation (B) CH (g) + 2O (g) C O ( g ) 4 2 2 + 2H O(l) ; H 2 (C) NaOH(s) + aq NaOH (aq) ; H (D) NaOH(aq) + HCl(aq) NaCl (aq) (s) Heat of combustion + H O(l) ; H 2 ASSERTION & REASON These questions contains, Statement I (assertion) and Statement II (reason). (A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I. (B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I (C) Statement-I is true, Statement-II is false (D) Statement-I is false, Statement-II is true 1 . Statement-I : Enthalpy of neutralization of CH3COOH by NaOH is less than that of HCl by NaOH. Because Statement-II : Enthalpy of neutralization of CH3COOH is less because of the absorption of heat in the ionization process. 2 . Statement-I : Enthalpy changes are positive when Na SO .10H O, CuSO .5H O and salts like NaCl, 24 2 42 KCl, etc., which do not form hydrates are dissolved in water. But enthalpy changes are negative when anhydrous salts capable of forming hydrates are dissolved in water. Because Statement-II : The difference in the behaviour is due to large differences in the molecular weight of hydrated and anhydrous salts. The substances with larger molecular weights usually show positive enthalpy changes on dissolution. 3 . Statement-I : Heat of neutralisation of HF (aq.), a weak acid, with NaOH (aq.) is more than 13.7 kcal, in an exothermic reaction. Because Statement-II : Some heat is lost in the ionisation of a weak acid.
4 . Statement-I : Enthalpy of formation of HCl is equal to bond energy of HCl. Because Statement-II : Enthalpy of formation and bond energy both involve the formation of one mole of HCl from the elements. 5 . Statement-I : Enthalpy of atomization is the heat of reaction H2O(l) H2O(g). Because Statement-II : Gaseous molecules are far apart of each other due to less attraction. 6 . Statement-I : The enthalpy of formation of H O(l) is greater than that of H O(g). 22 Because Statement-II : Enthalpy change is negative for the condensation reaction, H O(g) H O(l). 2 2 7 . Statement-I : As temperature increases, heat of reaction also increases for exothermic as well as for endothermic reactions. Because Statement-II : H varies with temperature as given by H2 (at T) = H1 (at T) + C ( T 2 – T) 2 1 1 P 8 . Statement-I : Heat of combustion is always negative. Because Statement-II : Heat of combustion is used to calculate of fuels. COMPREHENSION BASED QUESTIONS Comprehension # 1 The industrial preparation of a polymer, PTFE, is based on the synthesis of the monomer CF2 = CF2, which is produced according to reaction (i) below : 2CHClF2(g) CF2= CF2(g) + 2HCl ........(i) The monomer CF2= CF2 is also obtained by reaction (ii) below : .......(ii) 2CHF3(g) CF2= CF2(g) + 2HF(g) ; H = 198.1kJ/mol Consider the information below to answer the questions: Compound Hf MoleculeX–X (X–X) HCl(g) –92.3 F–F 154.7 Cl–Cl 246.7 CHClF2(g) –485.2 CF2= CF2(g) –658.3 CF4 –679.6 CCl4 –106.6 1 . The enthalpy change for reaction (i). (A) 100.23 kJ/mol (B) 127.5 kJ/mol (C) –127.5 kJ/mol (D) –100.23 kJ/mol
2 . (i) Use the expressions CX4(g) C(s) + 2X2(g) H= –Hf C(s) C(g) ; H=718 kJ/mol and 2X2(g) 4X(g) ; H = 2D(X – X) where X = F,Cl, to the enthalpy change for the two processes CX4(g) C(g) + 4X(g) What is the average C–X bond energies for the species CX4(g) (where X = F, Cl). (A) 329.5 kJ/mol, 426.75 kJ/mol (B) 426.75 kJ/mol, 329.5 kJ/mol (C) –329.5 kJ/mol, –426.75 kJ/mol (D) –426.75 kJ/mol, –329.5 kJ/mol (ii) Given that the C–H bond energy is 416.1 kJ/mol, the order of relative chemical reactivities of C–H, C–F, and C–Cl bonds. (A) C – H > C – Cl > C – F (B) C – F > C – Cl > C – H (C) C – Cl > C – H > C – F (D) C – Cl > C – F > C – H Comprehension # 2 The hydration enthalpy of anhydrous copper (II) sulphate is defined as the heat absorbed or evolved when one mole of anhydrous solid is converted to one mole of crystalline hydrated solid. CuSO4(s) + 5H2O(l) CuSO4.5H2O(s) It cannot be measured directly. In an experiment to determine the hydration enthalpy indirectly, 4 g of the anhydrous solid were added to 50 g of water and the rise in temperature was 8 degrees. When 4 g of the hydrated solid was added to 50 g of water the fall in temperature was 1.3 degrees. 1 . What is the heat produced when 4 g of anhydrous solid is added to 50 g of water. (A) 400 kJ (B) 1672 kJ (C) 200 kJ (D) 836 kJ 2 . What is the enthalpy of solution of anhydrous copper (II) sulphate in kJ/mol. (A) 69.9472 kJ/mol (B) 4054.375 kJ/mol (C) 139.8948 kJ/mol (D) 8108.750 kJ/mol 3 . Given that the enthalpy of solution of the hydrated copper (II) sulphate is +11.3 kJ/mol, what is the enthalpy of hydration of the anhydrous solid. (A) 97.321 kJ/mol (B) –97.321 kJ/mol (C) –77.971 kJ/mol (D) 77.971 kJ/mol Comprehension # 3 The Born Haber cycle below represents the energy changes occurring at 298K when KH is formed from its elements 2K(s) + H2(g) H 2KH(s) vx 2K(g) + 2H(g) z wy 2K+(g) + 2H–(g) Hatomisation K = 90 kJ/mol Hionisation K = 418 kJ/mol Hdissociation H = 436 kJ/mol Helectron affinity H = –78 kJ/mol Hlattice KH = –710 kJ/mol
1 . In terms of the letters v to z write down expressions for (i) H for the reaction 2K(s) + H2(g) 2KH(s) (ii) Hi of K (iii) Helectron affinity of H (iv) Hlattice of KH(s) 2 . Which of v to y is (i) The most exothermic ...................... . (ii) The most endothermic ...................... . 3 . Calculate the value of H showing all your working. 4 . Calculate the Hf of KH(s). 5 . Write a balanced equation for the reaction of KH with water. 6 . On complete reaction with water, 0.1 g of KH gave a solution requiring 25 cm3 of 0.1M HCl for neutralisation. Calculate the relative atomic mass of potassium from this information. MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE -3 True / False 1. F 2. T 3. T 4. T 5. T 6. F 7. F 8. F 9. F Fill in the Blanks 1. exothermic 2. endothermic 3. heat of transition 4. +33 kJ mol–1 7. –110.5 8. –3 RT 5. –18 kcal mol–1 6. –57.2 Match the Column 1. (A) - q, (B) - r, (C) - s, (D) - p, (E) - t 2. (A) - q, (B) - p, (C) - r, (D) - s 3. (A) - r, (B) - s, (C) - p, (D) - q Assertion - Reason Questions 1. (A) 2. (C) 3. (C) 4. (A) 5. (D) 6. (A) 7. (D) 8. (B) Comprehension Based Questions Comprehension # 1 : 1. (B) 2. (i) (B) (ii) (C) Comprehension #2 : 1. (B) 2. (A) 3. (C) Comprehension #3 : 1. (i) H = (v + w + x + y + z) (ii) w/2 (iii) y/2 (iv) z/2 2. (i) y (ii) w 3. –124 kJ/mol 4. –62 kJ/mol 6. 39
EXERCISE–04 [A] CONCEPTUAL SUBJECTIVE EXERCISE 1 . When 2 moles of C H are completely burnt 3120 kJ of heat is liberated. Calculate the heat of formation, 26 H°f for CH. Give H ° for CO (g) & H O(l) are –395 & –286 kJ respectively. 26 f 2 2 2 . The standard heats of formation of CH (g), CO (g) and H O(l) are –76.2, –398.8, –241.6 kJ mol–1. Calculate 42 2 amount of heat evolved by burning 1m3 of methane measured under normal (STP) conditions. 3. Calculate the enthalpy change when infinitely dilute solution of CaCl and Na CO mixed H°f for Ca2+(aq), 2 23 CO 2– (aq) and CaCO (s) are –129.80, –161.65, –288.5 kcal mol–1 respectively. 33 4 . The enthalpies of neutralization of NaOH & NH OH by HCl are –13680 calories and –12270 cal respectively. 4 What would be the enthalpy change if one gram equivalent of NaOH is added to one gram equivalent of NH Cl is solution ? Assume that NH OH and NaCl are quantitatively obtained. 44 5 . The heat of solution of anhydrous CuSO is –15.9 kcal and that of CuSO .5H O is 2.8 kcal. Calculate 4 42 the heat of hydration of CuSO . 4 11 6. The heat of reaction H (g) +2 Cl (g) HCl(g) at 27°C is –22.1 kcal. Calculate the heat of reaction 2 2 2 at 77°C. The molar heat capacities at constant pressure at 27°C for hydrogen, chlorine & HCl are 6.82, 7.70 & 6.80 cal mol–1 respectively. 7 . Calculate standard heats of formation of carbon-di-sulphide (l). Given the standard heat of combustion of carbon (s), sulphur (s) & carbon-di-sulphide (l) are : –393.3, –293.72 and–1108.76 kJ mol–1 respectively. 8 . The standard enthalpy of neutralization of KOH with (a) HCN (b) HCl in dilute solution is –2480 cal. geq–1 and –13.68 kcal, geq–1 respectively. Find the enthalpy of dissociation of HCN at the same temperature. 9 . At 300K, the standard enthalpies of formation of C H COOH(s), CO (g) & H O(l) are; –408, –393 & 65 22 –286 kJ mol–1 respectively. Calculate the heat of combustion of benzoic acid at : (i) constant pressure & (ii) constant volume. 1 0 . The heat liberated on complete combustion of 7.8 g of benzene is 327 kJ. This heat has been measured at constant volume & at 27°C. Calculate the heat of combustion of benzene at constant pressure. 1 1 . If the enthalpy of formation of HCl (g) and Cl– (aq) are –92.3 kJ/mole and –167.44 kJ/mol, find the enthalpy of solution of hydrogen chloride gas. 1 2 . 0.16 g of methane was subjected to combustion at 27°C in a bomb calorimeter. The temperature of calorimeter system (including water) was found to rise by 0.5°C. Calculate the heat of combustion of methane at (i) constant volume (ii) constant pressure. The thermal capacity of calorimeter system is 17.7 kJ K–1. (R = 8.313 mol–1K–1) 1 3 . W hen 1.0 g of fr uctose C H O (s) is burned in oxygen i n a bomb calorimeter, the temperature of the 6 12 6 calorimeter water increases by 1.56°C. If the heat capacity of the calorimeter and its contents is 10.0 kJ/°C. Calculate the enthalpy of combustion of fructose at 298 K. 1 4 . The enthalpy of dissociation of PH is 954 kJ/mol and that of P H is 1.485 M J mol–1. What is the bond 3 24 enthlapy of the P–P bond ? 1 5 . Using the bond enthalpy data given below, calculate the enthalpy change for the reaction, C H (g) + H (g) C H (g) 24 2 26 Data : Bond C–C C=C C–H H–H Bond Enthalpy 336.81 kJ/mol 606.68 kJ/mol 410.87 kJ/mol 431.79 kJ/mol
1 6 . The enthalpy change for the following process at 25°C and under constant pressure at 1 atm are as follows: CH (g) C(g) + 4H(g) H = 396 kcal/mole 4 r C H (g) 2C(g) + 6H(g) H = 676 kcal/mole 26 r Calculate C–C bond energy in C H & heat of formation of C H (g) 26 26 Given : sub C(s) = 171.8 kcal/mole B.E. (H–H) = 104.1 kcal/mole 1 7 . Find the enthalpy of S–S bond from the following data. (i) C H –S–C H (g) H°f = –147.2 kJ/mol 25 25 (ii) C H –S–S–C H (g) H°f = –201.9 kJ/mol 25 25 (iii) S(g) H°f = –222.8 kJ/mol 1 8 . Calculate the electron affinity of fluorine atom using the following data. Make Born–Haber's cycle. All the values are in kJ mol–1 at 25°C, H d is ( F ) = 160, H o (NaF(s)) = –571.I.E. [Na(g)] = 494, f s 2 H [Na(s)] = 101. Lattice energy of NaF(s) = –894. vap 1 9 . Cesium chloride is formed according to the following equation : Cs(s) + 0.5Cl (g) CsCl(s). 2 The enthalpy of sublimation of Cs, enthalpy of dissociation of chlorine, ionization energy of Cs & electron affinity of chlorine are 81.2, 243.0, 375.7 and –348.3 kJ mol–1. The energy change involved in the formation of CsCl is –388.6 kJ mol–1. Calculate the lattice energy of CsCl. 2 0 . The enthalpy of formation of ethane, ethylene and benzene from the gaseous atom are –2839.2, –2275.2 and –5506 kJ mol–1 respectively. Calculate the resonance energy of benzene. The bond enthalpy of C–H bond is given as equal to +410.87 kJ/mol. 2 1 . Two mole of ideal diatomic gas (C = 5/2 R) at 300 K and 5 atm expanded irreversly & adiabatically V,m to a final pressure of 2 atm against a constant pressure of 1 atm. Calculate q, w, H & U. 2 2 . A bomb containing 5.4g of Al and 15.97g of Fe2O3 is placed in an ice calorimeter containing initially 8 kg of ice and 8 kg of water. The reaction 2Al(s) + Fe O (s) Al O (s) + 2Fe(s) is set off by remote control and 23 23 it is then observed that the calorimeter contains 7.746 kg of ice and 8.254 kg of water. Find the H for the above reaction. Hfusion (ice) = 1.436 kcal/mole 2 3 . A sample of the sugar Dribose (C5H10O5) of mass 0.727g was placed in a calorimeter and then ignited in the presence of excess oxygen. The temperature raise by 0.910 K. In a separate experiment in the same calorimeter, the combustion of 0.825g of benzoic acid, for which the internal energy of combustion is 3251kJ mol1, gave a temperature rise of 1.940K. Calculate the internal energy of combustion of Dribose and its enthalpy of formation. 2 4 . The heat of combustion of formaldehyde(g) is 134 kcal mole1 and the heat of combustion of paraformaldehyde(s) is 122 kcal per (1/n) (CH2O)n. Calculate the heat of polymerization of formaldehyde to paraformaldehyde.
2 5 . The disaccharide maltose can be hydrolysed to glucose according to the equation C12H22O11(aq) + H2O(l) 2C6H12O6(aq) Using the following values, calculate the standard enthalpy change in this reaction: fH°(H2O, l) = 285.85 kJ·mol1 = 1263.1kJ·mol1 fH° (C H O , aq) = 2238.3 kJ·mol1 6 12 6 fH°(C12H22O11, aq) CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) 1. –88 kJ/mol 2. 35.973 MJ 3. 2.95 kcal 4. –1410 cal 5. –18.7 kcal 6. –22.123 kcal 7. 128.02 kJ 8. 11.2 kcal 9. (i) –3201 kJ/mol (ii) –3199.75 kJ/mol 10. –3273.77 kJ/mol 11. –75.14 kJ/mol 12. (i) –885 kJ/mol (ii) –889.980 kJ/mol 13. –2808 kJ/mol 14. 213 kJ/mol 15. –120.08 J/mol 16. B.E. (C–C) = 82 kcal/mol, fH [C H (g)] = –20.1 kcal/mol 17. 277.5 kJ/mol 26 18. E.A = –352 kJ mol–1 19. –618.7 kJ mol–1 20. –23.68 kJ/mol 21. q = 0, U = w = –1247.1 J, H = – 1745.94 J 22. H = 202.5 kcal 23. 2.13MJ mol1, 1.267MJ mol1 24. 12 kcal 25. 2.05 kJmol1
EXERCISE–04 [B] BRAIN STORMING SUBJECTIVE EXERCISE 1 . For the reaction cis-2-butene trans-2-butene and cis-2-butene 1-butene, H = – 950 and + 1771 cal/mol respectively. The heat of combustion of 1-butene is –649.8 kcal/mol. Determine the heat of combustion of trans-2-butene. Also calculate the bond energy of C=C bond in trans-2-butene. Given B.E of C = O = 196, O—H = 110, O = O = 118, C—C = 80 and C—H = 98 kcal/mol respectively. Hv(H2O) = 11 kcal/mol. 2 . Using the data (all values are in kJ/mol at 25°C) given below : (i) Enthalpy of polymerization of ethylene = – 72. (ii) Enthalpy of formation of benzene () = 49 (iii) Enthalpy of vaporization of benzene () = 30 (iv) Resonance energy of benzene () = – 152 (v) Heat of formation of gaseous atoms from the elements in their standard states H = 218, C = 715. Average bond energy of C—H = 415. Calculate the B.E. of C—C and C= C. [A : 331 and 590 kJ/mol] 3 . Determine resonance energy of benzene [C6H6 ()] from the following information : H°f of C6H6 () = + 49 kJ ; H°f of C2H2 (g) = +75 kJ H°v of C6H6 () = + 45 kJ B.E. C C = 930 kJ/mol ; C = C = 615 kJ/mol ; C — C = 348 kJ/mol 4 . Consider the following thermodynamic data : Enthalpy of formation of CaC2 (s) = – 60 kJ/mol ; Enthalpy of sublimation of Ca (s) = 179 kJ/mol ; Enthalpy of sublimation of C (s) = 718 kJ/mol ; First ionization energy of Ca (g) = 590 kJ/mol ; Second ionization energy of Ca (g) = 1143 kJ/mol ; Bond energy of C2 (g) = 614 kJ/mol ; First electron affinity of C2 (g) = –315 kJ/mol ; Second electron affinity of C2 (g) = +410 kJ/mol ; Draw a clear Born-Haber cycle and determine lattice energy of CaC2 (s). 5 . A swimmer breaths 20 times in one minute when swimming and inhale 200 mL of air in one breath. Inhalded air contain 20% O2 by volume and exhaled air contain 10% O2 by volume. If all oxygen are consumed in combustion of glucose in the body and 25% of energy obtained from combustion is available for muscular work. Determine the maximum distance this swimmer can swim in one hour if 100 kJ energy is required for 1.0 km swimming. Standard molar enthalpy of combustion of glucose is –2880 kJ/mol and body temperature is 37°C. 6 . Given the following standard molar enthalpies : H°f of CH3CN (g) = 88 kJ/mol, H°f of C2H6 = – 84 kJ/mol, H°Sublimation of C(gr) = 717 kJ/mol, bond dissociation energy of N2 (g) and H2 (g) are 946 and 436 kJ/mole respectively, B.E. (C—H) = 410 kJ/mol. Determine C—C and C N bond energies. 7 . By using the following data draw an appropriate energy cycle & calculate the enthalpy change of hydration of (i) the chloride ion ; (ii) the iodide ion. Comment on the difference in their values. * enthalpy change of solution of NaCl(s) = – 2 kJ/mol. * enthalpy change of solution of NaI(s) = + 2 kJ/mol. * enthalpy change of hydration of Na+ (g) = – 390 kJ/mol. * lattice energy of NaCl = – 772 kJ/mol. * lattice energy of NaI = – 699 kJ/mol.
8 . Use the following enthalpies of combustion in the calculations that follow. Element/Compound Enthalpy of Combustion C(s) – 394 H (g) – 286 2 – 876 – 2542 CH CO H(l) – 1561 32 – 1393 C H (g) 46 C H (g) 26 C H (g) 24 (i) Calculate the enthalpy change for the reaction : 2C(s) + 2H (g) + O (g) CH CO H(l) 2 2 32 (ii) Calculate the enthalpy of formation of buta–1, 3– diene C H . 46 (iii) Calculate the enthalpy of formation of ethene C H . 24 (iv) Calculate the enthalpy change for the conversion of ethene C H to ethane C H . 24 26 9 . Becker and Roth measured the heat evolved in the following processes at 20°C: (1) 1 mole of solid (COONH ) H O is burned in oxygen, (2) 1 mole of solid (COOH) (H O) is burned in oxygen, (3) 1 mole of 42 2 222 solid (COONH4)2H2O is dissolved in a large excess of water, (4) 1 mole of solid (COOH)2(H2O)2 is dissolved in a large excess of water, (5) 1 mole of oxalic acid in dilute solution is neutralized with gaseous ammonia. They found (1) 189.86 kcal, (2) 53.10 kcal, (3) 11.47 kcal, (4) 8.62 kcal, (5) 43.13 kcal; (1) and (2) were measured at constant volume, the others at constant pressure. The end products of (1) and (2) were nitrogen, carbon dioxide, and water. The heat of formation for 1 mole of water from the elements had previously been determined as 68.35 kcal at constant pressure and 20°C. Find the change in enthalpy when 1 mole of NH3 is formed from the elements at 20°C. 1 0 . Set up a thermodynamic cycle for determining the enthalpy of hydration of Ca2+ ions using the following data : Enthalpy of sublimation of Ca(s), +178.2kJ mol1; first and second ionization enthalpies of Ca(g), 589.7 kJ mol1 and 1145 kJ mol1; enthalpy of vaporization of bromine, + 30.91kJmol1; dissociation enthalpy of Br2(g), +192.9 kJmol1; electron gain enthalpy of Br(g), 331.0kJmol1; enthalpy of solution of CaBr2(s), 103.1 kJ mol1; enthalpy of hydration of Br (g), 337 kJmol1. BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) 1. H = –647.079 kcal, B.E. (C = C) = 159.921 kcal (iv) H = –118 kJ/mol 2. C – C = 343.67 kJ, C = C = 615.33 3. R.E. = –132 kJ/mol 4. L.E. = –2889 kJ 5. 1.1328 km 6. 366 kJ/mol, 1095 kJ/mol 7. for Cl– – 384 kJ mol–1, for I– – 307 kJ mol–1 8. (i) H = –484 kJ/mol (ii) H = 108 kJ/mol (iii) H = 33 9. –11.01 kcal 10. 1587kJmol1
EXERCISE–05(A) PREVIOUS YEARS QUESTIONS 1 . If at 298 K the bond energies of C–H, C–C, C=C and H–H bonds are respectively 414, 347, 615 and 435 kJ mol–1, the value of enthalpy change for the reaction : [AIEEE-2003] H C=CH (g) + H (g) H C–CH (g) at 298 K will be :- 22 2 33 (1) +125 kJ (2) –125 kJ (3) +250 kJ (4) –250 kJ 2 . The enthalpies of combustion of carbon and carbon monoxide are –393.5 and –283 kJ mol–1 respectively. The enthalpy of formation of carbon monoxide per mole :- [AIEEE-2004] (1) 110.5 kJ (2) 676.5 kJ (3) –676.5 kJ (4) –110.5 kJ 3. Consider the reaction : N + 3H 2NH carried out at constant temperature and pressure, if H and 2 2 3 U are the enthalpy and internal energy changes for the reaction, which of the following expressions is true? [AIEEE-2005] (1) H = U (2) H = 0 (3) H > U (4) H < U 4 . If the bond dissociation energies of XY, X and Y (all diatomic molecules) are in the ratio of 1 : 1 : 0.5 22 and H for the formation of XY is –200 kJ mol–1. The bond dissociation energy of X will be :- 2 f (1) 200 kJ mol–1 (2) 100 kJ mol–1 [AIEEE-2005] (3) 800 kJ mol–1 (4) 300 kJ mol–1 5. The standard enthlapy of formation (fH°) at 298K for methane, CH (g), is –74.8 kJ mol–1. The additional 4 information required to determine the average energy for C–H bond formation would be:- (1) Latent heat of vapourization of methane [AIEEE-2006] (2) The first four ionization energies of carbon and electron gain enthalpy of hydrogen (3) The dissociation energy of hydrogen molecule H [AIEEE-2006] 2 (4) The dissociation energy of H2 and enthalpy of sublimation of carbon 6 . The enthalpy changes for the following processes are listed below : Cl (g) = 2Cl(g), 242.3 kJ mol–1 2 151.0 kJ mol–1 211.3 kJ mol–1 I2(g) = 2I(g), ICl(g) = I(g) + Cl(g), I (s) = I (g), 62.76 kJ mol–1 22 Given that the standard states for iodine and chlorine are I2(s) and Cl2(g), the standard enthalpy of formation for ICl(g) is :- (1) –16.8 kJ mol–1 (2) +16.8 kJ mol–1 (3) +244.8 kJ mol–1 (4) –14.6 kJ mol–1
7 . (H – U) for the formation of carbon monoxide (CO) from its elements at 298 K is (R = 8.314 J K–1 mol–1) [AIEEE-2006] (1) 1238.78 J mol–1 (2) –2477.57 J mol–1 (3) 2477.57 J mol–1 (4) –1238.78 J mol–1 8 . Assuming that water vapour is an ideal gas, the internal energy change (U) when 1 mol of water is vapourised at 1 bar pressure and 100°C, (Given : Molar enthalpy of vapourisation of water at 1 bar and 373 K = 41 kJ mol–1 and R = 8.3 J mol–1 K–1 will be) :- [AIEEE-2007] (1) 4.100 kJ mol–1 (2) 3.7904 kJ mol–1 (3) 37.904 kJ mol–1 (4) 41.00 kJ mol–1 9 . Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below: 1 1 Cl (g) 2 dissH 22 Θ Cl(g) egH Cl–(g) hydH Cl–(aq) [AIEEE-2008] Θ Θ Θ Θ Θ 1 The energy involved in the conversion of 2 Cl2(g) to Cl–(aq) (using the data diss H Cl2 = 240 kJ mol–1, eg H Cl = –349 kJ mol–1, hyd H Cl = –381 kJ mol–1) will be:- (1) –610 kJ mol–1 (2) –850 kJ mol–1 (3) +120 kJ mol–1 (4) +152 kJ mol–1 1 0 . G 0 H 0 On the basis of the following thermochemical data : f aq H2O() H+(aq) + OH–(aq) ; H = 57.32 kJ H2(g) + 1 H2O () ; H = –286.20 kJ 2 O2(g) The value of enthalpy of formation of OH– ion at 25°C is :- [AIEEE-2009] (1) +228.88 kJ (2) –343.52 kJ (3) –22.88 kJ (4) –228.88 kJ 1 1 . The standard enthalphy of formation of NH3 is –46.0 kJ mol–1. If the enthalpy of formation of H2 from its atoms is –436 kJ mol–1 and that of N2 is –712 kJ mol–1, the average bond enthalpy of N–H bond in NH3 is :- [AIEEE-2010] (1) –1102 kJ mol–1 (2) –964 kJ mol–1 (3) + 352 kJ mol–1 (4) +1056 kJ mol–1 1 2 . The value of enthalpy change (H) for the reaction C2H5OH() + 3O2(g) 2CO2(g) + 3H2O() at 27°C is –1366.5 kJ mol–1. The value of internal energy change for the above reaction at this tem- perature will be :- [AIEEE-2011] (1) –1371.5 kJ (2) –1369.0 kJ (3) –1364.0 kJ (4) –1361.5 kJ
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