Chemistry 192 Problem Set 2 Spring, 2018 Solutions 1. The gas phase species NO 2 and N 2 O 4 equilibrate according to the reaction N 2 O 4(g) 2NO 2(g), and it is found that at 298K in a reaction vessel of fixed volume the equilibrium concentrations of each species are [NO 2 ]2.17 10 3 M and [N 2 O 4 ]3.64 10 4 M. Under different nonequilibrum conditions in the same vessel at the same temperature, NO 2 and N 2 O 4 are added such that their partial pressures are P NO2 0.456 bar and P N2 O 4 0.789 bar. Under the nonequilibrium conditions, predict whether the reaction will proceed to the right or left. K c [NO 2] 2 [N 2 O 4 ] (2.17 10 3 ) 2 1.29 10 2 3.64 10 4 K P K c (RT ) ngas (1.29 10 2 )(298)(0.08314) 0.321 Q P P 2 NO 2 P N2 O 4 (0.456)2 0.789 0.264 so the reaction proceeds to the right. Q P < K P 2. For the reaction between solid iodine, hydrogen gas and gas-phase hydrogen iodide I 2(s) + H 2(g) 2HI (g) at 298 K, the equilibrium partial pressures of hydrogen and hydrogen iodide are measured to be respectively P H2 32. bar and P HI 4.0 bar. Suppose an initial nonequilibrium mixture of solid iodine, hydrogen gas and hydrogen iodide is prepared at 298 K in a closed 10.0 L container containing 68. grams of hydrogen gas, 352. grams 1
of hydrogen iodide gas and 500. grams of solid iodine (sufficient to be in excess). Calculate the concentration equilibrium constant K c for the reaction, and predict whether the reaction will proceed spontaneously to the right or left under the non-equilibrium conditions. K P P HI 2 (4.0)2 0.50 P H2 32. K c K P (RT ) ngas 0.50(0.08314 298.) 1 2.0 10 2 ( ) mol 68. g 2.00 g [H 2 ] 3.4 M 10.0 L ( ) mol 352. g 127.9 g [HI] 0.275 M 10.0 L Q c > K c Q c [HI]2 [H 2 ] 0.2752 3.4 2.2 10 2 so reaction is spontaneous to the left 3. At 500.K and a total pressure of 5.57 bar, it is found that the degree of dissociation at equilibrium of gas-phase ammonia according to the reaction NH 3(g) 1 2 N 2(g) + 3 2 H 2(g) is α 0.564. In a separate experiment at 500.K, a non equilibrium mixture is prepared by combing 1.00 moles each of gas-phase NH 3, N 2 and H 2 in a 1.00 L flask. Calculate first K P and then K c for the ammonia dissociation reaction, and predict the direction of the reaction for the non equilibrium mixture. Let n 0 the initial number of moles of ammonia. n NH3 n N2 n H2 initial n 0 0 0 change αn 0 αn 0 /2 3αn 0 /2 equilibrium n 0 () αn 0 /2 3αn 0 /2 K P P 1/2 N 2 P 3/2 H 2 P NH3 n tot n 0 (1 + α) ( α/2 ) 1/2 ( 3α/2 ) 3/2 1 + α P 1 + α P 1 + α P 2
( ) 0.282 1/2 ( 0.846 1.564 1.564 0.436 1.564 ) 3/2 5.57 3.38 K c K P (RT ) ngas (3.38)[(0.08314)(500.)] 1 8.12 10 2 Q c 1 > K c 4. Consider the reaction at equilibrium reaction is spontaneous to the left PCl 5(g) PCl 3(g) + Cl 2(g). When a certain amount of PCl 5(g) is heated in a 10.0 liter flask at 275. C, analysis shows that at equilibrium the number of moles of each species is n P Cl5 0.151 mol, n P Cl3 0.400 mol and n Cl2 0.400 mol. Calculate K P and K c for the reaction at 275. C. ( ) 0.400 2 K c [PCl 3][Cl 2 ] ( 10.0 ) 1.06 10 1 [PCl 5 ] 0.151 10.0 K P K c (RT ) ngas (1.06 10 1 )[(0.08314)(548)] 1 4.83 5. For the reaction given in problem 4, if for all species at 275. C, 1.00 moles of each species were placed in the 10.0 L flask, determine whether the reaction would be favored to the left or right. ( ) 1.00 2 Q c ( 10.0 ) 0.100 1.00 10.0 Q c < K c to right 6. Consider the reaction at equilibrium between gas-phase iodine, hydrogen and hydrogen iodide H 2(g) + I 2(g) 2HI (g). At 450.C the concentration equilibrum constant for the reaction is K c 50.0. (a) Calculate K P at 450.C. K P K c (RT ) ngas K c (RT ) 0 50.0 3
(b) If 2.0 moles each of iodine vapor and hydrogen are placed in a 5.0 L flask, and the gases are allowed to come to equilibrium at 450.C, calculate the final total pressure and partial pressures of each gas in the container. H 2 I 2 HI initial 2.0 mol 2.0 mol 0 mol change -x mol -x mol 2x mol equilibrium (2.0-x) mol (2.0 x) mol 2x mol K c [HI]2 [H 2 ][I 2 ] 50.0 ( ) 2x 2 ( 5.0 ) ( ) 2.0 x 2.0 x 5.0 5.0 4x 2 4 4x + x 2 46x 2 200x + 200 0 x 200 ± (200) 2 4(46)(200) (2)(46) 200 ± 56.57 x 2.79, 1.55 92 The first solution is > 2 and consequently unphysical. Then n H2 n I2 2.0 mol 1.55 mol 0.44 mol n HI 2x 3.1 mol P H2 P I2 n H 2 RT V 7. For the reaction (0.44 mol)(0.08314 L bar mol 1 K 1 )(723 K) 5.0 L P HI (3.1 mol)(0.08314 L bar mol 1 K 1 )(723 K) 37. bar 5.0 L P tot P H2 + P I2 + P HI 48. bar N 2 O 4(g) 2NO 2(g) 5.3 bar it is found that K P 0.171 at 25.0 C. Calculate the degree of dissociation of N 2 O 4 at 25.0C and total pressures of 1.0 bar and 0.10 bar. N 2 O 4 NO 2 initial n 0 change -αn 2αn equilibrium ()n 2αn 4
K P (P NO 2 ) 2 (χ NO 2 P ) 2 P N2 O 4 χ N2 O 4 P n tot ()n + 2αn (1 + α)n χ NO2 n NO 2 n tot 2α 1 + α χ N2 O 4 1 + α 1.0 bar 0.10 bar ( 2α 1 + α P )2 ( 1 + α ) P 4α2 2 P 0.171 4α2 2 4.171α 2 0.171 α 0.20 4α 2 0.171 2 0.10 1.71 5.71α 2 1.71 α 0.55 8. The degree of dissociation (the fraction of the pure substance that dissociates at equilibrium) of NOCl (g) at 298K and a total pressure of 2.00 bar according to the reaction NOCl (g) NO (g) + 1 2 Cl 2(g) is α 0.00307. Calculate the pressure equilibrium constant, K P 298.K. Let n 0 be the initial number of moles of gas-phase NOCl. for the reaction at NOCl NO Cl 2 initial n 0 0 0 change -αn 0 αn 0 αn 0 /2 equilibrium n 0 () αn 0 αn 0 /2 5
n tot n 0 (1 + α/2) K P P NOP 1/2 Cl 2 (χ NOP )(χ Cl2 P ) 1/2 P NOCl χ NOCl P (α/(1 + α/2))((α/2)/(1 + α/2))1/2 ()/(1 + α/2) P α(α/2) 1/2 ()(1 + α/2) 1/2 P 0.00307(0.00307/2) 1/2 (1 0.00307)(1 + 0.00307/2) 1/2 2.00 1.70 10 4 9. Predict the effects of increased pressure and temperature on the following reactions at equilibrium. (a) CO (g) + H 2 O (g) CO 2(g) + H 2(g) H 41.8 kj mol 1 (b) Answer increase P - no change increase T - to left 2SO 2(g) + O 2(g) 2SO 3(g) H 196.5 kj mol 1 (c) Answer increase P - to right increase T - to left CaCO 3(s) Ca (s) + CO 2(g) H 179.7 kj mol 1 Answer increase P - to left increase T - to right 10. At 400.K the concentration equilibrium constant for the reaction HCONH 2(g) NH 3(g) + CO (g) is K c 4.84. If 10.0 grams of pure gas-phase HCONH 2 are placed in a 10.0 liter flask at 400.K, calculate the total pressure in the flask when equilibrium is reached. 10.0 g n initial 1 0.222 mol (12.00 + 16.00 + 14.00 + 3.00) g mol Let x be the number of moles of HCONH 2 that has reacted at equilibrium. 6
HCONH 2 NH 3 CO initial 0.222 mol 0 mol 0 mol change -x mol x mol x mol equilibrium (0.222-x) mol x mol x mol [NH 3 ][CO] [HCONH 2 ] (x/10.0) 2 (0.222 x)/10.0 4.84 x 2 0.222 x 48.4 x 2 + 48.4x 10.7 0 x 48.4 ± [(48.4)2 + 4(10.7)] 1/2 48.6, 0.220 0.220 2 n tot 0.222 mol x mol + 2x mol 0.222 mol + x mol 0.442 mol P n totrt V (0.442 mol)(0.08314 L bar mol 1 K 1 )(400. K) 10.0 L 1.47 bar 11. When metallic lead is added to a solution of Cr 3+ (aq) the following reaction occurs Pb (s) + 2Cr 3+ (aq) Pb 2+ (aq) + 2Cr2+ (aq) with associated equilibrium constant K c 3.2 10 10. (a) If solid lead is added to a 0.200 M Cr 3+ aqueous solution, find the final concentrations of all ions in the system. K c [Pb2+ ][Cr 2+ ] 2 [Cr 3+ ] 2 [Cr 3+ ] [Pb 2+ ] [Cr 2+ ] initial 0.200 M 0 M 0 M change -2x M x M 2x M equilibrium (0.200-2x) M x M 2x M x(2x) 2 3.2 10 10 (0.200 2x) 2 Because K c is small, we can ignore the term 2x in the denominator. Then Then 4x 3 1.3 10 11 x 1.5 10 4 M [Cr 3+ ] 0.200 M 2x 0.20 M to 2 significant figures, justifying our assumption. [Pb 2+ ] x 1.5 10 4 M 7 [Cr 2+ ] 2x 3.0 10 4 M
(b) If 100 ml of water are added to the solution at equilibrium, predict the direction that the equilibrium would be shifted. All concentrations are decreased, and by LeChâtelier s principle, the reaction shifts to the right. 12. At 145.K the the degree of dissociation of ozone in the gas-phase decomposition reaction O 3(g) 3 2 O 2(g) at a total pressure of P 1.00 bar is α 0.0221. Calculate the concentration equilibrium constant K c of the dissociation reaction. n O3 n O2 initial n 0 0 change -αn 0 (3/2)αn 0 equilibrium ()n 0 (3/2)αn 0 χ O2 (3/2)α 1 + α/2 ( (3/2)α 1 + α/2 P K P n tot n 0 (1 + α/2) ) 3/2 χ O3 1 + α/2 6.14 10 1 + α/2 P K c K P (RT ) ngas (6.14 10 3 )(0.08314 145.) 0.5 1.77 10 3 13. At a temperature of 452K gas-phase isopropyl alcohol dissociates into gas-phase acetone and hydrogen according to the reaction 3 (CH 3 ) 2 CHOH (g) (CH 3 ) 2 CO (g) + H 2(g). At P 2.00 bar and T 452K, the degree of dissociation isopropyl alcohol is measured to be α 0.182. Calculate the concentration equilibrium constant K c for the reaction. n (CH3 )2CHOH n (CH3 ) 2 CO n H2 initial n 0 0 0 change αn 0 αn 0 αn 0 equilibrium n 0 () αn 0 αn 0 8
n tot n 0 () + n 0 α + n 0 α n 0 (1 + α) K P P (CH 3 ) 2 COP H2 P (CH2 ) 2 CHOH ( )2 α 1 + α P ( ) 1 + α P α 2 P 2 2.00(0.182)2 1 (0.182) 2 0.0685 K c K P (RT ) n (0.0685)[(0.08314)(452)] 1 1.82 10 3 14. The gas-phase decomposition of antimony pentachloride into gas-phase antimony trichloride and chlorine gas SbCl 5(g) SbCl 3(g) + Cl 2(g) has a concentration equilibrium constant K C 2.50 10 2 at a temperature of 521. K. A sample of pure SbCl 5 gas is placed in a reaction vessel of fixed volume at T 521.K, and when the system attains equilibrium, the total pressure is measured to be 3.51 bar. Calculate the equilibrium degree of dissociation α (i.e. the fraction dissociated) of antimony pentachloride in the reaction vessel. K P K C RT (2.50 10 2 )(0.08314)(521.) 1.08 Let n 0 be the initial number of moles of SbCl 5. n SbCl5 n SbCl3 n Cl2 initial n 0 0 0 change αn 0 αn 0 αn 0 equilibrium n 0 () αn 0 αn 0 n tot n 0 (1 + α) K P P SbCl 3 P Cl2 χ SbCl 3 P χ Cl2 P P SbCl5 χ SbCl5 P α 1 + α P α 1 + α P α 2 1 + α P P 2 α 2 (3.51) 1.08 α 0.485 2 9
15. The equilibrium behavior of the gas-phase reaction P 4(g) 2P 2(g) is studied at 1325.K, and the concentration equilibrium constant is measured to be K C 9.08 10 4. Calculate the equilibrium degree of dissociation α of P 4 at 1325.K and a total pressure of 2.00 bar. K P K C (RT ) ngas (9.08 10 4 )[(0.08314)(1325)] 1 0.100 n P4 n P2 initial n 0 0 change -αn 0 2αn 0 equilibrium ()n 0 2αn 0 0.100 n tot n 0 (1 + α) ( )2 2α 1 + α P 1 + α P 4α 2 P 8.00α2 2 2 α 2 0.0125 α 0.111 2 10