Homework #5 Chapter 6 Chemical Equilibrium

Homework #5 Chapter 6 Chemical Equilibrium 2. Assume the reaction is A + B C + D. It is given that K9 and K [C][D]. At the start of [A][B] the reaction, before equilibrium is reached, there are 8 A molecules, 8 B molecules, and ( 0 V )(0 V ) [A][B] ( 8 V )(8 V no C or D molecules. Therefore, [C][D] 0. In order for Q to equal K and the ) system to be at equilibrium 6 A and B molecules must react to form 6 C and D molecules. Leaving 2 A and B molecules unreacted (K [C][D] ( 6 V )(6 V ) [A][B] ( 2 V )(2 V ) 9). 0. a) The reaction of interest is N 2+ 3H 2 2NH 3. Red NH 3 (This is the only line that increases with time.) Green H 2 (This line decreases faster that the N 2 line.) Blue N 2 b) Initially there is only H 2 and N 2 in the system therefore, in order to reach equilibrium some of the H 2 and N 2 will convert to NH 3. This results in the green, and blue lines (representing H 2 and N 2) decreasing as time goes on. Since there is no NH 3 to start with the red line (representing NH 3) must increase as time goes on. The slope of the lines represents how fast the levels of the substances change. The larger the coefficient in the chemical equation the steeper the slope will be, therefore, the green line (H 2) is the steepest because it has a coefficient of 3, followed by red (NH 3) which has a coefficient of 2, and finally blue line (N 2) which has a coefficient of. c) Equilibrium is reached when the concentration of the species is not changing which is represented by flat horizontal line. This happens about 2/3 of the way on the time axis.. At equilibrium the forward rate of the reaction equals the reveres rate of the reaction. The general expression for the equilibrium constant is K products therefore, the larger reactants the K the more product that are present at equilibrium. If K equals there will be equal amounts of products and reactants at equilibrium. Therefore, the reaction involving the NOCl as the reactant has lost of reactants at equilibrium and the reaction involving NO as the reactant has lots or products at equilibrium. 3. The equilibrium constant for both case is K [H 2 ][CO 2 ] [H 2 O][CO] Since the temperature, and volume are the same in both cases and the stoichiometric coefficients are all one (with mole of either both reactants or both products added) there will be only one way ( set of concentrations) to get the equilibrium constant to be the same for both cases. Therefore, at equilibrium both flasks will have the same composition of materials in them. 2. a) K [H 2O] K P [NH 3 ] 2 [CO 2 ] P H2O P2 NH P 3 CO 2

b) K [N 2 ][Br 2 ] 3 3 K P P N2 P Br2 c) K [O 2 ] 3 3 K P P O2 d) K [H 2O] [H 2 ] K P P H 2 O 24. K P P 2 SO3 P2 SO P 2 O 2 PV nrt 25. Given P H2 K [SO 3 ]2 [SO 2 ] 2 [O 2 ] ( P [SO3 ] RT )2 RT n v (concentration) ( [SO 2 ] RT )2 ( [O 2 ] RT ) RTK P (0.08206)(,00)(0.25) 23 At equilibrium [CH 3OH]0.5 M, [CO]0.24 M, [H 2]. M T327 C327+273.5600.K K [CO][H 2 ]2 [CH 3 OH] (0.24)(.)2.9 (0.5 ) PV nrt P n RT v K P P 2 COP H2 ([CO]RT)([H 2]RT) 2 K(RT) P CH3 OH [CH 2 (.9)((0.08206)(600. )) 2 4,700 3 OH]RT 26. Given: H 2(g) + Br 2(g) 2HBr(g) K P P 2 HBr P H 2 P Br2 a) HBr(g) ½H 2(g) + ½Br 2(g) K P2 P H2 2 P2 Br 2 P HBr K P b) 2HBr(g) H 2(g) + Br 2(g) 3.5 0 4 at,495k 3.5 0 4 0.0053 K P3 P H2 P Br2 P2 HBr K P 3.5 04 2.9 0 5 c) ½H 2(g) + ½Br 2(g) HBr(g) K P4 P HBr P 2 H P2 2 Br 2 27. 2N 2(g) + O 2(g) 2N 2O(g) K P 3.5 0 4 90 N 2 O 2 N 2O Equilibrium (mol) 2.80 0-4 mol 2.50 0-5 mol 2.00 0-2 mol Equilibrium (M n ) V.40 0-4 M.25 0-5 M.00 0-2 M K [N 2O] 2 [N 2 ] 2 [O 2 ] (.00 0 2 ) 2 (.40 0 4 ) 2 (.25 0 5 4.08 08 ) 2

Q [N 2O] 2 [N 2 ] 2 [O 2 ] (0.200) 2 (2.00 0 4 ) 2 4.08 08 (0.00245) Since QK the system is at equilibrium 28. N 2(g) + 3H 2(g) 2NH 3(g) K P P 2 NH 3 P N2 P (3. 0 2 ) 2 3 H2 (8.5 0 )(3. 0 3 3.8 04 ) 3 Q P P 2 NH 3 P N2 P (0.067) 2 3.2 03 H2 (0.525)(0.0076) 3 No the system is not at equilibrium. There are too many reactants, therefore, the reaction will proceed in the forward direction to reach equilibrium. 3. 3Fe(s) + 4H 2O(g) Fe 3O 4(s) + 4H 2(g) 4 K P P H 2 4 P H2 O Calculate pressure of H 2 P H2 O 5.0torr ( atm ) 0.097 atm P tot P H2 + P H2 O 36.3 torr ( atm ) 0.0478 atm P H2 P tot P H2 O 0.0478 atm 0.097 atm 0.028 atm 4 K P P H 2 4 (0.028)4 (0.097) 4 4.4 P H2 O 33. PCl 5(g) PCl 3(g) + Cl 2(g) P tot P PCl5 + P PCl3 + P Cl2 0.50 x + x + x 0.84 atm x 0.34 atm P PCl5 0.50 atm 0.34 atm 0.6 atm P PCl3 P Cl2 0.34 atm K P P PCl 3 P Cl2 (0.34)(0.34) 0.72 P PCl5 (0.6) K [PCl 3][O 2 ] [PCl 5 ] PV nrt K [PCl 3][O 2 ] [PCl 5 ] PCl 5 PCl 3 Cl 2 Initial 0.50 atm 0 atm 0 atm Change -x +x +x Equilibrium 0.50-x x x P PCl3 RT P n (concentration) RT v P Cl2 RT RT K P P PCl5 RT (0.72) 0.07 (0.08206)(523) 3

33. 2SO 3(g) 2SO 2(g) + O 2(g) SO 3 SO 2 O 2 Initial (mol) 2.0 mol 0 mol 0 mol Initial (M n ) V 4.00 M 0 M 0 M Change -2x +2x +x Equilibrium (mol)? 3.0 mol? Equilibrium (M n ) 3.0 mol V 4.00-2x 3.0 L.0 M 2x x From SO 2 we see that x 0.50 M Therefore calculate the concentrations at equilibrium [SO 3 ] 4.00 2x 4.00 2(0.50) 3.0 M [SO 2 ] 2x 2(0.50).0 M [O 2 ] x 0.5 M K [SO 2] 2 [O 2 ] [SO 3 ] 2 (.0)2 (0.50) (3.0) 2 0.056 37. H 2O(g) + Cl 2O(g) 2HOCl(g) K [HOCl]2 [H 2 O][Cl 2 O] 0.0900 a) Calculate the molarities [HOCl] n V [H 2 O] n V.0 mol.0 L 0.0 mol.0 L.0 M 0.0 M [Cl 2 O] n 0.0 mol 0.0 M V.0 L Q [HOCl]2 [H 2 O][Cl 2 O] (.0) 2.0 02 (0.0)(0.0) Q >K therefore, there are too many products and the reverse reaction will occur to reach equilibrium (or reaction will go to the left). b) Calculate the molarities Q [HOCl] n V [H 2 O] n V 0.084 mol 2.0 L 0.98 mol 2.0 L 0.042 M 0.49 M [Cl 2 O] n 0.080 mol 0.040 M V 2.0 L [HOCl]2 [H 2 O][Cl 2 O] (0.042)2 (0.040)(0.49) 0.090 System is at equilibrium. c) Calculate the molarities [HOCl] n V [H 2 O] n V [Cl 2 O] n V 0.24 mol 3.0 L 0.56 mol 3.0 L 0.000 mol 3.0 L 0.083 M 0.9 M 3.3 0 4 M 4

Q [HOCl]2 [H 2 O][Cl 2 O] (0.083) 2 (0.9)(3.3 0 4 ) 0 Q >K therefore, there are too many products and the reverse reaction will occur to reach equilibrium (or reaction will go to the left). 38. H 2O(g) + Cl 2O(g) 2HOCl(g) P 2 HOCl K K P 0.0900 P H2 OP Cl2 O 2 a) Q P P HOCl (.00)2.00 P H 2OP Cl (.00)(.00) 2O Q P > K P, therefore, there are too many products and the reverse reaction will occur to reach equilibrium (or reaction will go to the left). b) P HOCl 2.0 torr( atm ) 0.0276 atm P H2 O 200. torr ( atm ) 0.263 atm P Cl2 O 49.8 torr ( atm ) 0.0655 atm P 2 HOCl (0.0276) 2 (0.263)(0.0655) 0.0442 Q P P H2 OP Cl2 O Q P < K P, therefore, there are too many reactants and the forward reaction will occur to reach equilibrium (or reaction will go to the right). Note: Pressures were converted to atm. If you plugged in the pressure in torr you would have gotten the same answer. This works for this problem because there are the same number of moles of gas on the product and the reactants side of the equation. If there were different number of moles on the product and the reactants side of the reaction you would have to plug the pressure in in atm. c) P HOCl 20.0 torr( atm ) 0.0263 atm P H2 O 296 torr ( atm ) 0.389 atm P Cl2 O 5.0 torr ( atm ) 0.098 atm P 2 HOCl (0.0263) 2 (0.389)(0.098) 0.0898 Q P P H2 OP Cl2 O System is essentially at equilibrium 43. H 2O(g) + Cl 2O(g) 2HOCl(g) K a) [HOCl]2 [H 2 O][Cl 2 O] 0.090 H 2O Cl 2O HOCl Initial (g).0 g 2.0 g 0 Initial (n m ) M x 0.055 mol 0.023 mol 0 mol Initial (M n V ) 0.055 M 0.023 M 0 M Change -x -x +2x Equilibrium 0.055-x 0.023-x 2x 5

K [HOCl]2 [H 2 O][Cl 2 O] (2x) 2 (0.0550 x)(0.023 x) 0.090 b) 4x 2 0.003 0.078x + x 2 0.090 4x 2.7 0 4 0.0070x + 0.090x 2 3.9x 2 0.0070x +.7 0 4 0 x b ± b2 4ac 0.0070 ± 4.9 0 5 0.008 7.82 0.0064 and 0.0046 The molarity cannot be negative, therefore, the x 0.0046 M With guess and check x 0.00457 Concentrations at equilibrium [H 2 O] 0.055 M 0.0046 M 0.050 M [Cl 2 O] 0.023 M 0.0046 M 0.08 M [HOCl] 2(0.0046 M) 0.0092 M H 2O Cl 2O HOCl Initial (mol) 0 mol 0 mol.0 mol Initial M 0 M 0 M 0.50 M n V Change X X -2x Equilibrium X X 0.50-2x K [HOCl]2 (0.50 2x)2 0.090 [H 2 O][Cl 2 O] (x)(x) 4.0x 2 2.0x + 0.25 x 2 0.090 4.0x 2 2.0x + 0.25 0.0904x 2 3.9x 2 2.0x + 0.25 0 x b ± b2 4ac 2.0 ± 4.0 3.9 0.29 and 0.22 7.82 Since the concentration of HOCl was only 0.50 M to start with you cannot subtract more than 0.50 M. Therefore, the answer must be 0.22 If you use guess and check x 0.27, 0.22 with correct significant figures Concentrations at equilibrium [H 2 O] [Cl 2 O] 0.022 M [HOCl] 0.50 M 2(0.22 M) 0.06 M 44. K [HF]2 (0.400) 2 320. [H 2 ][F 2 ] (0.0500)(0.000) Calculate Q after adding 0.200 mol of F 2 Q [HF]2 [H 2 ][F 2 ] Concentration of F 2 added 0.200 mol [F 2 ] 0.0400 M 5.00 L Calculate the initial F 2 concentration 6

0.200 mol [F 2 ] 0.0400 M 5.00 L [F 2 ] 0.000 M + 0.0400 M 0.0500 M (0.400) 2 Q (0.0500)(0.0500) 64.0 Q<K, therefore, reaction will proceed toward products H 2 F 2 HF Initial 0.0500 M 0.0500 M 0.400 M Change -x -x +2x Equilibrium 0.0500-x 0.0500-x 0.400+2x K [HF]2 [H 2 ][F 2 ] (0.400 + 2x) 2 (0.0500 x)(0.0500 x) 320. (0.400 + 2x)2 (0.0500 x) 2 320. 0.400 + 2x 0.0500 x 7.9 0.400 + 2x 0.895 7.9x 9.9x 0.495 x 0.0249 [H 2 ] [F 2 ] 0.0500 M 0.0249 M 0.025 M [HF] 0.400 M + 2(0.0249 M) 0.450 M 45. 2SO 2(g) + O 2(g) 2SO 3(g) K P K P P 2 SO3 2 P O 2 0.25 P SO 2 SO 2 O 2 SO 3 Initial 0.50 atm 0.50 atm 0 atm Change -2x -x +2x Equilibrium 0.50-2x 0.50-x 2x P 2 SO3 P2 SO P 2 O 2 (2x) 2 (0.50 2x) 2 (0.50 x) 0.25 Using guess and check x 0.062 atm, 0.062 with correct significant figures Partial pressures at equilibrium P SO2 0.50 atm 2(0.062 atm) 0.38 atm P O2 0.50 atm 0.062 atm 0.44 atm P SO3 2(0.062 atm) 0.2 atm 46. N 2(g) + O 2(g) 2NO(g) K [N 2O] 2 [N 2 ][O 2 ] 0.050 P NO 2 K P P N2 P O2 PV nrt K P P RT n v (concentration) P NO 2 ([N 2 O]RT) 2 K 0.050 P N2 P O2 ([N 2 ]RT)([O 2 ]RT) 7

N 2 O 2 NO Initial 0.80 atm 0.20 atm 0 atm Change -x -x +2x Equilibrium 0.80-x 0.20-x 2x K P P NO (2x) 2 P N2 P O2 (0.80 x)(0.20 x) 0.050 4x 2 x 2.0x + 0.6 0.050 4x 2 0.050x 2 0.050x + 0.0080 3.95x 2 0.050x + 0.0080 0 x b ± b2 4ac 0.050 ± 0.0025 0.26 0.052 and 0.039 7.90 The pressure cannot be negative therefore the x 0.039 Using guess and check x 0.0392, 0.039 with correct significant figures Calculate the pressure of NO P 2x 2 0.039atm 0.078atm NO 49. 2CO 2(g) 2CO(g) + O 2(g) K [CO]2 [O 2 ] [CO 2 ] 2 2.0 0 6 CO 2 CO O 2 Initial (mol) 2.0 mol 0 mol 0 mol Initial (M n ) V 0.40 M 0 M 0 M Change -2x +2x +x Equilibrium 0.40-2x 2x x K [CO]2 [O 2 ] (2x) 2 x [CO 2 ] 2 2.0 0 6 (0.40 2x) 2 This is a very small equilibrium constant therefore try assuming that 2x is much less than 0.40 therefore assume 0.40-2x 0.40 K 4x3 2.0 0 6 0.402 x 0.0043 M Test to make sure that this is a good assumption (5% rule) Amount Gained/Lost 2(0.0043) 00% 2.5% Original Conentration 0.40 Since it is less than 5% the assumption was good. Concentrations at Equilibrium [CO 2 ] 0.40 M 2(0.0043 M) 0.39 M [CO] 2(0.0043 M) 0.0086 M [O 2 ] 0.0043 M 8

50. N 2O 4(g) 2NO 2(g) 2 K P P NO 2 0.25 a) N 2O 4 NO 2 Initial 4.5 atm 0 atm Change -x +2x Equilibrium 4.5-x 2x b) K P P 2 NO 2 (2x)2 (4.5 x) 0.25 4x 2. 0.25x 4x 2 + 0.25x. 0 x b ± b2 4ac 0.25 ± 0.063 8 0.50 and 0.56 8 The pressure cannot be negative, therefore, x 0.50 atm Using guess and check x 0.500, 0.50 with correct significant figures Partial pressures at equilibrium P NO2 2(0.50 atm).0 atm 4.5 atm 0.5 atm 4.0 atm N 2O 4 NO 2 Initial 0 atm 9.0 atm Change +x -2x Equilibrium x 9.0-2x K P P 2 NO 2 (9.0 2x)2 0.25 x 4x 2 36x + 8 0.25x 4x 2 36.25x + 8 0 x b ± b2 4ac 36.25 ±,34,296 5.0 and 4.0 8 If x5.0 the pressure of NO 2 at equilibrium would be negative, therefore, x 4.0 Using guess and check x 4.00 atm, 4.0 with correct significant figures Partial pressures at equilibrium P NO2 9.0 atm 2(4.0 atm).0 atm 4.0 atm c) No it does not matter which direction equilibrium is reached. d) If the volume is decreased by ½ then the pressure will increase by 2. N 2O 4 NO 2 Initial 9 atm 0 atm Change -x +2x Equilibrium 9-x 2x K P P 2 NO 2 (2x)2 (9 x) 0.25 4x 2 2.25 0.25x 9

4x 2 + 0.25x 2.25 0 x b ± b2 4ac 0.25 ± 0.063 36 0.72 and 0.78 8 The pressure cannot be negative, therefore, x 0.72 atm Using guess and check x 0.720, 0.72 with correct significant figures Partial pressures at equilibrium P NO2 2(0.72 atm).4 atm 9.0 atm 0.72 atm 8.3 atm 54. Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) K [FeSCN2+ ] [Fe 3+ ][SCN. 03 ] Change n M V Fe 3+ SCN - FeSCN 2+ 0.020 M 0.0 M 0 M Change -x -x +x Equilibrium 0.020-x 0.0-x x K [FeSCN2+ ] [Fe 3+ ][SCN ] x. 03 (0.020 x)(0.0 x) x x 2. 03 0.2x + 0.0020 x. 0 3 x 2 32x + 2.2. 0 3 x 2 33x + 2.2 0 x b ± b2 4ac 33 ± 7,700 9,680 0.0 and 0.020 2,200 The molarity cannot be negative, therefore, x 0.020 Using guess and check x 0.0978, 0.020 with correct significant figures Concentrations at equilibrium [Fe 3+ ] 0.020 M 0.020 M 0 M [SCN ] 0.0 M 0.020 M 0.08 M [FeSCN 2+ ] 0.020 M 56. If the volume of the reaction vessel is increased the overall pressure of the system drops causing the partial pressures of all the gases to drop. To compensate for this, equilibrium shifts in the direction that has the most gas molecules. If both sides have the same number of gas molecules, then there will be no change in equilibrium. If an inert gas is added to the system it causes the total pressure to go up by whatever the partial pressure of the inert gas is. However, the partial pressures of the gases in the reaction remain unchanged, causing the system to remain at equilibrium. 58. a) Adding water increases volume and shifts equilibrium to the side with the great number of moles of aqueous solution. Equilibrium shifts toward reactants (left). b) Adding AgNO 3 will cause AgSCN to precipitate out of solution, therefore, more reactants will form. Equilibrium shifts toward reactants (left). 0

c) Adding FeOH will cause Fe(OH) 3 to precipitate out of solution, therefore, more reactants will form. Equilibrium shifts toward reactants (left). d) Add Fe(NO 3) 3 causes more Fe 3+ ion to be in solution which shifts equilibrium away from reactants. Equilibrium shifts toward products (right). 59. When sodium hydroxide is added to the beaker it produces OH - ions in solution, which bond with the H + ions to form H 2O. Removing H + ions from solution causes equilibrium to shift to the products (right). Since CrO 4 2- is yellow in color, as more CrO 4 2- forms the solution turns yellow. 6. a) Equilibrium shifts to the products (right). b) Equilibrium shifts to the products (right). c) When an inert gases is added and the volume remains the same the partial pressures of the other gases in the system do not change. Therefore, there is no shift in equilibrium. d) For exothermic reactions (heat in the products side) when the temperature is increased the equilibrium shifts to the reactants (left). e) When the volume of the container is decreased equilibrium shifts to the side of the reaction with the overall fewer number of gas molecule. If the same number of gas molecules are present on both sides no shift in equilibrium occurs. Therefore, no shift in equilibrium occurs. 62. a) Equilibrium shifts to the reactants (left). b) For endothermic reactions (heat on reactants side) when the temperature is increased equilibrium shifts to products (right). c) Adding an inert gas does not affect the partial pressure of the gases. Therefore, there is no shift in equilibrium. d) Equilibrium shifts to the products (right). e) When the volume of the container is increased equilibrium shifts to the side of the reaction with the overall larger number of gas molecules. Therefore, equilibrium shifts to the products (right). 63. a) Equilibrium shifts toward reactants (left). b) Equilibrium shifts toward products (right). c) Equilibrium shifts toward reactants (left). d) Adding an inert gas does not affect the partial pressure of the gases. Therefore, there is no shift in equilibrium. e) When the volume of the container is increased, equilibrium shifts to the side of the reaction with the larger number of gas molecules. If the same number of gas molecules are present on both sides no shift in equilibrium occurs. Therefore, no shift in equilibrium occurs. f) For an exothermic reaction the heat is on the products side. Therefore, decreasing the heat will cause the equilibrium to shift to the products side. Equilibrium shifts toward products (right)

65. a) Equilibrium shifts to reactants (left) more SO 3 forms. b) When the pressure is decreased equilibrium shifts to the toward the side with the overall fewer gas molecules. Equilibrium shifts to the reactants (left) more SO 3 forms. c) Adding an inert gas does not affect the partial pressure of the gases. Therefore, there is no shift in equilibrium; the amount of SO 3 stays the same. d) For endothermic reactions (heat on reactants side) when the temperature is decreased equilibrium shifts to reactants (left); more SO 3 forms. e) Equilibrium shifts to the products (right); moles of SO 3 decreases. 72. a) Na 2O(s) 2Na(l) + ½O 2(g) K 2Na(l) + O 2(g) Na 2O 2(s) Na 2O(s) + ½O 2(g) Na 2O 2(s) K3 K 2 0 25 K 3 5 0 b) NaO(g) Na(l) + ½ O 2(g) K 2 Na 2O(s) 2Na(l) + ½ O2(g) K 2Na(l) + O 2(g) Na 2O 2(s) K3 29 4 03 NaO(g) + Na 2O(s) Na 2O 2(s) + Na(l) K K 2 (2 0 25 )(2 0 5 ) 0.08 K 3 5 0 29 c) NaO(g) Na(l) + ½ O 2(g) K 2 NaO(g) Na(l) + ½ O 2(g) K 2 2Na(l) + O 2(g) Na 2O 2(s) 2NaO(g) Na 2O 2(s) 73. NO(g) + O(g) NO 2(g) K NO 2(g) + O 2(g) O 3(g) + NO(g) K 2 O 2(g) + O(g) O 3(g) K K K 2 K3 K 2 K 2 K 3 (2 0 5 )(2 0 5 ) 5 0 29 8 0 8 2.5 (6.8 0 49 )(5.8 0 34 08 ) 75. 3H 2(g) + N 2(g) 2NH 3(g) H 2 N 2 NH 3 Equilibrium 5.0 M 8.0 M 4.0 M Initial x y 0 M Change -3z -z +z Equilibrium x-3z5.0m y-z8.0m 2z4.0M Looking at the change in NH 3 concentration will allow for z to be solved for 2z 4.0 M z 2.0 M Calculate the initial concentration of N 2 y - z 8.0 y - 2.0 8.0 y 0.0 M Calculate the initial concentration of H 2 x - 3z 8.0 x 3(2.0) 5.0 x.0 M 2

76. 3H 2(g) + N 2(g) 2NH 3(g) 2 K P P NH 3 3 5.3 05 P N2 P H2 H 2 N 2 NH 3 Initial 0 M 0 M x Change +3z +z -2z Equilibrium 3z z x-2z x 0.50x 2 z x 4 Equilibrium 0.75x 0.25x 0.50x Note when the amount of NH 3 drops to 50% so does the partial pressure K P P 2 NH 3 3 P N2 P (0.5x) 2 5.3 05 H2 (0.25x)(0.75x) 3 2.37 5.3 05 x2 x 0.002 The original partial pressure of NH 3 is 0.002atm 79. PCl 5(g) PCl 3(g) + Cl 2(g) P tot P PCl5 + P PCl3 + P Cl2 358.7torr( atm ) 0.4720 atm (at equilibrium) a) PCl 5 PCl 3 Cl 2 Initial (m) 2.456 g 0 g 0 g b) Initial (n m ) Mx 0.060 mol 0 mol 0 mol Initial (P nrt ) V 0.24894 atm 0 atm 0 atm Change -x +x +x Equilibrium 0.24894-x x x Therefore P tot P PCl5 + P PCl3 + P Cl2 0.24894 x + x + x 0.4720 x 0.223 P PCl5 0.2489 atm x 0.2489 atm 0.223 atm 0.0258 atm P PCl3 P Cl2 x 0.223 atm K P P PCl 3 P Cl2 (0.223)(0.223).93 P PCl5 0.0258 PCl 5 PCl 3 Cl 2 Equilibrium 0.0258 atm 0.223 atm 0.223 atm Add (n) 0 mol 0 mol 0.250 mol Add(P nrt ) V 0 atm 0 atm 5.36 atm Initial 0.0258 atm 0.223 atm 5.58 atm Change +x -x -x Equilibrium 0.0258+x 0.223-x 5.58-x 3

K P P PCl 3 P Cl2 (0.223 X)(5.58 X).93 P PCl5 0.0258 + x x 2 5.80x +.24.93 0.0258 + x x 2 5.80x +.24 0.0498 +.93x x 2 7.73x +.9 0 x b ± b2 4ac 7.73 ± ( 7.63)2 4()(.9) 7.57 and 0.56 2() Since the pressure cannot be negative, therefore, x 0.56 Using guess and check x 0.578 with significant figures x 0.58 Partial pressure at equilibrium P PCl5 0.0258 atm + x 0.0258 atm + 0.56 atm 0.82 atm P PCl3 0.223 atm x 0.223 atm 0.56 atm 0.067 atm P Cl2 5.58 atm x 5.58 atm 0.56 atm 5.42 atm 8. Deteremine the equation of interest N 2O 4(g) 2NO 2(g) Determine the expresion and value of K P 2 K P P NO 2 (.20)2 0.34 4.2 Determine the concentrations after volume doubled PV nrt P nrt V Therefore, if the volume is doubled the pressure is cut in half Initial (pressure after doubling volume) P NO2.20 atm 0.34 atm Determine the direction that the reaction will go Le Chatelier principle states that if the volume is increased the reaction will proceed in the direction with the greater number of moles of gas (forward reaction). Make ICE table N 2O 4 NO 2 Initial 0.7 0.600 Change -x +2x Equilibrium 0.7-x 0.600+2x Calculate the partial pressures after doubling volume K P P 2 NO 2 (0.600 + 2x)2 4.2 0.7 x 4x 2 + 2.40x + 0.360 0.7 4.2x 4x 2 + 6.6x + 0.350 0 x b ± b2 4ac 6.6 ± 43.6 ( 5.6) 0.052 and.7 8 Pressure cannot be negative, therefore, x 0.052 Partial pressures at equilibrium P NO2 0.600atm + 2x 0.600atm + 2(0.052 atm) 0.704 atm 0.7 atm x 0.7 atm 0.052 atm 0.2 atm 4

72. a) n PCl5 2.450 g ( mol PCl 5 ) 0.077 mol PCl 208.22 g PCl 5 5 P nrt L atm (0.077 mol)(0.08206 )(600. K) V 0.500 L b) PCl 5(g) PCl 3(g) + Cl 2(g) K [PCl 3 ][Cl 2 ] [PCl 5 ] P PCl 3 RT P PCl5 RT P PCl 2 RT K P (.5) (0.08206 L atm mol K RT K P.5 ) (600K) 566 PCl 5 PCl 3 Cl 2 Initial.6 atm 0 atm 0 atm Change -x +x +x Equilibrium.6-x x x xx K P.6 x 566 x 2 657 566x x 2 + 566x 657 0 x b ± b2 4ac.6 atm 566 ± 5662 4()( 657).58 and 567.2 2() Pressure cannot be negative, therefore, x.58 Using guess and check x.5763, with correct significant figures x.5 Calculate P PCl5 at equilibrium.6 atm.5 atm 0.0 atm c) Calculate partial pressure of other gases P PCl3.5 atm P Cl2.5 atm Calculate P tot P tot P PCl5 + P PCl3 + P Cl2 0.0 atm +.5 atm +.5 atm 2.3 atm d) Initial:.6 atm Final: 0.0 atm Amount Dissociated.6 atm 0.0 atm.5 atm Percent Dissociated (.5 ) 00% 99.%.6 5