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chapter14_part2

# chapter14_part2 - Chapter 14 Chemical Equilibrium 657 52...

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Chapter 14: Chemical Equilibrium 657 52. Answer: 2.19% Strategy and Explanation: Use a method similar to that described in the answer to Question 44(b). (conc. N 2 ) = (conc. H 2 ) = 1.00 mol N 2 O 4 /5.00 L = 0.200 M N 2 (g) + 3 H 2 (g) 2 NH 3 (g) initial conc. (M) 0.200 0.200 0 change as reaction occurs (M) – x – 3x + 2x equilibrium conc. (M) 0.200 – x 0.200 – 3x 2x At equilibrium K c = NH 3 [ ] 2 N 2 [ ] H 2 [ ] 3 = (2x) 2 (0.200 " x)(0.200 " 3x) 3 = 5.97 × 10 –2 The reactant-favored reaction will have larger reactant concentrations than product concentrations, so let’s assume x is small, such that subtraction from the reactant concentrations is negligible: 0.200 – x 0.200, and 0.200– 3x 0.200. (2x) 2 (0.200)(0.200) 3 = 5.97 × 10 –2 However, this value of x is not very small, so the assumption may not be a good one. Using the method described in Appendix A.7, plug this approximate value of x back into the equation in the places where we ignored it, to obtain a more accurate value. Repeat the procedure until the value of x stops changing: x = 5.97 " 10 # 2 0.200 # 0.00489 ( ) 0.200 # 3(0.00489) ( ) 3 4 = 0.00431 x = 5.97 " 10 # 2 0.200 # 0.00431 ( ) 0.200 # 3(0.00431) ( ) 3 4 = 0.00437 x = 5.97 " 10 # 2 0.200 # 0.00437 ( ) 0.200 # 3(0.00437) ( ) 3 4 = 0.00437 Percentage N 2 converted = amount of N 2 converted initial amount of N 2 " 100 % = 0.00437 M 0.200 " 100 % = 2.19 % Reasonable Answer Check: Substituting the equilibrium concentrations into the equilibrium expression will reproduce K c : [N 2 ] = 0.200 – x = 0.196 M, [H 2 ] = 0.100 – 3x = 0.187 M, [NH 3 ] = 2x = 0.00874 M K c = (0.00874) 2 (0.196)(0.187) 3 = 5.96 × 10 –2 53 . Answer: (a) No (b) proceeds toward products Strategy and Explanation: Given an equation for a reaction, the initial moles of the gaseous reactant in a known volume, and the value of the equilibrium constant, determine if the system is at equilibrium and, if it is not, determine which direction it must proceed to reach equilibrium. Follow the procedure given in Section 14.5. Calculate Q and compare it with K. If Q = K, then the system is at equilibrium. If Q < K, the reaction proceeds toward products to reach equilibrium; if Q > K, the reaction proceeds toward reactants to reach equilibrium. K c = 3.58 × 10 –3 (conc. SO 3 ) = 0.15 mol/10.0 L = 0.015 M (conc. SO 2 ) = 0.015 mol/10.0 L = 0.0015 M (conc. O 2 ) = 0.0075 mol/10.0 L = 0.00075 M

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Chapter 14: Chemical Equilibrium 658 Q c = (conc. SO 2 ) 2 (conc. O 2 ) (conc. SO 3 ) 2 = (0.0015) 2 (0.00075) (0.015) 2 = 7.5 " 10 # 6 (a) Q c K c , so the reaction is not at equilibrium. (b) Q c < K c , so the reaction must proceeds toward products to reach equilibrium. Reasonable Answer Check: Considering the powers of 10, the size of Q makes sense (10 " 3 ) 2 (10 " 4 ) (10 " 2 ) 2 = (10 " 6 )(10 " 4 ) (10 " 4 ) = 10 " 6 54. Answer: (a) No (b) proceeds toward products Strategy and Explanation: Follow the method given in the solution to Question 53. K c = 9.4 × 10 –1 (conc. CH 4 ) = 0.25 mol/10.0 L = 0.025 M (conc. H 2 O) = 0.15 mol/10.0 L = 0.015 M (conc. CO) = 0.25 mol/10.0 L = 0.025 M (conc. H 2 ) = 0.15 mol/10.0 L = 0.015 M Q c = (conc. CO)(conc. H 2 ) 3 (conc. CH 4 )(conc. H 2 O) = (0.025)(0.015) 3 (0.025)(0.015) = 2.3 " 10 # 4 (a) Q c K c , so the reaction is not at equilibrium. (b) Q c < K c , so the reaction must proceeds toward products to reach equilibrium. Reasonable Answer Check: Considering the powers of 10, the size of Q makes sense: (10 " 2 )(10 " 2 ) 3 (10 " 2 )(10 " 2 ) = (10 " 8 ) (10 " 4 ) = 10 " 4 55 . Answer: (a) No (b) proceed toward reactants (c) [N 2 ] = [O 2 ] = 0.002 M; [NO] = 0.061 M Strategy and Explanation: Given an equation for a reaction, the initial moles of the gaseous reactant in a known
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chapter14_part2 - Chapter 14 Chemical Equilibrium 657 52...

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