chapter14_part2

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

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Unformatted text preview: 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 lets 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 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....
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This note was uploaded on 07/15/2009 for the course CHE 131 taught by Professor Kerber during the Fall '08 term at SUNY Stony Brook.

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

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