Chem 162-2010 homework 5th week

# Chem 162-2010 - Hill Petrucci Chapter 14 Homework Problems 5th Week CHAPTER 14 Equilibrium Constant Relationships 19 Determine the values of Kc

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th Week CHAPTER 14 Equilibrium Constant Relationships 19. Determine the values of K c that correspond to the following values of K p . (a) SO 2 Cl 2 (g) SO 2 (g) + Cl 2 (g) K p = 2.9 x 10 -2 at 303K (b) 2NO 2 (g) 2NO(g) + O 2 K p = 0.275 at 700K (c) CO(g) + Cl 2 (g) COCl 2 (g) K p = 22.5 at 395 o C K p inatm = K c inM (RT) ∆ngas K c inM = K p inatm /(RT) ∆ngas (a) Kc = (2.9 x 10 -2 )/((0.08205 x 303) (2-1) ) = 1.17 x 10 -3 (b) Kc = (0.275)/((0.08205 x 700) (3-2) ) = 4.79 x 10 -3 (c) Kc = (22.5)/((0.08205 x 668) (1-2) ) = 1.23 x 10 3 25. What is the value of K p at 298K for the reaction ½ CH 4 (g) + H 2 O(g) ½ CO 2 (g) + 2H 2 (g), given the following data at 298K? CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) K p = 9.80 x 10 -6 CO(g) + 3H 2 (g) CH 4 (g) + H 2 O(g) K p = 8.00 x 10 24 Use Hess’ Law: Reverse and take half of the top equation; Take the inverse square root of Kp. ½ CO(g) + ½ H 2 O(g) ½ CO 2 (g) + ½ H 2 (g) K p = (1/(9.80 x 10 -6 )) 0.5 = (3.19 x 10 2 ) Reverse and take half of the top equation; Take the inverse square root of Kp. ½ CH 4 (g) + ½ H 2 O(g) ½ CO(g) + 3/2 H 2 (g) K p = (1/(8.00 x 10 24 )) 0.5 = (3.54 x 10 -13 ) Add the equations and multiply the K’s. ½ CH 4 (g) + H 2 O(g) ½ CO 2 (g) + 2H 2 (g) Kp = (3.19 x 10 2 ) x (3.54 x 10 -13 ) = 1.13 x 10 -10 27. Determine K c at 298K for the reaction 2CH 4 (g) C 2 H 2 (g) + 3H 2 (g), given the following data at 298K (1) CH 4 (g) + H 2 O(g) CO(g) + 3H 2 (g) K p = 1.2 x 10 -25 (2) 2C 2 H 2 (g) + 3O 2 (g) 4CO(g) + 2H 2 O(g)K p = 1.1 x 10 2 1

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2 + ½ O 2 (g) H 2 O(g) K p = 1.1 x 10 40 Double the first equation and square the K (1) 2CH 4 (g) + 2H 2 O(g) 2CO(g) + 6H 2 (g) K p = (1.2 x 10 -25 ) 2 = 1.44 x 10 -50 Halve and reverse the second equation, and take the square root of the inverse of K. (2) 2CO(g) + H 2 O(g) C 2 H 2 (g) + 3/2O 2 (g) K p = (1/(1.1 x 10 2 )) 0.5 = 9.53 x 10 -2 Triple the third equation, and take the cube of K. (3) 3H 2 + 3/2 O 2 (g) 3H 2 O(g) K p = (1.1 x 10 40 ) 3 = 1.33 x 10 120 Add the equations and multiply the K’s. 2CH 4 (g) C 2 H 2 (g) + 3H 2 (g) Kp = 1.83 x 10 69 K cinM = K pinatm /(RT) ∆ngas Kc = (1.83 x 10 69 )/(0.08205 x 298) (4-2) = 3.06 x 10 66 35. For the reaction H 2 S(g) + I 2 (s) S(s) + 2HI(g), K p = 1.33 x 10 -5 at 333K. What will be the total pressure of the gases above an equilibrium mixture of P HI = 0.010 x P H2S ? H 2 S(g) + I 2 (s) S(s) + 2HI(g) H 2 S(g) + I 2 (s) S(s) + 2HI(g) Initial Change Equilibrium X 0.010X Kp = ([HI] 2 )/[H 2 S] = 1.33 x 10 -5 ([0.010X] 2 )/[X] = 1.33 x 10 -5 X = 0.133 = P H2S 0.010X = 0.00133 = P HI Ptotal = 0.133 + 0.00133 = 0.134 atm 79. An analysis of the gaseous phase [S 2 (g) and CS 2 (g)] present at equilibrium at 1009 o C in the reaction C(s) + S 2 (g) CS 2 (g) shows it to be 13.71% C and 86.29% S, by mass.
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## This document was uploaded on 11/02/2011 for the course GEN CHEM 162 at Rutgers.

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Chem 162-2010 - Hill Petrucci Chapter 14 Homework Problems 5th Week CHAPTER 14 Equilibrium Constant Relationships 19 Determine the values of Kc

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