Consider the reaction: CO(g)+H2O(g)<-->CO2(g)+H2 (g) Kc= 102 at 500K A reaction
ID: 799696 • Letter: C
Question
Consider the reaction: CO(g)+H2O(g)<-->CO2(g)+H2 (g) Kc= 102 at 500K A reaction mixture initially contains 0.140 M CO and 0.140 M H2O What will be the equilibrium concentration of CO? H2O? CO2? H2? Consider the reaction: CO(g)+H2O(g)<-->CO2(g)+H2 (g) Kc= 102 at 500K A reaction mixture initially contains 0.140 M CO and 0.140 M H2O What will be the equilibrium concentration of CO? H2O? CO2? H2?Explanation / Answer
CO(g) + H2O(g) -----> CO2(g) + H2(g)
Kc=102
0.14M 0.14M -------> 0 0
-x ...........-x ..................+x ...........+x
0.14-x ..0.14-x ..........x .............x
Ka = [CO2] [H2] / ([CO] [H2O])
102 = x^2 / (0.14-x)^2.... take the square root of both sides
10.10 = x / (0.14-x) .... solve for x
x = 0.1274
[CO] = [H2O] = 0.0126M
[CO2] = [H2] = 0.1274M