Asked by Anonymous
A mixture consisting of 1 mol of H2O (g) and 1 mol CO (g) is placed in a 11 L reaction vessel at 800 K. At equilibrium 0.647 mol CO2 (g) is present as a result of the reaction:
CO(g) + H2O (g) --> CO2(g) + H2
What is K(equilibrium constant) at 800 K?
CO(g) + H2O (g) --> CO2(g) + H2
What is K(equilibrium constant) at 800 K?
Answers
Answered by
DrBob222
...........CO + H2O ==> H2 + CO2
initial....1.....1.......0....0
change.....-x....-x.....x.....x
equil.....1-x....1-x.....x....0.647
You know x = 0.647 which allows you to calculate moles CO, H2O, H2, and CO2 at equil. Then concn = moles/L, and substitute into Keq expression to solve for Keq.
initial....1.....1.......0....0
change.....-x....-x.....x.....x
equil.....1-x....1-x.....x....0.647
You know x = 0.647 which allows you to calculate moles CO, H2O, H2, and CO2 at equil. Then concn = moles/L, and substitute into Keq expression to solve for Keq.
Answered by
Mr. ShaW
DrBob222 the correct answer is Kc = 69 moles per solution of pk
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