Asked by Gabriella
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, 0.422 mol CO2(g) is present as a result of the reaction
CO(g) + H2O(g)<->CO2(g) + H2(g).
What is Kc at 800 K?
CO(g) + H2O(g)<->CO2(g) + H2(g).
What is Kc at 800 K?
Answers
Answered by
DrBob222
(H2O) = 1/11 = 0.0909M
(CO) = 1/11 = 0.0909
(0.422/11) = 0.0384
......CO(g) + H2O(g)<->CO2(g) + H2(g)
I...0.0909...0.0909.....0
C....-x........-x.......x
E....0.0909-x.0.0909-x..x
The problems tells you that x = 0.0384.
Use that to determine (CO) and (H2O) ast equilibrium, substitute into Kc expression and solve for Kc.
(CO) = 1/11 = 0.0909
(0.422/11) = 0.0384
......CO(g) + H2O(g)<->CO2(g) + H2(g)
I...0.0909...0.0909.....0
C....-x........-x.......x
E....0.0909-x.0.0909-x..x
The problems tells you that x = 0.0384.
Use that to determine (CO) and (H2O) ast equilibrium, substitute into Kc expression and solve for Kc.
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