Question

# The equilibrium constant of the equation, $$H_2 O(g) + CO(g) \rightleftharpoons H_2(g) + CO_2(g)$$ is $$0.44$$ at $$1259\ K$$. The value of equilibrium constant of the equation, $$H_2(g) + CO_2(g) \rightleftharpoons H_{2}O(g) + CO(g)$$ will be:

A
0.22
B
10.44
C
10.44
D
0.44

Solution

## The correct option is B $$\dfrac {1}{0.44}$$GivenEquilibrium constant is 0.44Solution$$Kc=\frac{[H_{2}]*[CO_{2}]}{[H_{2}O]*[CO]}$$As we can see reaction is reversed Therefore Equilibirium constant become 1/0.44The correct option is BChemistry

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