CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
10.44
loader
C
10.44
loader
D
0.44
loader

Solution

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

Chemistry

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image