Equilibrium constants for the following reactions at 1200 K are given :
`2H_2O(g)hArr2H_2(g)+O_2(g):K_1=6.4xx10^(-8)`
`2CO(g)hArr2CO(g)+O_2(g):K_2=1.6xx10^(-6)`
The equilibrium constant for the reaction `H_2(g)+CO_2(g)hArrCO(g)+H_2O(g)` at 1200K will be-

Options

A. 0.05
B. 20
C. 0.2
D. 5

Explanation

`2H_2O(g) `2CO_2(g) Reverse reaction of equation (1)
`2H_2(g)+O_2(g) `K_3=1/(K_1)=1/(6.4xx10^(-8))`
Adding equation (2) and (3)
`2H_2(g)+2CO_2(g) `K_4=K_3xxK_2=(1xx1.6xx10^(-6))/(6.4xx10^(-8))=(100)/(4)=25`
Dividing equation (4) by 2
`2H_2(g)+CO_2(g) `K_5=sqrt(K_4)=sqrt(25)=5`