Calculate the equilibrium constant for the reaction
`H_(2)(g)+CO_(2)(g) hArr H_(2)O(g)+CO(g)` at `1395K`
If the equilibrium constants at `1395 K` for the following are:
`2H_(2)O(g) hArr 2H_(2)+O_(2)(g), K_(1)=2.1xx10^(-13)`
`2CO_(2)(g) hArr 2CO(g)+O_(2)(g), K_(2)=1.4xx10^(-12)`

Answer

`2.58`

Explanation

`H_(2)(g)+CO_(2)(g)hArrH_(2)O(g)+CO(g)`.
`2H_(2)O(g)hArr2H_(2)(g)+O_(2)(g) K_(1)=2.1xx10^(-13)` ....(1)
`2CO_(2)(g) hArr2CO(g)+O_(2)(g) K_(2)=1.4xx10^(-12)` ....(2)
`(1)/(2)"eq".(2)-(1)/(2)"eq".(1)`
`CO_(2)(g)-H_(2)O(g)hArrCO(g)+(1)/(2)O_(2)(g)`.
`CO_(2)(g)+H_(2)(g)hArrH_(2)O(g)+CO(g)`.
`(1)/(2)[eq. (2)-eq. (1)] rArr K=((K_(2))/K_(1))^(1//2)=((1.4xx10^(-12))/(2.1xx10^(-13)))^(1//2)=((14)/2.1)^(1//2)=2.58`.