Calculate the equilibrium constant for the reaction,
`H_(2(g))+CO_(2(g))hArrH_(2)O_((g))+CO_((g))` at `1395 K`, if the
equilibrium constants at `1395 K` for the following are:
`2H_(2)O_((g))hArr2H_(2)+O_(2(g))` (`K_(1)=2.1xx10^(-13)`)
`2CO_(2(g))hArr2CO_((g))+O_(2(g))` (`K_(2)=1.4xx10^(-12)`)

Explanation

For `2H_(2)O_((g))hArr2H_(2(g))+O_(2(g))`
`K_(1)=([H_(2)]^(2)[O_(2)])/([H_(2)O]^(2))` ….(`1`)
For `2CO_(2(g))hArr2CO_((g))+O_(2(g))`
`K_(2)=([CO]^(2)[O_(2)])/([CO_(2)]^(2))` ….(`2`)
For `CO_(2(g))+H_(2(g))hArrH_(2)O_((g))+CO_((g))`
`K=([H_(2)O][CO])/([CO_(2)][H_(2)])` ... (`3`)
By dividing eq. (`2`) by eq. (`1`),
`(K_(2))/(K_(1))=([CO]^(2)[O_(2)])/([CO_(2)]^(2))xx([H_(2)O]^(2))/([H_(2)]^(2)[O_(2)])`
`(K_(2))/(K_(1))=([CO]^(2)[H_(2)O]^(2))/([CO_(2)]^(2)[H_(2)]^(2))=K^(2)`
or `K=sqrt(((K_(2))/K_(1)))=sqrt(((1.4xx10^(-12))/(2.1xx10^(-13))))=2.58s`