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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)`)

Text Solution

Verified by Experts

For `2H_(2)O hArr 2H_(2)+O_(2)`
`K_(1)=([H_(2)]^(2)[O_(2)])/([H_(2)O]^(2)) …(i)`
For `2CO_(2) hArr 2CO+O_(2)`
`K_(2)=([CO]^(2)[O_(2)])/([CO_(2)]^(2)) …(ii)`
For `CO_(2)+H_(2) hArr H_(2)O+CO`
`K=([H_(2)O][CO])/([CO_(2)][H_(2)]) ...(iii)`
Thus, dividing equations (ii) by (i), we get
`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)` [By equation (iii)]
or `K=(K_(2)/K_(1))^(1//2)=((1.4xx10^(-12))/(2.1xx10^(-13)))^(1//2)=2.58`
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