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For the water gas reaction, C(s) +H(2)...

For the water gas reaction,
`C(s) +H_(2)O(g) hArr CO(g)+H_(2)(g)`
the standard Gibbs free energy of reaction (at `1000K)` is `-8.1 kJ mol^(-1)`. Calculate its equilibrium constant.

Text Solution

Verified by Experts

We know that
`K="antilog"((-DeltaG^(@))/(2.303RT))` . . . .(i)
Given that, `DeltaG^(@)=-8.1kJ//mol`
`R=8.314xx10^(-3)kJ" "K^(-1)mol^(-1)`
Substituting these values in eq. (i), we get
`K="antilog"[(-(-8.1))/(2.303xx8.314xx10^(-3)xx1000)]`
`K=2.65`
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