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The equilibrium constant for the reactio...

The equilibrium constant for the reaction
`CO_(2)(g) +H_(2)(g) hArr CO(g) +H_(2)O(g) at 298 K` is `73`. Calculate the value of the standard free enegry change `(R =8.314 J K^(-1)mol^(-1))`

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The correct Answer is:
`-10.63 kJ "mol"^(-1)`
`DeltaG^(@) = -2.300 RT "log K`
` K = 73, T = 273 + 25 = 298 K`
`DeltaG^(@) = -2.303 xx (8.314 J K^(-1) "mol"^(-1)) xx (298 K) log 73`
` = - 5705.8 xx 1.863 = - 10.36 kJ "mol"^(-1)`.
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