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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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We know, `DeltaG^(Theta) =- 2.303 RT log K_(c)`
Given, `K_(c) = 70, R = 8.314 J K^(-1) mol^(-1), T = 298 K`
Therefore, form the above equation
`DeltaG^(Theta) = - 2.303 xx 8.314 xx 298 log_(10)70 =- 10527 kJ`
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