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For the equilibrium reaction : 2H(2)(g) ...

For the equilibrium reaction : `2H_(2)(g) + O_(2)(g) iff 2H_(2)O(l)` at 298 K, `DeltaG^(@) = - 474.78 kJ "mol"^(-1)`. Calculate log K for it.
`(R = 8.314 J K^(-1) "mol"^(-1))`.

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Verified by Experts

The correct Answer is:
`83.21`
`DeltaG^(@) = -2.303 RT "log K`
or `log K = -(DeltaG^(@))/(2.303 RT)`
`DeltaG^(@) = -474.78 kJ "mol"^(-1), R = 8.314 J K^(-1) "mol"^(-1)`
T = 298 K
`therefore log K = (-474.78 xx 10^(3)J "mol"^(-1))/(2..3030 xx 8.314 xx 298)`
` = 83.21` .
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