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

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

Text Solution

Verified by Experts

The correct Answer is:
83.21 at `25^@C`
`log K=-(DeltaG^@)/"2.303 RT"`
`=-(-474.78xx10^3)/(2.303xx8.314xx298)`
=83.21
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