For the equilibrium, `2H_(2)(g)+O_(2)(g)hArr2H_(2)O(l)` at `25^(@)C,DeltaG^(@)` is -474.78kJ `mol^(-1)`. Calculate log K for it (R=8.314 `JK^(-1)mol^(-1)`)

For the equilibrium, `2H_(2)(g)+O_(2)(g)hArr2H_(2)O(l)` at `25^(@)C,De

1.	For the equilibrium, `2H_(2)(g)+O_(2)(g)hArr2H_(2)O(l)` at `25^(@)C,DeltaG^(@)` is -474.78kJ `mol^(-1)`. Calculate log K for it (R=8.314 `JK^(-1)mol^(-1)`)
Answer» Correct Answer - 83.2 `-DeltaG^(@)=2.303RT" log K"," "474780=2.303xx8.314xx298" log K or log K"=83.2`

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For the equilibrium, `2H_(2)(g)+O_(2)(g)hArr2H_(2)O(l)` at `25^(@)C,DeltaG^(@)` is -474.78kJ `mol^(-1)`. Calculate log K for it (R=8.314 `JK^(-1)mol^(-1)`)

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