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At 25^(@)C, the standard free energy cha...

At `25^(@)C`, the standard free energy change for a reaction is 5.4 kJ. Calculate the value of equilibrium constant of the reaction at the temperature.

Text Solution

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We know,, `DeltaG^(0)=-RTlnK` .
We have, `DeltaG^(0)=5.4kJ=5.4xx10^(3)J and T=298K`
`therefore 5.4xx10^(3)=-8.314xx298xx2.303logK`
or `logK=-0.95" "therefore K=0.113`
Therefore, at `25^(@)C` the value of equilibrium constant for the gvien reaction will be 0.113.
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