Calculate the value of Δ*G*° = -RTln*K* where T = 298K and lnK = 3.19

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`DeltaG^o` is the Gibbs free energy.

`DeltaG^o = -RTlnK`

`R = 0.0083144621 kJK^(-1)mol^(-1)`

`DeltaG^o = -0.0083144621 kJK^(-1)mol^(-1) xx 298 K xx 3.19`

`DeltaG^o = -7.903 kJmol^(-1)`

**Therefore, the gibbs free energy difference is -7.903 kJ/mol**

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