The equilibrium constant for a reaction is `10`. What will be the value of `DeltaG^(Θ)`? `R=8.314 J K^(-1) mol^(-1), T=300 K`.A. `+5.527 KJ mol^(-1)`B. `-5.527 KJ mol^(-1)`C. `+55.27 KJ mol^(-1)`D. `-55.27 KJ mol^(-1)`

The equilibrium constant for a reaction is `10`. What will be the valu

1.	The equilibrium constant for a reaction is `10`. What will be the value of `DeltaG^(Θ)`? `R=8.314 J K^(-1) mol^(-1), T=300 K`.A. `+5.527 KJ mol^(-1)`B. `-5.527 KJ mol^(-1)`C. `+55.27 KJ mol^(-1)`D. `-55.27 KJ mol^(-1)`
Answer» Correct Answer - B `Delta G^(@)=-2.303` RT log K `=-2.303xx8 5 10^(-3)xx300xxlog 10` `Delta G^(@)=-5.527 kJ mol^(-1)`

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The equilibrium constant for a reaction is `10`. What will be the value of `DeltaG^(Θ)`? `R=8.314 J K^(-1) mol^(-1), T=300 K`.A. `+5.527 KJ mol^(-1)`B. `-5.527 KJ mol^(-1)`C. `+55.27 KJ mol^(-1)`D. `-55.27 KJ mol^(-1)`

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