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What is the value of equilibrium constant for a reaction when \Delta G^0= -21.28 kj

Option: 1

e^{3.73}


Option: 2

10^{3.73}


Option: 3

10^{2.59}


Option: 4

e^{2.59}


Answers (1)

best_answer

As we have learned

Standard Gibbs Energy -

 \Delta _{r}G^{0} = -RT\ ln\ k

- wherein

K= equilibrium constant of the reaction

 

 \Delta G^0= -RTlnK

-21.28*10^{3}= -8.314 * 298 * 2.303 log k

K = 10^{3.73}

Posted by

jitender.kumar

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