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Question: For a reversible reaction at 298 K the equilibrium constant K is 200. What is the value of \(\Delta...

For a reversible reaction at 298 K the equilibrium constant K is 200. What is the value of ΔG0\Delta G ^ { 0 } at 298 K?

A

−13.13kcal- 13.13 \mathrm { kcal }

B

−0.13kcal- 0.13 \mathrm { kcal }

C

D

−0.413kcal- 0.413 \mathrm { kcal }

Answer

Explanation

Solution

: Applying ΔGo=−2.303RT×log⁡K\Delta \mathrm { G } ^ { \mathrm { o } } = - 2.303 \mathrm { RT } \times \log \mathrm { K } ,

=−2.303×2×298×log⁡200= - 2.303 \times 2 \times 298 \times \log 200

=−3158.4cal=−3.158kcal= - 3158.4 \mathrm { cal } = - 3.158 \mathrm { kcal }