The equilibrium constant of a reaction is 10. What will be the value of $Δ G^{0}$ at 300 K?

- 5.74 kJ

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- 574 kJ

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+ 11.48 kJ

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+5.74 kJ

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Solution

The correct option is D - 5.74 kJ
$Δ G$ = -2.303 RT log K
= $- 2.303 \times 8.314 \times 300 l o g 10$
=-5.74 kJ

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The equilibrium constant of a reaction is 10. What will be the value of ΔG0 at 300 K?

The correct option is D - 5.74 kJΔG= -2.303 RT log K = −2.303×8.314×300 log10=-5.74 kJ

The equilibrium constant of a reaction is 10. What will be the value of $Δ G^{0}$ at 300 K?

The correct option is D - 5.74 kJ
$Δ G$ = -2.303 RT log K
= $- 2.303 \times 8.314 \times 300 l o g 10$
=-5.74 kJ