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← IIT-JEE 2007 · Paper 1
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IIT-JEE 2007 · Paper 1, Chemistry question 31

The value of log⁡10K\log_{10}K for a reaction A⇌B\ce{A <=> B} is

(Given: ΔrH298K∘=−54.07\Delta_rH^\circ_{298K}=-54.07 kJ mol⁻¹, ΔrS298K∘=10\Delta_rS^\circ_{298K}=10 J K⁻¹ mol⁻¹ and R=8.314R=8.314 J K⁻¹ mol⁻¹; 2.303×8.314×298=57052.303\times8.314\times298=5705)

Choose one option
Answer and worked solution

Official answer: B

Gibbs energy. ΔG∘=ΔH∘−TΔS∘=−54070−298(10)=−57050\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ=-54070-298(10)=-57050 J mol⁻¹.

Equilibrium constant. ΔG∘=−2.303RTlog⁡K\Delta G^\circ=-2.303RT\log K:
log⁡K=570505705=10.\log K=\frac{57050}{5705}=10.

(B)

Topics:Equilibrium and Gibbs energyGibbs energy and spontaneity
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