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← JEE Advanced 2018 · Paper 2
ChemistryNumerical+3Question 31 of 54

JEE Advanced 2018 · Paper 2, Chemistry question 13

Consider the following reversible reaction,

A(g)+B(g)⇌AB(g).\mathrm{A(g) + B(g) \rightleftharpoons AB(g)}.

The activation energy of the backward reaction exceeds that of the forward reaction by 2RT2RT (in J mol−1^{-1}). If the pre-exponential factor of the forward reaction is 4 times that of the reverse reaction, the absolute value of ΔG⊖\Delta G^\ominus (in J mol−1^{-1}) for the reaction at 300 K is ____.

(Given; ln(2) = 0.7, RTRT = 2500 J mol−1^{-1} at 300 K and GG is the Gibbs energy)

Answer and worked solution

Official answer: 8430

Equilibrium constant.
K=kfkb=AfAbe(Eb−Ef)/RT=4e2RT/RT=4e2.K=\frac{k_f}{k_b}=\frac{A_f}{A_b}e^{(E_b-E_f)/RT}=4e^{2RT/RT}=4e^2.

Gibbs energy.
ΔG∘=−RTln⁡K=−2500(ln⁡4+2)=−2500(1.4+2)=−8500 J mol−1.\Delta G^\circ=-RT\ln K=-2500(\ln4+2)=-2500(1.4+2)=-8500\ \mathrm{J\,mol^{-1}}.

Absolute value: 8500.

Answer: 8500 (key: 8430 to 8570)