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← JEE Advanced 2023 · Paper 1
ChemistryInteger+4Question 44 of 51

JEE Advanced 2023 · Paper 1, Chemistry question 10

The plot of log⁡kf\log k_f versus 1/T1/T for a reversible reaction A(g)⇌P(g)\ce{A(g) <=> P(g)} is shown.

Pre-exponential factors for the forward and backward reactions are 1015 s−110^{15}\,\mathrm{s^{-1}} and 1011 s−110^{11}\,\mathrm{s^{-1}}, respectively. If the value of log⁡K\log K for the reaction at 500 K is 6, the value of ∣log⁡kb∣|\log k_b| at 250 K is _____.

[KK = equilibrium constant of the reaction; kfk_f = rate constant of forward reaction; kbk_b = rate constant of backward reaction]

Straight line of log k_f (y-axis) against 1/T in K^-1 (x-axis), sloping downward. A dashed guide marks the point on the line at 1/T = 0.002 K^-1, where log k_f = 9. No other values are labelled.
Answer and worked solution

Official answer: 5

Write log⁡k=log⁡A−E2.303R⋅1T\log k=\log A-\dfrac{E}{2.303R}\cdot\dfrac1T.

Forward reaction. At 1/T=0.0021/T=0.002, log⁡kf=9\log k_f=9:
9=15−0.002 Ef2.303R  ⟹  Ef2.303R=3000.9=15-0.002\,\frac{E_f}{2.303R}\implies\frac{E_f}{2.303R}=3000.

Backward reaction at 500 K. K=kf/kbK=k_f/k_b, so log⁡kb=9−6=3\log k_b=9-6=3. Then
3=11−0.002 Eb2.303R  ⟹  Eb2.303R=4000.3=11-0.002\,\frac{E_b}{2.303R}\implies\frac{E_b}{2.303R}=4000.

At 250 K (1/T=0.0041/T=0.004):
log⁡kb=11−4000×0.004=−5  ⟹  ∣log⁡kb∣=5.\log k_b=11-4000\times0.004=-5\implies|\log k_b|=5.

Topics:Arrhenius equationLaw of mass action and Kc, Kp
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