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Chemical Equilibrium question

2023 · Shift 1 · Q10
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Chemical Equilibrium question

2023 · Shift 1 · Q10

JEE AdvancedChemistryChemical EquilibriumNumerical+4 / −1
The plot of log⁡kf\log k_flogkf​ versus 1/T1 / T1/T for a reversible reaction A(g)⇌P(g)\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{P}(\mathrm{g})A(g)⇌P(g) is shown. JEE Advanced 2023 Paper 1 Online Chemistry - Chemical Equilibrium Question 1 English Pre-exponential factors for the forward and backward reactions are 1015 s−110^{15} \mathrm{~s}^{-1}1015 s−1 and 1011 s−110^{11} \mathrm{~s}^{-1}1011 s−1, respectively. If the value of log⁡K\log KlogK for the reaction at 500 K500 \mathrm{~K}500 K is 6 , the value of ∣log⁡kb∣\left|\log k_b\right|∣logkb​∣ at 250 K250 \mathrm{~K}250 K is ‾\underline{\hspace{2cm}}​. [K= equilibrium constant of the reaction kf= rate constant of forward reaction kb= rate constant of backward reaction ]\begin{aligned} & {[K=\text { equilibrium constant of the reaction }} \\\\ & k_f=\text { rate constant of forward reaction } \\\\ & \left.k_b=\text { rate constant of backward reaction }\right] \end{aligned}​[K= equilibrium constant of the reaction kf​= rate constant of forward reaction kb​= rate constant of backward reaction ]​
Numerical answer
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Correct answer: 5

  1. Use Arrhenius form for forward and backward reactions

For a reversible reaction,

kf=Afe−Ef/RT,kb=Abe−Eb/RT k_f = A_f e^{-E_f/RT}, \qquad k_b = A_b e^{-E_b/RT}kf​=Af​e−Ef​/RT,kb​=Ab​e−Eb​/RT

with Af=1015 s−1,Ab=1011 s−1.A_f=10^{15}\ \text{s}^{-1}, \qquad A_b=10^{11}\ \text{s}^{-1}.Af​=1015 s−1,Ab​=1011 s−1.

Also, K=kfkb.K=\frac{k_f}{k_b}.K=kb​kf​​.

  1. Extract slope information from the graph of log⁡kf\log k_flogkf​ vs 1/T1/T1/T

For Arrhenius equation in base-10 logarithm,

log⁡kf=log⁡Af−Ef2.303R⋅1T.\log k_f = \log A_f - \frac{E_f}{2.303R}\cdot \frac{1}{T}.logkf​=logAf​−2.303REf​​⋅T1​.

From the given straight-line plot, the line passes through:

  • at 1/T=01/T=01/T=0, log⁡kf=15\log k_f = 15logkf​=15 (since intercept is log⁡Af=15\log A_f = 15logAf​=15),
  • and from the graph, at 1/T=0.004 K−11/T = 0.004\ \text{K}^{-1}1/T=0.004 K−1, log⁡kf=3\log k_f = 3logkf​=3.

Hence slope is

m=3−150.004−0=−120.004=−3000.m = \frac{3-15}{0.004-0} = \frac{-12}{0.004} = -3000.m=0.004−03−15​=0.004−12​=−3000.

Thus,

log⁡kf=15−3000(1T).\log k_f = 15 - 3000\left(\frac{1}{T}\right).logkf​=15−3000(T1​).
  1. Find log⁡kf\log k_flogkf​ at 500 K500\,\text{K}500K

At T=500 KT=500\,\text{K}T=500K,

1T=0.002.\frac{1}{T}=0.002.T1​=0.002.

So,

log⁡kf=15−3000(0.002)=15−6=9.\log k_f = 15 - 3000(0.002)=15-6=9.logkf​=15−3000(0.002)=15−6=9.
  1. Use the given value of log⁡K\log KlogK at 500 K500\,\text{K}500K

Given,

log⁡K=6.\log K = 6.logK=6.

Since

K=kfkb,K=\frac{k_f}{k_b},K=kb​kf​​,

we have

log⁡K=log⁡kf−log⁡kb.\log K = \log k_f - \log k_b.logK=logkf​−logkb​.

Therefore at 500 K500\,\text{K}500K,

6=9−log⁡kb6 = 9 - \log k_b6=9−logkb​

which gives

log⁡kb=3.\log k_b = 3.logkb​=3.
  1. Determine activation term for backward reaction

For backward reaction,

log⁡kb=log⁡Ab−Eb2.303R⋅1T.\log k_b = \log A_b - \frac{E_b}{2.303R}\cdot \frac{1}{T}.logkb​=logAb​−2.303REb​​⋅T1​.

Since log⁡Ab=11\log A_b=11logAb​=11, at 500 K500\,\text{K}500K,

3=11−Eb2.303R⋅1500.3 = 11 - \frac{E_b}{2.303R}\cdot \frac{1}{500}.3=11−2.303REb​​⋅5001​.

So,

Eb2.303R⋅1500=8\frac{E_b}{2.303R}\cdot \frac{1}{500}=82.303REb​​⋅5001​=8

which implies

Eb2.303R=4000.\frac{E_b}{2.303R}=4000.2.303REb​​=4000.

Hence,

log⁡kb=11−4000(1T).\log k_b = 11 - 4000\left(\frac{1}{T}\right).logkb​=11−4000(T1​).
  1. Find log⁡kb\log k_blogkb​ at 250 K250\,\text{K}250K

At T=250 KT=250\,\text{K}T=250K,

1T=0.004.\frac{1}{T}=0.004.T1​=0.004.

Therefore,

log⁡kb=11−4000(0.004)=11−16=−5.\log k_b = 11 - 4000(0.004)=11-16=-5.logkb​=11−4000(0.004)=11−16=−5.

So,

∣log⁡kb∣=5.|\log k_b|=5.∣logkb​∣=5.
  1. Final answer
5\boxed{5}5​
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