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Chemical Kinetics and Nuclear Chemistry question

2019 · 12 Jan · Shift 2 · Q10
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Chemical Kinetics and Nuclear Chemistry question

2019 · 12 Jan · Shift 2 · Q10

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
For a reaction consider the plot of ℓ\ellℓ n k versus 1/T given in the figure. If the rate constant of this reaction at 400 K is 10–5 s–1, then the rate constant at 500 K is – JEE Main 2019 (Online) 12th January Evening Slot Chemistry - Chemical Kinetics and Nuclear Chemistry Question 143 English
  1. A
    10 −-− 4 s −-− 1
  2. B
    4 ×\times× 10 −-− 4 s −-− 1
  3. C
    10 −-− 6 s −-− 1
  4. D
    2 ×\times× 10 −-− 4 s −-− 1
View written solutionFree

Correct answer: A

  1. Use Arrhenius equation in logarithmic form

For a reaction,

ln⁡k=ln⁡A−EaR⋅1T\ln k = \ln A - \frac{E_a}{R}\cdot \frac{1}{T}lnk=lnA−REa​​⋅T1​

So, a plot of ln⁡k\ln klnk versus 1/T1/T1/T is a straight line with slope

−EaR-\frac{E_a}{R}−REa​​
  1. Read the slope from the graph

From the given straight line graph, the slope is

−2000-2000−2000

Hence,

−EaR=−2000-\frac{E_a}{R} = -2000−REa​​=−2000

So,

EaR=2000\frac{E_a}{R} = 2000REa​​=2000
  1. Relate the two rate constants

Using

ln⁡(k2k1)=−EaR(1T2−1T1)\ln \left(\frac{k_2}{k_1}\right)= -\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)ln(k1​k2​​)=−REa​​(T2​1​−T1​1​)

Given:

  • k1=10−5 s−1k_1 = 10^{-5}\,\text{s}^{-1}k1​=10−5s−1 at T1=400 KT_1=400\,\text{K}T1​=400K
  • T2=500 KT_2=500\,\text{K}T2​=500K
  • EaR=2000\dfrac{E_a}{R}=2000REa​​=2000

Therefore,

ln⁡(k210−5)=−2000(1500−1400)\ln \left(\frac{k_2}{10^{-5}}\right)= -2000\left(\frac{1}{500}-\frac{1}{400}\right)ln(10−5k2​​)=−2000(5001​−4001​)
  1. Calculate the temperature term
1500−1400=4−52000=−12000\frac{1}{500}-\frac{1}{400} = \frac{4-5}{2000} = -\frac{1}{2000}5001​−4001​=20004−5​=−20001​

So,

ln⁡(k210−5)=−2000(−12000)=1\ln \left(\frac{k_2}{10^{-5}}\right)= -2000\left(-\frac{1}{2000}\right)=1ln(10−5k2​​)=−2000(−20001​)=1

Thus,

k210−5=e\frac{k_2}{10^{-5}} = e10−5k2​​=e

Hence,

k2=e×10−5≈2.718×10−5 s−1k_2 = e\times 10^{-5} \approx 2.718\times 10^{-5}\,\text{s}^{-1}k2​=e×10−5≈2.718×10−5s−1
  1. Match with the nearest option

This value is approximately

2.7×10−5 s−12.7\times 10^{-5}\,\text{s}^{-1}2.7×10−5s−1

which does not match any option exactly.

Given the available choices, none is correct. If the slope were about −4600-4600−4600 instead of −2000-2000−2000, then option A would result. So the stored answer appears inconsistent with the graph-based calculation.

Derived answer: None of the given options matches; calculated value is 2.7×10−5 s−12.7\times 10^{-5}\,\text{s}^{-1}2.7×10−5s−1.

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