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Chemical Kinetics and Nuclear Chemistry question

2021 · 25 Feb · Shift 1 · Q23
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Chemical Kinetics and Nuclear Chemistry question

2021 · 25 Feb · Shift 1 · Q23

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
For the reaction, aA + bB →\to→ cC + dD, the plot of log k vs 1T{1 \over T}T1​ is given below : JEE Main 2021 (Online) 25th February Morning Shift Chemistry - Chemical Kinetics and Nuclear Chemistry Question 117 English The temperature at which the rate constant of the reaction is 10-4 s-1 is ‾\underline{\hspace{2cm}}​ K. (Rounded off to the nearest integer) [Given : The rate constant of the reaction is 10-5 s-1 at 500 K.]
Numerical answer
View written solutionFree

Correct answer: 526

  1. Use the Arrhenius equation in logarithmic form

For a reaction,

log⁡k=log⁡A−Ea2.303R⋅1T\log k = \log A - \frac{E_a}{2.303R}\cdot \frac{1}{T}logk=logA−2.303REa​​⋅T1​

So, a plot of log⁡k\log klogk vs 1T\dfrac{1}{T}T1​ is a straight line with slope

−Ea2.303R-\frac{E_a}{2.303R}−2.303REa​​
  1. Read the slope from the given graph

From the graph, the straight line gives slope approximately

m=−3000m = -3000m=−3000

Hence,

log⁡k=log⁡A−3000(1T)\log k = \log A - 3000\left(\frac{1}{T}\right)logk=logA−3000(T1​)
  1. Use the two-temperature form of Arrhenius equation

For two states,

log⁡(k2k1)=−Ea2.303R(1T2−1T1)\log\left(\frac{k_2}{k_1}\right)= -\frac{E_a}{2.303R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)log(k1​k2​​)=−2.303REa​​(T2​1​−T1​1​)

Since slope magnitude is 300030003000,

log⁡(k2k1)=−3000(1T2−1T1)\log\left(\frac{k_2}{k_1}\right)= -3000\left(\frac{1}{T_2}-\frac{1}{T_1}\right)log(k1​k2​​)=−3000(T2​1​−T1​1​)
  1. Substitute the given data

Given:

  • k1=10−5 s−1k_1 = 10^{-5}\,\text{s}^{-1}k1​=10−5s−1 at T1=500 KT_1=500\,\text{K}T1​=500K
  • k2=10−4 s−1k_2 = 10^{-4}\,\text{s}^{-1}k2​=10−4s−1 at T2=?T_2=?T2​=?

Then,

log⁡(10−410−5)=log⁡(10)=1\log\left(\frac{10^{-4}}{10^{-5}}\right)=\log(10)=1log(10−510−4​)=log(10)=1

So,

1=−3000(1T2−1500)1 = -3000\left(\frac{1}{T_2}-\frac{1}{500}\right)1=−3000(T2​1​−5001​) 1T2−1500=−13000\frac{1}{T_2}-\frac{1}{500} = -\frac{1}{3000}T2​1​−5001​=−30001​ 1T2=1500−13000\frac{1}{T_2} = \frac{1}{500} - \frac{1}{3000}T2​1​=5001​−30001​ 1T2=6−13000=53000=1600\frac{1}{T_2} = \frac{6-1}{3000} = \frac{5}{3000} = \frac{1}{600}T2​1​=30006−1​=30005​=6001​

Thus,

T2=600 KT_2 = 600\,\text{K}T2​=600K
  1. Comparison with stored answer

My derived answer is 600 K600\,\text{K}600K, whereas the stored correct answer is 526 K526\,\text{K}526K.

If the graph actually has a different slope than −3000-3000−3000, then the answer could change. In fact, for the answer to be 526 K526\,\text{K}526K, the slope would need to be about −5000-5000−5000. So the result depends critically on the graph reading.

Using the standard interpretation that the graph slope is −3000-3000−3000, the correct temperature is:

600 K600\,\text{K}600K
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