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Chemical Kinetics and Nuclear Chemistry question

2022 · 27 Jun · Shift 2 · Q21
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Chemical Kinetics and Nuclear Chemistry question

2022 · 27 Jun · Shift 2 · Q21

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
It has been found that for a chemical reaction with rise in temperature by 9 K the rate constant gets doubled. Assuming a reaction to be occurring at 300 K, the value of activation energy is found to be ‾\underline{\hspace{2cm}}​ kJ mol −-− 1. [nearest integer] (Given ln10 = 2.3, R = 8.3 J K −-− 1 mol −-− 1, log 2 = 0.30)
Numerical answer
View written solutionFree

Correct answer: 59

  1. Use Arrhenius equation in two-temperature form

For a reaction,

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

Here:

  • rate constant doubles, so k2k1=2\dfrac{k_2}{k_1}=2k1​k2​​=2
  • initial temperature T1=300 KT_1=300\,\text{K}T1​=300K
  • rise in temperature =9 K=9\,\text{K}=9K, so T2=309 KT_2=309\,\text{K}T2​=309K

Thus,

ln⁡2=EaR(1300−1309)\ln 2=\frac{E_a}{R}\left(\frac{1}{300}-\frac{1}{309}\right)ln2=REa​​(3001​−3091​)
  1. Simplify the temperature term
1300−1309=309−300300×309=992700=110300\frac{1}{300}-\frac{1}{309}=\frac{309-300}{300\times 309}=\frac{9}{92700}=\frac{1}{10300}3001​−3091​=300×309309−300​=927009​=103001​

So,

ln⁡2=EaR⋅110300\ln 2=\frac{E_a}{R}\cdot \frac{1}{10300}ln2=REa​​⋅103001​

Hence,

Ea=R (ln⁡2) (10300)E_a=R\,(\ln 2)\,(10300)Ea​=R(ln2)(10300)
  1. Evaluate ln⁡2\ln 2ln2 from given data

Given:

ln⁡10=2.3,log⁡2=0.30\ln 10=2.3,\qquad \log 2=0.30ln10=2.3,log2=0.30

Since,

ln⁡2=(log⁡2)(ln⁡10)=0.30×2.3=0.69\ln 2=(\log 2)(\ln 10)=0.30\times 2.3=0.69ln2=(log2)(ln10)=0.30×2.3=0.69
  1. Substitute values
Ea=8.3×0.69×10300 J mol−1E_a=8.3\times 0.69\times 10300\,\text{J mol}^{-1}Ea​=8.3×0.69×10300J mol−1

First,

8.3×0.69=5.7278.3\times 0.69=5.7278.3×0.69=5.727

Then,

Ea=5.727×10300=58988.1 J mol−1E_a=5.727\times 10300=58988.1\,\text{J mol}^{-1}Ea​=5.727×10300=58988.1J mol−1

Convert to kJ mol−1^{-1}−1:

Ea≈58.99 kJ mol−1E_a\approx 58.99\,\text{kJ mol}^{-1}Ea​≈58.99kJ mol−1

Nearest integer:

59\boxed{59}59​
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