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Chemical Kinetics and Nuclear Chemistry question

2013 · Shift 0 · Q14
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Chemical Kinetics and Nuclear Chemistry question

2013 · Shift 0 · Q14

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such a reaction will be:(R = 8.314 JK–1 mol–1 and log 2 = 0.301)
  1. A
    48.6 kJ mol–1
  2. B
    58.5 kJ mol–1
  3. C
    60.5 kJ mol–1
  4. D
    53.6 kJ mol–1
View written solutionFree

Correct answer: D

  1. Use Arrhenius equation in two-temperature form

For a reaction,

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

Given:

  • k2/k1=2k_2/k_1 = 2k2​/k1​=2 (rate doubles)
  • T1=300 KT_1 = 300\,\text{K}T1​=300K
  • T2=310 KT_2 = 310\,\text{K}T2​=310K
  • R=8.314 J K−1mol−1R = 8.314\,\text{J K}^{-1}\text{mol}^{-1}R=8.314J K−1mol−1
  • log⁡2=0.301\log 2 = 0.301log2=0.301
  1. Substitute the values
0.301=Ea2.303×8.314(1300−1310)0.301 = \frac{E_a}{2.303\times 8.314}\left(\frac{1}{300}-\frac{1}{310}\right)0.301=2.303×8.314Ea​​(3001​−3101​)
  1. Evaluate the temperature term
1300−1310=310−300300×310=1093000=19300\frac{1}{300}-\frac{1}{310} = \frac{310-300}{300\times 310} = \frac{10}{93000} = \frac{1}{9300}3001​−3101​=300×310310−300​=9300010​=93001​

So,

0.301=Ea2.303×8.314×193000.301 = \frac{E_a}{2.303\times 8.314}\times \frac{1}{9300}0.301=2.303×8.314Ea​​×93001​
  1. Solve for EaE_aEa​
Ea=0.301×2.303×8.314×9300E_a = 0.301 \times 2.303 \times 8.314 \times 9300Ea​=0.301×2.303×8.314×9300

Now,

2.303×8.314≈19.1472.303 \times 8.314 \approx 19.1472.303×8.314≈19.147 0.301×19.147≈5.7630.301 \times 19.147 \approx 5.7630.301×19.147≈5.763 Ea≈5.763×9300≈53595.9 J mol−1E_a \approx 5.763 \times 9300 \approx 53595.9\,\text{J mol}^{-1}Ea​≈5.763×9300≈53595.9J mol−1 Ea≈53.6 kJ mol−1E_a \approx 53.6\,\text{kJ mol}^{-1}Ea​≈53.6kJ mol−1
  1. Match with options

The correct option is:

D: 53.6 kJ mol−1\boxed{\text{D: }53.6\,\text{kJ mol}^{-1}}D: 53.6kJ mol−1​
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