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Chemical Kinetics and Nuclear Chemistry question

2017 · 9 Apr · Shift 1 · Q17
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Chemical Kinetics and Nuclear Chemistry question

2017 · 9 Apr · Shift 1 · Q17

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is : (Assume activation energy and preexponential factor are independent of temperature; ln 2 = 0.693; R = 8.314 J mol−1 K−1)
  1. A
    107.2 kJ mol −-− 1
  2. B
    53.6 kJ mol −-− 1
  3. C
    26.8 kJ mol −-− 1
  4. D
    214.4 kJ mol −-− 1
View written solutionFree

Correct answer: A

  1. Use Arrhenius equation

For a reaction rate constant,

k=Ae−Ea/(RT)k = A e^{-E_a/(RT)}k=Ae−Ea​/(RT)

Taking ratio at two temperatures T1T_1T1​ and T2T_2T2​:

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​)

Since rate quadruples, assuming rate ∝k\propto k∝k:

k2k1=4\frac{k_2}{k_1}=4k1​k2​​=4

with

T1=300 K,T2=310 KT_1=300\,\text{K}, \quad T_2=310\,\text{K}T1​=300K,T2​=310K

  1. Substitute values

ln⁡4=2ln⁡2=2(0.693)=1.386\ln 4 = 2\ln 2 = 2(0.693)=1.386ln4=2ln2=2(0.693)=1.386

Also,

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,

1.386=Ea8.314⋅193001.386 = \frac{E_a}{8.314}\cdot \frac{1}{9300}1.386=8.314Ea​​⋅93001​

  1. Solve for EaE_aEa​

Ea=1.386×8.314×9300E_a = 1.386 \times 8.314 \times 9300Ea​=1.386×8.314×9300

First,

8.314×9300=77320.28.314 \times 9300 = 77320.28.314×9300=77320.2

Then,

Ea=1.386×77320.2≈107156 J mol−1E_a = 1.386 \times 77320.2 \approx 107156\,\text{J mol}^{-1}Ea​=1.386×77320.2≈107156J mol−1

Ea≈107.2 kJ mol−1E_a \approx 107.2\,\text{kJ mol}^{-1}Ea​≈107.2kJ mol−1

  1. Match with options

The correct option is:

A: 107.2 kJ mol−1107.2\,\text{kJ mol}^{-1}107.2kJ mol−1

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