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Chemical Kinetics and Nuclear Chemistry question

2021 · 16 Mar · Shift 1 · Q22
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Chemical Kinetics and Nuclear Chemistry question

2021 · 16 Mar · Shift 1 · Q22

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 K is 1.0 ×\times× 10 −-− 3 s −-− 1 and the activation energy Ea = 11.488 kJ mol −-− 1, the rate constant at 200 K is ‾×\underline{\hspace{2cm}}\times​× 10 −-− 5 s −-− 1. (Round off to the Nearest Integer). (Given : R = 8.314 J mol −-− 1 K −-− 1)
Numerical answer
View written solutionFree

Correct answer: 10

  1. Use the Arrhenius relation in two-temperature form:
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=1.0×10−3 s−1,T1=300 Kk_1 = 1.0\times 10^{-3}\ \text{s}^{-1},\quad T_1=300\ \text{K}k1​=1.0×10−3 s−1,T1​=300 K T2=200 K,Ea=11.488 kJ mol−1=11488 J mol−1T_2=200\ \text{K},\quad E_a=11.488\ \text{kJ mol}^{-1}=11488\ \text{J mol}^{-1}T2​=200 K,Ea​=11.488 kJ mol−1=11488 J mol−1 R=8.314 J mol−1K−1R=8.314\ \text{J mol}^{-1}\text{K}^{-1}R=8.314 J mol−1K−1
  1. Substitute into the equation:
ln⁡(k210−3)=−114888.314(1200−1300)\ln\left(\frac{k_2}{10^{-3}}\right)= -\frac{11488}{8.314}\left(\frac{1}{200}-\frac{1}{300}\right)ln(10−3k2​​)=−8.31411488​(2001​−3001​)

First calculate:

1200−1300=3−2600=1600\frac{1}{200}-\frac{1}{300}=\frac{3-2}{600}=\frac{1}{600}2001​−3001​=6003−2​=6001​

Also,

114888.314≈1381.77\frac{11488}{8.314}\approx 1381.778.31411488​≈1381.77

So,

ln⁡(k210−3)=−1381.77×1600\ln\left(\frac{k_2}{10^{-3}}\right)= -1381.77\times \frac{1}{600}ln(10−3k2​​)=−1381.77×6001​ ln⁡(k210−3)≈−2.303\ln\left(\frac{k_2}{10^{-3}}\right)\approx -2.303ln(10−3k2​​)≈−2.303
  1. Exponentiate:
k210−3=e−2.303≈0.10\frac{k_2}{10^{-3}}=e^{-2.303}\approx 0.1010−3k2​​=e−2.303≈0.10

Hence,

k2=10−3×0.10=1.0×10−4 s−1k_2 = 10^{-3}\times 0.10 = 1.0\times 10^{-4}\ \text{s}^{-1}k2​=10−3×0.10=1.0×10−4 s−1
  1. Express in the required form:
1.0×10−4=10×10−51.0\times 10^{-4} = 10\times 10^{-5}1.0×10−4=10×10−5

So the required integer is:

10\boxed{10}10​
  1. Comparison with stored answer:

Stored correct answer = 101010.

Our derived answer matches the stored answer.

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