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Chemical Kinetics and Nuclear Chemistry question

2021 · 27 Aug · Shift 2 · Q13
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Chemical Kinetics and Nuclear Chemistry question

2021 · 27 Aug · Shift 2 · Q13

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
The first order rate constant for the decomposition of CaCO3CaCO_3CaCO3​ at 700 K is 6.36 ×\times× 10 −-− 3s −-− 1 and activation energy is 209 kJ mol −-− 1. Its rate constant (in s −-− 1) at 600 K is x ×\times× 10 −-− 6. The value of x is ‾\underline{\hspace{2cm}}​. (Nearest integer) [Given R = 8.31 J K −-− mol −-− 1; log 6.36 ×\times× 10 −-− 3 = −-− 2.19, 10 −-− 4.79 = 1.62 ×\times× 10 −-− 5]
Numerical answer
View written solutionFree

Correct answer: 16

  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​)

We are given:

  • k2=6.36×10−3 s−1k_2 = 6.36\times 10^{-3}\,s^{-1}k2​=6.36×10−3s−1 at T2=700 KT_2=700\,KT2​=700K
  • T1=600 KT_1=600\,KT1​=600K
  • Ea=209 kJ mol−1=209000 J mol−1E_a=209\,kJ\,mol^{-1}=209000\,J\,mol^{-1}Ea​=209kJmol−1=209000Jmol−1
  • R=8.31 J K−1 mol−1R=8.31\,J\,K^{-1}\,mol^{-1}R=8.31JK−1mol−1

We need k1k_1k1​ at 600 K600\,K600K.

  1. Substitute into the formula
log⁡(k700k600)=2090002.303×8.31(1600−1700)\log\left(\frac{k_{700}}{k_{600}}\right)=\frac{209000}{2.303\times 8.31}\left(\frac{1}{600}-\frac{1}{700}\right)log(k600​k700​​)=2.303×8.31209000​(6001​−7001​)

Now,

1600−1700=700−600600×700=100420000=14200\frac{1}{600}-\frac{1}{700} = \frac{700-600}{600\times 700} = \frac{100}{420000}=\frac{1}{4200}6001​−7001​=600×700700−600​=420000100​=42001​

Also,

2.303×8.31≈19.132.303\times 8.31 \approx 19.132.303×8.31≈19.13

So,

log⁡(k700k600)=20900019.13×4200\log\left(\frac{k_{700}}{k_{600}}\right)=\frac{209000}{19.13\times 4200}log(k600​k700​​)=19.13×4200209000​ 19.13×4200≈8034619.13\times 4200 \approx 8034619.13×4200≈80346

Thus,

log⁡(k700k600)≈20900080346≈2.60\log\left(\frac{k_{700}}{k_{600}}\right)\approx \frac{209000}{80346}\approx 2.60log(k600​k700​​)≈80346209000​≈2.60

Hence,

log⁡k600=log⁡k700−2.60\log k_{600}=\log k_{700}-2.60logk600​=logk700​−2.60
  1. Use the given logarithm

Given:

log⁡(6.36×10−3)=−2.19\log(6.36\times 10^{-3})=-2.19log(6.36×10−3)=−2.19

Therefore,

log⁡k600=−2.19−2.60=−4.79\log k_{600} = -2.19 - 2.60 = -4.79logk600​=−2.19−2.60=−4.79

So,

k600=10−4.79k_{600}=10^{-4.79}k600​=10−4.79

Given:

10−4.79=1.62×10−510^{-4.79}=1.62\times 10^{-5}10−4.79=1.62×10−5

Thus,

k600=1.62×10−5 s−1=16.2×10−6 s−1k_{600}=1.62\times 10^{-5}\,s^{-1}=16.2\times 10^{-6}\,s^{-1}k600​=1.62×10−5s−1=16.2×10−6s−1
  1. Find xxx

Since rate constant is written as x×10−6x\times 10^{-6}x×10−6,

x=16.2x=16.2x=16.2

Nearest integer:

16\boxed{16}16​
  1. Comparison with stored answer

Stored correct answer = 16.

Our derived answer matches it.

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