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Chemical Kinetics and Nuclear Chemistry question

2022 · 29 Jun · Shift 1 · Q14
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Chemical Kinetics and Nuclear Chemistry question

2022 · 29 Jun · Shift 1 · Q14

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
The activation energy of one of the reactions in a biochemical process is 532611 J mol −-− 1. When the temperature falls from 310 K to 300 K, the change in rate constant observed is k300 = x ×\times× 10 −-− 3 k310. The value of x is ‾\underline{\hspace{2cm}}​. [Given : ln⁡10=2.3\ln 10 = 2.3ln10=2.3, R = 8.3 J K −-− 1 mol −-− 1]
Numerical answer
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Correct answer: 1

  1. Use the Arrhenius relation for two temperatures:
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​)

Here,

  • Ea=532611 J mol−1E_a = 532611\,\text{J mol}^{-1}Ea​=532611J mol−1
  • R=8.3 J K−1mol−1R = 8.3\,\text{J K}^{-1}\text{mol}^{-1}R=8.3J K−1mol−1
  • T1=310 KT_1 = 310\,\text{K}T1​=310K with rate constant k310k_{310}k310​
  • T2=300 KT_2 = 300\,\text{K}T2​=300K with rate constant k300k_{300}k300​

So,

ln⁡(k300k310)=−5326118.3(1300−1310)\ln\left(\frac{k_{300}}{k_{310}}\right)= -\frac{532611}{8.3}\left(\frac{1}{300}-\frac{1}{310}\right)ln(k310​k300​​)=−8.3532611​(3001​−3101​)
  1. First calculate:
5326118.3=64170\frac{532611}{8.3}=641708.3532611​=64170

And,

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​

Thus,

ln⁡(k300k310)=−64170×19300\ln\left(\frac{k_{300}}{k_{310}}\right)= -64170\times \frac{1}{9300}ln(k310​k300​​)=−64170×93001​ =−6.9= -6.9=−6.9
  1. Convert from natural log to base 10 using ln⁡10=2.3\ln 10 = 2.3ln10=2.3:
−6.9=−3×2.3=ln⁡(10−3)-6.9 = -3\times 2.3 = \ln(10^{-3})−6.9=−3×2.3=ln(10−3)

Hence,

k300k310=10−3\frac{k_{300}}{k_{310}} = 10^{-3}k310​k300​​=10−3

So,

k300=1×10−3k310k_{300} = 1\times 10^{-3} k_{310}k300​=1×10−3k310​

Therefore,

x=1x=1x=1
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