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  • Two reactions of the same order have equal exponential factors, but their activation energies differ by  <img alt="24.9 \mathrm{~kJ} / \mathrm{mol}" src="https://entrancecorner.oncodecogs

Two reactions of the same order have equal exponential factors, but their activation energies differ by  24.9 \mathrm{~kJ} / \mathrm{mol} . Calculate the ratio of the rate constants of this reaction at  27 ^o\mathrm{C}\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{K-1 \textrm {mol } ^ { - 1 } )}\right..

Option: 1

2.1\times10^4


Option: 2

2.1\times10^3


Option: 3

2.1


Option: 4

None of these


Answers (1)

best_answer

We know that,
\mathrm{\log _{10} k=\log _{10} A-\frac{E}{2.303 R T} }
So, \mathrm{\log _{10} k_2=\log _{10} A-\frac{E_2}{2.303 R T} \ldots eqn(1) }

and   \mathrm{\log _{10} k_1=\log _{10} A-\frac{E_1}{2.303 R T} \ldots eqn(2) } 

Substracting equation (2) from (1), we get,
\mathrm{ \log _{10}\left(\frac{k_2}{k_1}\right)=\frac{\left(E_1-E_2\right)}{2.303 R T} } 
\mathrm{ \log _{10}\left(\frac{k_2}{k_1}\right)=\frac{24.9 \times 1000}{2.303 \times 8.314 \times 300} }
\mathrm{ \frac{k_2}{k_1}=2.14 \times 10^4 }.

Posted by

Gautam harsolia

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