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Two recreations \mathrm{R_1} and \mathrm{R_2} have identical pro -exponential factors. Activation energy of \mathrm{R_1} rate constants for reactions \mathrm{R_1}and \mathrm{R_2} respectively at \mathrm{ 250 \, k}, the \mathrm{ \ln \left(k_2 / k_1\right)} in equal to.

Option: 1

4.48


Option: 2

15.67


Option: 3

9.62


Option: 4

20.64


Answers (1)

best_answer

For Arrhenius Equation
\mathrm{\begin{aligned} & k=A_0 e^{\frac{-E a}{R T}} \\ & k_1=A_0 e^{-E a_1 / R T}\text {...(i) } \\ & k_2=A_0 e^{-E a_2 / R T} \text {...(ii) } \end{aligned}}

Ondiving eq (ii) by i
\mathrm{ \begin{aligned} \ln \left(\frac{k_2}{k_1}\right)=\frac{E a_1-E a_2}{R_T} & =\frac{20,000}{8.314 \times 250} \\ & =9.62 \end{aligned} }

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himanshu.meshram

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