Question

For a first order reaction $A\to P$, the tem[erature (T) dependent rate constant (k) was found to follow the equation $\log k=-\left(2000\right)\frac{1}{T}+6$. The pre exponential factor A and the activation energy $E_a$ respectively, are:

• A. $1.0\times {10}^6{\text{s}}^{-1}$ and $9.7{\text{kJ mol}}^{-1}$
• B. $6.0\times {\text{s}}^{-1}$ and $16.6{\text{kJ mol}}^{-1}$
• C. $1.0\times {10}^6{\text{s}}^{-1}$ and $16.6{\text{kJ mol}}^{-1}$
• D. $1.0\times {10}^6{\text{s}}^{-1}$ and $38.29{\text{kJ mol}}^{-1}$

Answer 1

The correct option is D $1.0\times {10}^6{\text{s}}^{-1}$ and $38.29{\text{kJ mol}}^{-1}$

According to Arrhenius equation,
$k=A{e}^{-\frac{E_a}{RT}}$
$\Rightarrow lnk=lnA-\frac{E_a}{RT}$
$\Rightarrow \log k=\log A-\frac{E_a}{2.303RT}...\left(1\right)$
Given,
$\log k=6.0-\left(2000\right)\frac{1}{T}...\left(2\right)$
On comparing equation $\left(1\right)$ and $\left(2\right)$,
$\log A=6.0\Rightarrow A={10}^6{\text{s}}^{-1}$
and $\frac{E_a}{2.303R}=2000\Rightarrow E_a=2000\times 2.303\times 8.314=38.29{\text{kJ mol}}^{-1}$