For a first order reaction, $A \rightarrow P$, the temperature $(T)$ dependent rate constant $(k)$ was found…
For a first order reaction, $A \rightarrow P$, the temperature $(T)$ dependent rate constant $(k)$ was found to follow the equation, $\log k=-(2000) / T+6.0$
The pre-exponential factor $A$ and the activation energy $\left(E_a\right)$, respectively, are
$1.0 \times 10^6 \mathrm{~s}^{-1}$ and $9.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$6.0 \mathrm{~s}^{-1}$ and $16.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$1.0 \times 10^6 \mathrm{~s}^{-1}$ and $16.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$1.0 \times 10^6 \mathrm{~s}^{-1}$ and $38.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
Comparing the slope and intercept of the given equation with the following Arrhenius equation :
$
\log k=-\frac{E_a}{2303 R T}+\log A
$
Hence, $\log A=6$ i.e. $A=10^6 \mathrm{~s}^{-1}$.
Comparing slope gives $E_a=38.3 \mathrm{~kJ} / \mathrm{mol}$