Question

For a first order reaction $A\to P$, the temperature (T) dependent rate constant $\left(k\right)$ was found to follow the equation $\log k=-\left(2000\right)\frac{1}{T}+6.0$ The pre-exponential factor $A$ and the activation energy $E_a$, respectively, are:

• A. $1.0\times {10}^6{s}^{-1}$ and $9.2kJmo{1}^{-1}$
• B. $1.0\times {10}^6{s}^{-1}$ and $16.6kJmo{1}^{-1}$
• C. $6.0s^{-1}$ and $16.6kJmo{1}^{-1}$
• D. $1.0\times {10}^6{s}^{-1}$ and $38.3kJmo{1}^{-1}$

Answer 1

The correct option is D $1.0\times {10}^6{s}^{-1}$ and $38.3kJmo{1}^{-1}$

Given, $\log K=6-\frac{2000}{T}$
Since, $\log K=\log A-\frac{Ea}{2.303RT}$
$\Rightarrow logA=6$ and $\frac{-E_a}{2.303R}=-2000$

So, $A={10}^6{\sec }^{-1}$ and $Ea=38.3kJ/mole$