Question

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

• A. $1.0\times10^6 s^{-1}$ and $9.2\, kJ\, mol^{-1}$
• B. $6.0\, s^{-1}$ and $16.6 \,kJ\, mol^{-1}$
• C. $1.0\times10^6 s^{-1}$ and $16.6\, kJ mol^{-1}$
• D. $1.0\times10^6 s^{-1}$ and $38.3 \,kJ \,mol^{-1}$

Answer 1

Answer: (D)

The logarithmic form of Arrhenius equation is
$\log k=\log A-\frac{E_a}{2.303 RT}$
Given, $\log k=6-\frac{2000}{T}$
Comparing the above two equations :
$\log A=6\Rightarrow A=10^6$
and $\frac{E_a}{2.303 R}=2000$
$\Rightarrow E_a = 2000 \times 2.303 \times 8.314\, J$
$=38.3 \,kJ\, mol^{-1}$