Question

For a first order reaction, $A\to P$, the temperature $\left(T\right)$ dependent on the rate constant $\left(k\right)$ was formed to follow the equation $lo{g}_{10}k=-\frac{2000}{T}+6.0$. The pre-exponential factor $A$ and activation energy $E_a$, respectively are:

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

Answer 1

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

The Arrhenius equation is
$logk=-\frac{E_a}{2.303RT}+logA$
But
$lo{g}_{10}k=-\frac{2000}{T}+6.0$
Hence,
$logA=6.0$
The pre-exponential factor $A=1.0\times {10}^6{s}^{-1}$
Also
$-\frac{E_a}{2.303RT}=-\frac{2000}{T}$
$-\frac{E_a}{2.303R}=-2000$
$E_a=2000\times 2.303R$
$E_a=2000\times 2.303\times 8.314$
$E_a=38294Jmo{l}^{-1}$
The activation energy $E_a=38.3kJmo{l}^{-1}$