Question

For a first order reaction, $A\to P$, the temperature (T) dependent rate constant (k) 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 $38.3kJmo{l}^{-1}$

$lo{g}_{e}K=\frac{-Ea}{RT}+lo{g}_{e}A$
$lo{g}_{10}K=\frac{-Ea}{2.303RT}+lo{g}_{10}A$
$lo{g}_{10}A=6$
$A={10}^6$
Also, $\frac{Ea}{R\times 2.303}=2000$
$\therefore Ea=38.3\times {10}^{3}Jmo{l}^{-1}$