Question

For a first order reaction, $A⟶P$, the temperature $\left(T\right)$ 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. $1\times {10}^6{s}^{-1}$ and $9.2$ $kJ$ ${mol}^{-1}$
• B. $6s^{-1}$ and $16.6$ $kJ$ ${mol}^{-1}$
• C. $1\times {10}^6{s}^{-1}$ and $16.6$ $kJ$ ${mol}^{-1}$
• D. $1\times {10}^6{s}^{-1}$ and $38.3$ $kJ$ ${mol}^{-1}$

Answer 1

Answer: (D)

$1\times {10}^6{s}^{-1}$ and $38.3$ $kJ$ ${mol}^{-1}$

$k={Ae}^{-E_a/RT}\log \left(k\right)=\log \left(A\right)-\frac{E_a}{2.303R}.\frac{1}{T}\therefore \log \left(A\right)=6.0A=1\times {10}^6{s}^{-1}\therefore \frac{E_a}{R}=2.303\times 2000=4.606kJ{mol}^{-1}{E}_a=4.606\times {10}^3\times RJ/mol=38.3kJmo{l}^{-1}$