Question

The activation energy of a reaction is $94.14$ kj/mole, and the value of rate constant at $313$K is $1.8\times {10}^{-1}se{c}^{-1}$. Calculate the frequency factor A.

• A. $\left[A\right]=8.25\times {10}^{10}{s}^{-4}$.
• B. $\left[A\right]=7.923\times {10}^{10}{s}^{-4}$.
• C. $\left[A\right]=9.194\times {10}^{10}{s}^{-4}$.
• D. none of these

Answer 1

The correct option is C $\left[A\right]=9.194\times {10}^{10}{s}^{-4}$.

By using arrhenius equation,
$⟹\log k=\frac{-E_a}{2.303RT}+\log A$

We get
$⟹\log A=\log (1.8\times {10}^{-5})+\frac{94140}{2.303\times 8.314\times 313}$
$⟹\left(\log 1.8\right)-5+15.7082)$
$⟹0.2553-5+15.7082=10.9635$

$\therefore \log A=\left(10.9634\right)=9.914\times {10}^{10}$