Question

The rate constant of a reaction is $1.5\times {10}^7{s}^{-1}$ at ${50}^{\circ }C$ and $4.5\times {10}^7{s}^{-1}$ at ${100}^{\circ }C.$ Calculate the Arrhenius parameter A.
Take $\log \left(1.5\right)=0.18$

• A. $5.42\times {10}^{10}{s}^{-1}$
• B. $5.42\times {10}^5{s}^{-1}$
• C. $2.12\times {10}^{10}{s}^{-1}$
• D. $2.12\times {10}^5{s}^{-1}$

Answer 1

The correct option is A $5.42\times {10}^{10}{s}^{-1}$

$lo{g}_{10}\frac{k_2}{k_1}=\frac{E_a}{2.303R}(\frac{1}{T_1}-\frac{1}{T_2})$
$lo{g}_{10}\frac{4.5\times {10}^7}{1.5\times {10}^7}=\frac{E_a}{2.303\times 8.314}(\frac{1}{323}-\frac{1}{373})$

$E_a=2.2\times {10}^{4}J/mol$

We know that,
$lo{g}_{10}k=lo{g}_{10}A-\frac{E_a}{2.303RT}$
$lo{g}_{10}(1.5\times {10}^7)=lo{g}_{10}A-\frac{2.2\times {10}^4}{2.303\times 8.314\times 323}$
$A=5.42\times {10}^{10}{s}^{-1}$