Question

The slope of Arrhenius Plot $\left(lnkv/s\frac{1}{T}\right)$ of first order reaction is $–5\times {10}^{3}K$. The value of $E_a$ of the reaction is. Choose the correct option for your answer.
[Given $R=8.314J{K}^{–1}mo{l}^{–1}$]

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Answer 1

From Arrhenius equation,

$lnk=lnA-\frac{E_a}{RT}$

For $lnkvs\frac{1}{T}$ graph

$Slope=-\frac{E_a}{R}$

$-5\times {10}^3=\frac{-E_a}{8.314}$

$E_a=5\times {10}^3\times 8.314$
$E_a=41.57\approx 41.5kJmo{l}^{-1}$