Question

When the energy of activation is $90kJmo{l}^{-1}$, the rate constant of a reaction at $303K$ and $273K$ are $k_2$ and $6.5\times {10}^{-8}{s}^{-1}$ respectively.
Find the value of $lo{g}_{10}{k}_2$.
Take,
$log\left(6.5\right)\approx 0.8$

• A. $+6.2$
• B. $-21.22$
• C. $-10.32$
• D. $-5.5$

Answer 1

The correct option is D $-5.5$

Expression for rate constant at two different temperature is given by
$lo{g}_{10}\frac{k_2}{k_1}=\frac{E_a}{2.303R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$

Given $k_1=6.5\times {10}^{-8}{s}^{-1};E_a=90kJmo{l}^{-1};$
$R=8.314\times {10}^{-3}kJmo{l}^{-1}{K}^{-1};$
$T_1=273K$ and $T_2=303K$
Substituting the values in the above equation,
$lo{g}_{10}\frac{k_2}{6.5\times {10}^{-8}}=\frac{90\times 30}{2.303\times 8.314\times {10}^{-3}\times 303\times 273}$

$lo{g}_{10}\frac{k_2}{6.5\times {10}^{-8}}=1.704$
$lo{g}_{10}{k}_2-lo{g}_{10}(6.5\times {10}^{-8})=1.704$
$lo{g}_{10}{k}_2=1.704-7.2=-5.5$