Question

Calculate the value of $lo{g}_{10}\left(k_2\right)$ where $k_2$ is the rate constant of a reaction at $293K$ when the energy of activation is $103kJmo{l}^{-1}$
Also, the rate constant at $273Kis7.87\times {10}^{-7}{s}^{-1}$
$lo{g}_{10}\left(7.87\right)\approx 0.9$

• A. $-10.32$
• B. $-4.75$
• C. $-0.2$
• D. None of these

Answer 1

The correct option is B $-4.75$

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 $k_1=7.87\times {10}^{-7}{s}^{-1};E_a=103kJmo{l}^{-1};$
$R=8.314\times {10}^{-3}kJmo{l}^{-1}{K}^{-1};$
$T_1=273K$ and $T_2=293K$
Substituting the values in the above equation,
$lo{g}_{10}\frac{k_2}{7.87\times {10}^{-7}}=\frac{103\times 20}{2.303\times 8.314\times {10}^{-3}\times 293\times 273}$

$lo{g}_{10}\frac{k_2}{7.87\times {10}^{-7}}=1.345$
$lo{g}_{10}{k}_2-lo{g}_{10}(7.87\times {10}^{-7})=1.345$
$lo{g}_{10}{k}_2=1.345-6.1=-4.75$