Question

The rate of a gaseous reaction triples when temperature is increased by $10^{\circ }C$ from $25^{\circ }C$. The energy of activation of the reaction (in $kJmo{l}^{-1}$ ) will be

• A. 40
• B. 70
• C. 83.8
• D. 200

Answer 1

Answer: (C)

Given, $T_1=25^{\circ }C=25+273=298K$
$T_2=\left(25+10^{\circ }C\right)=35^{\circ }C=308K$
Rate $\propto k$
$\Rightarrow \frac{(\text{ Rate }{)}_{25^{\circ }}}{(\text{ Rate }{)}_{35^{\circ }}}=\frac{k_{25^{\circ }}}{k_{35^{\circ }}}$
$\frac{k_{25^{\circ }}}{k_{35^{\circ }}}=\frac{1}{3}$
From Arrhenius equation,
$\log \frac{k_{25^{\circ }}}{k_{35^{\circ }}}=\frac{E_a}{2.303R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$
$\log 3=\frac{E_a}{2.30\times 8.314\times 10^{-3}}\left[\frac{1}{298}-\frac{1}{308}\right]$
$0.477=\frac{E_a}{2.30\times 8.314\times 10^{-3}}\left[\frac{10}{298\times 308}\right]$
$\therefore E_a=83.8kJmo{l}^{-1}$