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 $k J \,m o l^{-1}$ ) will be

• A. 40
• B. 70
• C. 83.8
• D. 200

Answer 1

Answer: (C)

Given, $T _{1}=25^{\circ} C =25+273=298\, K$
$T_{2}=\left(25+10^{\circ} C\right)=35^{\circ} C=308\, K$
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,
$\log \frac{k_{25^{\circ}}}{k_{35^{\circ}}}=\frac{E_{a}}{2.303 R}\left[\frac{1}{T_{1}}-\frac{1}{T_{2}}\right]$
$\log 3=\frac{E_{a}}{2.30 \times 8.314 \times 10^{-3}}\left[\frac{1}{298}-\frac{1}{308}\right]$
$0.477=\frac{E_{a}}{2.30 \times 8.314 \times 10^{-3}}\left[\frac{10}{298 \times 308}\right]$
$\therefore E_{a}=83.8 \,kJ\, mol ^{-1}$