Question

The rate of a gaseous reaction triples when temperature is increased by $\text{1}{\text{0}}^{\text{o}}\text{C}$ from $\text{2}{\text{5}}^{\text{o}}\text{C}\text{.}$ The energy of activation of the reaction(in $\text{kJ}\text{mo}{\text{l}}^{-1}$ ) will be

• A. 40
• B. 70
• C. 83.8
• D. 200

Answer 1

Answer: (C)

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