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=298\,K $ $ {{T}_{2}}=(25+10{{\,}^{o}}C)=35{{\,}^{o}}C=308\,K $ $ 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}}=83.8\,kJ\,mo{{l}^{-1}} $