The rate of a particular reaction doubles when temperature changes from 27°C to 37°C. Calculate the activation energy of such reaction.

Solution

According to the Arrhenius equation

\log \frac{k_{2}}{k_{1}} = \frac{Ea}{2.303 R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}}) \log \frac{1}{2} = \frac{Ea}{2.303 \times 8.314} \times (\frac{310 - 300}{310 \times 300}) Ea = \frac{19.147 \times 93000 \times 0.3010}{10} = \frac{19.147 \times 9300 \times 0.3010}{1000} Ea = 52.99 kJ {mol}^{- 1} .

Some More Questions From Chemical Kinetics Chapter

For Daily Free Study Material Join wiredfaculty