Ncert Exercises & Solutions

The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

From Arrhenius equation, we obtain

\begin{aligned} \log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}}) \\ It is given that, k_{2} = 4 k_{1} \\ \begin{matrix} T_{1} = 293 K \\ T_{2} = 313 K \\ Therefore, \log \frac{4 k_{1}}{k_{2}} = \frac{E_{a}}{2.303 \times 8.314} (\frac{313 - 293}{293 \times 313}) \\ \Rightarrow 0.6021 = \frac{20 \times E_{a}}{2.303 \times 8.314 \times 293 \times 313} \\ \Rightarrow E_{a} = \frac{0.6021 \times 2.303 \times 8.314 \times 293 \times 313}{20} \\ = 52863.33 J {mol}^{- 1} \\ = 52.86 kJ {mol}^{- 1} \end{matrix} \end{aligned}

Hence, the required energy of activation is 52.86 kJmol

^{- 1}

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