The rate constants for a reaction at 400 K and 500 K are 2 . 60 × 10 - 5 s - 1 and 2 . 60 × 10 - 3 s - 1 ,respectively. The activation energy of the reaction in kJmol - 1 is ______

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Aicharya · Accepted Answer

To find the activation energy ( $E_a$ ), use the Arrhenius equation in its logarithmic form: $\ln \left(\frac{k_2}{k_1}ight)=-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}ight)$ Given: $\cdot$ $k_1=2.60 \times 10^{-5} \mathrm{~s}^{-1}$ at $T_1=400 \mathrm{~K}$ - $k_2=2.60 \times 10^{-3} \mathrm{~s}^{-1}$ at $T_2=500 \mathrm{~K}$ 1. Calculate $\frac{b_2}{k_1}$ : $\frac{k_2}{k_1}=\frac{2.60 \times 10^{-3}}{2.60 \times 10^{-5}}=100$ 2. Calculate $\frac{1}{T_2}-\frac{1}{T_2}$ : $\frac{1}{500}-\frac{1}{400}=-0.0005 \mathrm{~K}^{-1}$ 3. Substitute and solve for $E_a$ : $\begin{gathered} \ln (100)=-\frac{E_a}{8.314} \times(-0.0005) \ 4.605=\frac{E_a \times 0.0005}{8.314} \ E_a=\frac{4.605 \times 8.314}{0.0005} \approx 76.6 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{gathered}$ The activation energy $E_a$ is approximately $76.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

The rate constants for a reaction at 400 K and 500 K are 2 . 60 × 10 - 5 s - 1 and 2 . 60 × 10 - 3 s - 1 …

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