Q.The activation energy of a certain reaction is 100 kJ/mole. What is the change in the rate constant of the reaction if the temperature is changed from $25^{\circ}C$ to $35^{\circ}C$ ? Let the rate constants at $25^{\circ}C$ and $35^{\circ}C$ be $K_1$ and $K_2$ respectively.

Applying the Arrhenius two-temperature equation with $E_a=100$ kJ/mol over a rise from 298 K to 308 K gives $K_2/K_1 \approx 3.7$.

Given:

• Activation energy, $E_a = 100$ kJ/mol $= 1\times10^5$ J/mol
• $T_1 = 25^{\circ}C = 298$ K
• $T_2 = 35^{\circ}C = 308$ K
• $R = 8.314$ J K$^{-1}$mol$^{-1}$

Arrhenius two-temperature relation:

$\ln\frac{K_2}{K_1} = \frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

Step 1 - Compute $\dfrac{1}{T_1}-\dfrac{1}{T_2}$:

$\frac{1}{298} = 0.0033557\ K^{-1}, \qquad \frac{1}{308} = 0.0032468\ K^{-1}$

$\frac{1}{T_1}-\frac{1}{T_2} = 0.0033557 - 0.0032468 = 0.0001089\ K^{-1}$

Step 2 - Compute $E_a/R$:

$\frac{E_a}{R} = \frac{100000}{8.314} = 12028\ K$

Step 3 - Compute $\ln(K_2/K_1)$:

$\ln\frac{K_2}{K_1} = 12028 \times 0.0001089 = 1.310$

Step 4 - Solve for $K_2/K_1$:

$\frac{K_2}{K_1} = e^{1.310} \approx 3.71$

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