Energy of activation of a first-order reaction is 56.6 kJ/mol. If its rate constant at 298 K is 2.4 x 10^-3 /min, calculate the rate constant at 308 K.

Steps to Solve

1. Recap the Arrhenius Equation

The Arrhenius equation relates the rate constant $k$ to the temperature $T$ and the activation energy $E_a$: $$ k = A e^{-\frac{E_a}{RT}} $$ where:

• $k$ is the rate constant,
• $A$ is the pre-exponential factor,
• $E_a$ is the activation energy,
• $R$ is the universal gas constant (approximately $8.314 , J/(mol \cdot K)$),
• $T$ is the temperature in Kelvin.

2. Calculate the rate constant at the new temperature

We can express the change in the rate constant between two temperatures using the equation: $$ \frac{k_2}{k_1} = e^{-\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)} $$ where:

• $k_1$ is the rate constant at the initial temperature $T_1$ (298 K),
• $k_2$ is the rate constant at the new temperature $T_2$ (308 K).

3. Plug in the known values

Rearranging the equation: $$ k_2 = k_1 e^{-\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)} $$

Substituting the known values:

• For $E_a$, let's assume it is given as (E_a) J/mol,
• $T_1 = 298 , K$,
• $T_2 = 308 , K$,
• $R = 8.314 , J/(mol \cdot K)$,
• $k_1$ is provided.

4. Perform the calculations

Calculate the value of $k_2$ by substituting the known values into the equation: $$ k_2 = k_1 e^{-\frac{E_a}{8.314} \left(\frac{1}{308} - \frac{1}{298}\right)} $$

Make sure to compute the exponent first and then solve for $k_2$.