Question

The rate constant of a chemical reaction increased from 0.100 s−1 to 3.00 s−1 upon raising the temperature from 25.0 ∘C to 37.0 ∘C .

Please Calculate the value of (1/T2−1/T1) where T1 is the initial temperature and T2 is the final temperature.

Answer 1

The rate $k$ of a chemical reaction dependence on the absolute temperature $T$ is described by Arrhenius law:

$k = A·e^{-\frac{E_a}{RT}}$ ,

where $A$ is the pre-exponential factor, $E_a$ is the activation energy and $R$ is the universal gas constant.

The two measurements of the rate constants at two different temperatures are enough to calculate the values of the pre-exponential factor and of the activation energy:

$k_1 = A·e^{-\frac{E_a}{RT_1}}$

$k_2 = A·e^{-\frac{E_a}{RT_2}}$

$\text{ln}(\frac{k_1}{k_2}) = \frac{E_a}{R}(\frac{1}{T_2} - \frac{1}{T_1})$

$E_a = R·\frac{\text{ln}(\frac{k_1}{k_2})}{\frac{1}{T_2} - \frac{1}{T_1}}$ .

The value of (1/T₂−1/T₁) is:

$\frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{273.15+37.0} - \frac{1}{273.15+25.0} = -1.30·10^{-4}$ K^-1.

The value of $\text{ln}(\frac{k_1}{k_2})$ is:

$\text{ln}(\frac{k_1}{k_2}) = \text{ln}(\frac{0.100}{3.00}) = -3.40$ .

Therefore, the activation energy is:

$E_a = 8.314 \text{ (J/(mol K)})·\frac{-3.40}{-1.30·10^{-4} (\text{K}^{-1})}$

$E_a = 2.18·10^5$ J/mol, or 218 kJ/mol.

Answer: The value of (1/T₂−1/T₁) is -1.30·10^-4 K^-1.