Question

When the temperature of a reaction A + B → C is increased from 300 K to 310 K the rate constant increases by 12%. What is the Activation energy of the reaction?

• A. 192.17 kJ/mol
• B. 8.76 kJ/mol
• C. 85.69 kJ/mol
• D. 163.9 kJ/mol

Answer 1

Answer: (B)

8.76 kJ/mol

Answer 2

The relationship between the rate constant ($k$) and temperature ($T$) is given by the Arrhenius equation. For two different temperatures $T_1$ and $T_2$, the rate constants $k_1$ and $k_2$ are related by the integrated Arrhenius equation:

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

where $E_a$ is the activation energy and $R$ is the ideal gas constant ($8.314 \text{ J mol}^{-1}\text{ K}^{-1}$).

Given:

$T_1 = 300 \text{ K}$

$T_2 = 310 \text{ K}$

The rate constant increases by 12%, which means $k_2 = k_1 + 0.12k_1 = 1.12k_1$. Therefore, $\frac{k_2}{k_1} = 1.12$.

Substitute the given values into the Arrhenius equation:

$$\ln(1.12) = \frac{E_a}{8.314 \text{ J mol}^{-1}\text{ K}^{-1}}\left(\frac{310 \text{ K} - 300 \text{ K}}{300 \text{ K} \times 310 \text{ K}}\right)$$

Calculate the terms:

$\ln(1.12) \approx 0.1133$

$\frac{310 - 300}{300 \times 310} = \frac{10}{93000} \approx 0.00010752688 \text{ K}^{-1}$

Now, substitute these values back into the equation:

$$0.1133 = \frac{E_a}{8.314} \times 0.00010752688$$

Solve for $E_a$:

$$E_a = \frac{0.1133 \times 8.314}{0.00010752688}$$ $$E_a = \frac{0.9420022}{0.00010752688}$$ $$E_a \approx 8760.3 \text{ J/mol}$$

To convert the activation energy from joules per mole to kilojoules per mole, divide by 1000:

$$E_a = \frac{8760.3}{1000} \text{ kJ/mol}$$ $$E_a \approx 8.76 \text{ kJ/mol}$$