The rate constant of a reaction is found to increase by a factor of 4 when the temperature is increased from 300 K to 360 K. Calculate the activation energy (Ea) of the reaction, a... The rate constant of a reaction is found to increase by a factor of 4 when the temperature is increased from 300 K to 360 K. Calculate the activation energy (Ea) of the reaction, assuming the Arrhenius equation is valid. (R = 8.314 J/mol·K) Read more

Steps to Solve

1. Write down the Arrhenius equation

The Arrhenius equation relates the rate constant of a reaction to the activation energy and temperature:

$$k = Ae^{-\frac{E_a}{RT}}$$

where: $k$ is the rate constant $A$ is the pre-exponential factor $E_a$ is the activation energy $R$ is the gas constant $T$ is the temperature

2. Set up the ratio of rate constants at two different temperatures

Let $k_1$ be the rate constant at temperature $T_1$ and $k_2$ be the rate constant at temperature $T_2$. We are given $T_1 = 300 \text{ K}$ and $T_2 = 360 \text{ K}$, and that $k_2 = 4k_1$. Write Arrhenius equation for both temperatures:

$$k_1 = Ae^{-\frac{E_a}{RT_1}}$$ $$k_2 = Ae^{-\frac{E_a}{RT_2}}$$

Divide the second equation by the first:

$$\frac{k_2}{k_1} = \frac{Ae^{-\frac{E_a}{RT_2}}}{Ae^{-\frac{E_a}{RT_1}}} = e^{-\frac{E_a}{RT_2} + \frac{E_a}{RT_1}} = e^{\frac{E_a}{R}(\frac{1}{T_1} - \frac{1}{T_2})}$$

3. Solve for $E_a$

We know that $\frac{k_2}{k_1} = 4$. Substituting this into the equation above gives:

$$4 = e^{\frac{E_a}{R}(\frac{1}{T_1} - \frac{1}{T_2})}$$

Take the natural logarithm of both sides:

$$\ln{4} = \frac{E_a}{R}(\frac{1}{T_1} - \frac{1}{T_2})$$

Now, solve for $E_a$:

$$E_a = \frac{R \ln{4}}{\frac{1}{T_1} - \frac{1}{T_2}}$$

4. Plug in the values and calculate $E_a$

We are given $R = 8.314 \text{ J/mol K}$, $T_1 = 300 \text{ K}$, and $T_2 = 360 \text{ K}$.

$$E_a = \frac{8.314 \text{ J/mol K} \cdot \ln{4}}{\frac{1}{300 \text{ K}} - \frac{1}{360 \text{ K}}}$$

$$E_a = \frac{8.314 \cdot 1.3863}{\frac{360 - 300}{300 \cdot 360}} = \frac{11.525}{ \frac{60}{108000}} = \frac{11.525}{0.0005556} \text{ J/mol}$$

$$E_a \approx 20724 \text{ J/mol} = 20.724 \text{ kJ/mol}$$