© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: State the Known Data
The reaction rate doubles when temperature changes from
$300\,\text{K}$ to $310\,\text{K}$. So, $k_2 / k_1 = 2$.
The gas constant $R = 8.314\,\text{J\,K}^{-1}\text{mol}^{-1}$.
We are also given $\log(2) = 0.301.$
Step 2: Write the Relevant Arrhenius Equation
From the Arrhenius equation in logarithmic form, the relation between
rate constants and activation energy $E_a$ is:
$$
\log\left(\frac{k_2}{k_1}\right)
= \frac{E_a}{2.303\,R}
\left(\frac{1}{T_1} - \frac{1}{T_2}\right)
$$
Here, $T_1 = 300\,\text{K}$ and $T_2 = 310\,\text{K}.$
Step 3: Substitute the Known Values
We have:
$$
\log\left(\frac{k_2}{k_1}\right) = \log(2) = 0.301
$$
So,
$$
0.301 = \frac{E_a}{2.303 \times 8.314}
\left(\frac{1}{300} - \frac{1}{310}\right)
$$
Step 4: Calculate the Temperature Difference Factor
Compute the difference:
$$
\frac{1}{300} - \frac{1}{310}
= \frac{310 - 300}{300 \times 310}
= \frac{10}{93000}
= \frac{1}{9300} \approx 1.075 \times 10^{-4}
$$
Step 5: Solve for Activation Energy $E_a$
Rewriting the equation:
$$
0.301 = \frac{E_a}{2.303 \times 8.314}
\times 1.075 \times 10^{-4}
$$
Therefore,
$$
E_a = 0.301
\times (2.303 \times 8.314)
\,\Big/\,
(1.075 \times 10^{-4}).
$$
Carry out the multiplication and division to find $E_a$ in J/mol, then convert to kJ/mol:
$$
E_a \approx 53.6\,\text{kJ\,mol}^{-1}.
$$
Final Answer
The activation energy of the reaction is approximately
$53.6\,\text{kJ\,mol}^{-1}$.
