The rate of a reaction quadruples when temperature changes from 27°C to 57°C. Calculate the energy of activation.

Given R = 8.314 J K^–1 mol^–1 , log4 = 0.6021

The correct option is 38.04 kJ/mol.

To find the energy of activation ($E_a$), we use the Arrhenius equation in its logarithmic form, which relates the rate constants of a reaction at two different temperatures to the activation energy:

$\log \left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303\times R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

Let's identify the given values from the problem statement:
1. The rate of reaction quadruples, which means:
$\frac{k_2}{k_1}=4$
2. The initial temperature ($T_1$) in Kelvin:
$T_1=27°C=27+273=300\text{K}$
3. The final temperature ($T_2$) in Kelvin:
$T_2=57°C=57+273=330\text{K}$
4. Universal gas constant, $R=8.314{\text{J K}}^{-1}{\text{mol}}^{-1}$
5. $\log 4=0.6021$

Now, substitute these values into the Arrhenius equation:

$\log \left(4\right)=\frac{E_a}{2.303\times 8.314}\left(\frac{1}{300}-\frac{1}{330}\right)$

Simplify the temperature term inside the parentheses:

$\frac{1}{300}-\frac{1}{330}=\frac{330-300}{300\times 330}=\frac{30}{99000}=\frac{1}{3300}$

Substitute this back into the relation:

$0.6021=\frac{E_a}{19.147}\times \frac{1}{3300}$

Solve for $E_a$:

$E_a=0.6021\times 19.147\times 3300$

$E_a\approx 38044\text{J/mol}$

To convert the activation energy into kilojoules per mole (kJ/mol), divide by 1000:

$E_a=\frac{38044}{1000}\text{kJ/mol}=38.04\text{kJ/mol}$

Thus, the activation energy of the reaction is 38.04 kJ/mol.