Question

The rate of a particular reaction quadruples when the temperature increases from 300 K to 320 K. Calculate the energy of activation \( (E_a) \) of the reaction assuming that it does not change with temperature.
(Given : \( \log 4 = 0 \cdot 60, R = 8 \cdot 314 \text{ JK}^{-1} \text{ mol}^{-1} \))

Hint:

Ensure temperatures are in Kelvin.
Check the units of \( R \); using 8.314 gives \( E_a \) in Joules.
Most activation energies are expressed in kJ/mol.

Answer 1

Step 1: Set up the two-temperature Arrhenius relation.
When a reaction's rate constant changes from $k_1$ at $T_1$ to $k_2$ at $T_2$, the activation energy connects them through \[ \log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) \] Here $T_1 = 300$ K, $T_2 = 320$ K, and since the rate quadruples, $k_2/k_1 = 4$, so $\log(k_2/k_1) = 0.60$.
Step 2: Work out the two reciprocal temperature terms separately.
\[ \frac{1}{T_1} = \frac{1}{300}, \qquad \frac{1}{T_2} = \frac{1}{320} \] \[ \frac{1}{300} - \frac{1}{320} = \frac{320 - 300}{300 \times 320} = \frac{20}{96000} = \frac{1}{4800}\ \text{K}^{-1} \]
Step 3: Substitute everything into the Arrhenius relation.
\[ 0.60 = \frac{E_a}{2.303 \times 8.314} \times \frac{1}{4800} \] \[ 0.60 = \frac{E_a}{19.147 \times 4800} \]
Step 4: Solve for $E_a$ and convert units.
\[ E_a = 0.60 \times 19.147 \times 4800 = 55143.36\ \text{J/mol} \] Dividing by 1000 to convert joules to kilojoules, \[ E_a \approx 55.14\ \text{kJ/mol} \]
\[ \boxed{E_a = 55.14\ \text{kJ/mol}} \]