Question

The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction, assuming that it does not change with temperature. \[ (\log 2 = 0.30,\; \log 4 = 0.60) \quad [R = 8.314\; J\, K^{-1}\, mol^{-1}] \]

Hint:

Always use the Arrhenius relation for comparing rates at two temperatures to find activation energy.

Answer 1

Step 1: Use the Arrhenius equation in the logarithmic form:
\[ \log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303\, R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \] Step 2: Given: \( \frac{k_2}{k_1} = 4 \), \( T_1 = 293\, K \), \( T_2 = 313\, K \)
\[ \log 4 = 0.60 = \frac{E_a}{2.303 \times 8.314} \left( \frac{20}{293 \times 313} \right) \] Step 3: Calculate denominator:
\[ 293 \times 313 = 91709 \] \[ \Rightarrow 0.60 = \frac{E_a}{19.147} \times \frac{20}{91709} \] Step 4: Rearranging,
\[ 0.60 = \frac{E_a \times 20}{19.147 \times 91709} \] \[ E_a = \frac{0.60 \times 19.147 \times 91709}{20} \] \[ E_a = \frac{0.60 \times 19.147 \times 91709}{20} \] \[ E_a = \frac{0.60 \times 19.147 \times 91709}{20} = \frac{0.60 \times 1757134.423}{20} \] \[ = \frac{1054280.65}{20} = 52714.03\; J/mol \] Step 5: So, the activation energy is:
\[ E_a \approx 52.7\; kJ/mol \]