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Q3.30:
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.

Solution :

Given Variables & Initial Setup

To determine the Activation Energy ($E_a$), we utilize the Arrhenius equation in its integrated logarithmic form, which relates the rate constants ($k$) at two different temperatures ($T_1$ and $T_2$).

  • Initial Temperature ($T_1$): $293\text{ K}$
  • Final Temperature ($T_2$): $313\text{ K}$
  • Rate Constant Ratio: $\frac{k_2}{k_1} = 4$ [Given: Rate quadruples]
  • Universal Gas Constant ($R$): $8.314\text{ J K}^{-1}\text{ mol}^{-1}$

Step 1: The Arrhenius Equation Formulation

The Arrhenius equation for two temperatures is defined as [Per the Arrhenius Principle]:

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

Step 2: Substitution of Numerical Values

Substitute the given variables into the equation:

$\log(4) = \frac{E_a}{2.303 \times 8.314} \left( \frac{313 - 293}{293 \times 313} \right)$

Knowing $\log(4) \approx 0.6021$:

$0.6021 = \frac{E_a}{19.147} \left( \frac{20}{91709} \right)$

Step 3: Algebraic Isolation of Activation Energy ($E_a$)

Rearranging the equation to solve for $E_a$:

$E_a = \frac{0.6021 \times 19.147 \times 91709}{20}$

$E_a = \frac{1056721.2}{20}$

$E_a = 52836.06\text{ J mol}^{-1}$

Step 4: Final Conversion to Standard Units

Converting Joules to Kilojoules for standard thermodynamic reporting [By standard SI convention]:

$E_a = 52.84\text{ kJ mol}^{-1}$

Reaction Coordinate Energy Reactants Eₐ

Final Solution: The energy of activation for the reaction is $52.84\text{ kJ mol}^{-1}$.


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