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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 form for two different temperatures. The following variables are provided:

  • Initial temperature ($T_1$) = $293\text{ K}$
  • Final temperature ($T_2$) = $313\text{ K}$
  • Ratio of rate constants: $\frac{k_2}{k_1} = 4$ [As the rate quadruples, the rate constant quadruples]
  • Universal gas constant ($R$) = $8.314\text{ J K}^{-1}\text{ mol}^{-1}$

Step 1: Applying the Arrhenius Equation

The Arrhenius equation relating rate constants at two temperatures is defined as:

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

[Per the logarithmic form of the Arrhenius Equation]

Step 2: Substituting Numerical Values

Substitute the known quantities into the equation:

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

Step 3: Calculating Intermediate Values

  • $\log(4) \approx 0.6021$
  • $T_2 - T_1 = 20\text{ K}$
  • $T_1 \times T_2 = 293 \times 313 = 91709\text{ K}^2$
  • $2.303 \times 8.314 \approx 19.147\text{ J K}^{-1}\text{ mol}^{-1}$

Step 4: Algebraic Rearrangement for $E_a$

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

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

$E_a = \frac{1057036.7}{20}$

$E_a = 52851.8\text{ J mol}^{-1}$

Step 5: Final Conversion to Standard Units

To express the result in $\text{kJ mol}^{-1}$, divide by $1000$:

$E_a \approx 52.85\text{ kJ mol}^{-1}$

Reaction Coordinate Energy E_a Activation Energy Profile

Final Solution: The activation energy of the reaction is approximately $52.85\text{ kJ mol}^{-1}$.


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