Question:

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, log 4=0.6021

Updated On: Mar 26, 2025
  • 38.04 KJ/mol
  • 380.4 KJ/mol
  • 3.80 KJ/mol
  • 3804 KJ/mol
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The Correct Option is A

Solution and Explanation

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, log 4 = 0.6021

Solution:

We can use the Arrhenius equation to solve this problem:

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

Where:

  • k1 and k2 are the rate constants at temperatures T1 and T2, respectively.
  • Ea is the activation energy.
  • R is the ideal gas constant (8.314 J K-1 mol-1).
  • T1 and T2 are the temperatures in Kelvin.

Given:

  • T1 = 27°C = 27 + 273.15 = 300.15 K
  • T2 = 57°C = 57 + 273.15 = 330.15 K
  • k2 / k1 = 4 (rate quadruples)

Substituting the values into the Arrhenius equation:

$$ \ln(4) = \frac{E_a}{8.314}\left(\frac{1}{300.15} - \frac{1}{330.15}\right) $$

We know that ln(4) = 2.303 * log(4) = 2.303 * 0.6021 = 1.3862963

$$ 1.3862963 = \frac{E_a}{8.314}\left(\frac{330.15 - 300.15}{300.15 \times 330.15}\right) $$

$$ 1.3862963 = \frac{E_a}{8.314}\left(\frac{30}{99099.225}\right) $$

$$ 1.3862963 = \frac{E_a}{8.314}(3.02727 \times 10^{-4}) $$

$$ E_a = \frac{1.3862963 \times 8.314}{3.02727 \times 10^{-4}} $$

$$ E_a = \frac{11.5257}{3.02727 \times 10^{-4}} $$

$$ E_a = 38073.47 \text{ J/mol} $$

$$ E_a = 38.07347 \text{ kJ/mol} $$

Therefore, the activation energy is approximately 38.04 kJ/mol.

Option 1: 38.04 KJ/mol

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