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

• A. 38.04 KJ/mol
• B. 380.4 KJ/mol
• C. 3.80 KJ/mol
• D. 3804 KJ/mol

Answer 1

The Correct Option is A

The reaction rate increases by a factor of four when the temperature is raised from 27°C to 57°C. Determine the activation energy.

Given R = 8.314 J K^-1 mol^-1, log 4 = 0.6021

Solution:

The Arrhenius equation is applied 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 represent the rate constants at temperatures T₁ and T₂, respectively.
• E_a denotes the activation energy.
• R is the ideal gas constant, valued at 8.314 J K^-1 mol^-1.
• T₁ and T₂ are the temperatures expressed in Kelvin.

Provided data:

• T₁ = 27°C = 300.15 K
• T₂ = 57°C = 330.15 K
• The ratio of rate constants, k₂ / k₁, is 4.

Substituting the given values into the Arrhenius equation:

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

It is known that ln(4) = 2.303 * log(4) = 2.303 * 0.6021 = 1.3862963

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

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

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

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

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

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

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

The calculated activation energy is approximately 38.04 kJ/mol.

Option 1: 38.04KJ/mol