The rate of a reaction quadruples when temperature changes from 27^oC…

The correct answer is option 1. \(38.04\, \ kJ/mol\).

To solve this problem, we will use the Arrhenius equation, which relates the rate constant \((k)\) of a chemical reaction to the temperature \((T)\) and the activation energy \((E_a)\). The equation in its logarithmic form is:

\(log k = log A - \frac{E_a}{2.303RT}\)

where:

\(k\) is the rate constant,

\(A\) is the pre-exponential factor (frequency factor),

\(E_a\) is the activation energy,

\(R\) is the universal gas constant,

\(T\) is the temperature in Kelvin,

According to the problem, the rate of the reaction quadruples \(k_2 = 4k_1\) when the temperature increases from \(27^oC\text{ 300 K}\) to \(57^oC\text{ 330 K}\). Using the logarithmic form of the Arrhenius equation, for two different temperatures we have:

\(log k_1 = log A - \frac{E_a}{2.303 \times 8.314 \times 300} ----(i)\)

and \(log k_2 = log A - \frac{E_a}{2.303 \times 8.314 \times 330}\)

or, \(log 4k_1 = log A - \frac{E_a}{2.303 \times 8.314 \times 330}----(ii)\)

Subtracting equation (i) from equation (ii), we get

\(log 4k_1 - log k_1 = \left(log A - \frac{E_a}{2.303 \times 8.314 \times 330}\right) - \left(log A - \frac{E_a}{2.303 \times 8.314 \times 300}\right)\)

\(⇒ log 4 = \frac{E_a}{2.303 \times 8.314}\left(\frac{1}{300} - \frac{1}{330}\right)\)

\(⇒ 0.602 = \frac{E_a}{2.303 \times 8.314}\left(\frac{330 - 300}{99000}\right)\)

\(⇒ 0.602 = \frac{E_a}{2.303 \times 8.314}\left(\frac{30}{99000}\right)\)

\(⇒ 0.602 = \frac{E_a}{2.303 \times 8.314}\left(\frac{1}{3300}\right)\)

\(⇒ E_a = 0.602 \times 2.303 \times 8.314 \times 3300\)

\(⇒ E_a = 38037\, \ J/mol\)

\(⇒ E_a = 38.037\, \ kJ/mol\)

\(⇒ E_a \approx 38.04\, \ kJ/mol\)