The rate constant for a first order reaction becomes 2 times when the temperature is raised from 330K to 380K. Then energy of activation for the reaction will be

14.45 kJ/mol

The correct answer is (1) 14.45 kJ/mol\)

Arrhenius equation is used to calculate the energy of activation of the reaction

\(log\frac{k_2}{k_1} = \frac{E_a}{2.303}\left[\frac{1}{T_1} - \frac{1}{T_2}\right] -------(1)\)

Given, that the rate constant of a first-order reaction becomes \(2\) times when the temperature is raised from \(330K\) to \(380 K\). So,

\(\frac{k_2}{k_1} = 2\), \(T_1 = 330K\) and \(T_2 = 380 K\)

Substituting the values in the equation (1), we get

\(log 2 = \frac{E_a}{2.303 × 8.314} \left[\frac{1}{330} - \frac{1}{380}\right]\)

\(⇒ 0.3010 = \frac{E_a}{2.303 × 8.314} \left[\frac{50}{125400}\right]\)

\(⇒ E_a = \frac{0.3010 × 2.303 × 8.314 ×125400}{50}\)

\(⇒ E_a =14454.33 J/mol\)

\(⇒ E_a =14.45 kJ/mol\)