Question

For a reaction A $\rightarrow$ B, the rate doubles when temperature increases from 298 K to 308 K. What is the activation energy (Ea)?

• A. 52 kJ/mol
• B. 48 kJ/mol
• C. 62 kJ/mol
• D. 66 kJ/mol

Hint:

Use the Arrhenius equation to determine the activation energy from the temperature dependence of the reaction rate.

Answer 1

The Correct Option is B

Activation energy is computed using the Arrhenius equation: \[ k = A e^{-\frac{E_a}{RT}} \] Here, \(k\) represents the rate constant, \(A\) the pre-exponential factor, \(E_a\) the activation energy, \(R\) the gas constant (8.314 J/mol·K), and \(T\) the temperature in Kelvin. Given that the rate doubles: \[ \frac{k_2}{k_1} = 2 \] Applying the logarithmic form of the Arrhenius equation: \[ \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] Substituting the known values: \[ \ln 2 = \frac{E_a}{8.314} \left( \frac{1}{298} - \frac{1}{308} \right) \] The calculated activation energy is: \[ E_a = 48 \, \text{kJ/mol} \]