Question

The rate constants for the first order reaction are \(x\) s\({}^{-1}\) at \(300\) K and \(y\) s\({}^{-1}\) at \(320\) K. Calculate the activation energy of the reaction.
\([R=8.314\) J K\({}^{-1}\) mol\({}^{-1}\) and \(y/x=4]\)

• A. \(55.34\) kJ.
• B. \(27.67\) kJ.
• C. \(83.1\) kJ.
• D. \(110.68\) kJ.

Hint:

Apply the two-temperature form of the Arrhenius equation with the ratio of rate constants equal to 4.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The Arrhenius equation at two temperatures is
\(\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{T_2-T_1}{T_1T_2}\right)\).

Step 2: Data
\(k_1 = x\) at \(T_1 = 300\text{ K}\) and \(k_2 = y\) at \(T_2 = 320\text{ K}\).
\(\dfrac{y}{x} = 4\), so \(\log 4 = 0.602\).

Step 3: Calculation
\(E_a = \dfrac{2.303\times 8.314\times 0.602\times 300\times 320}{20}\)
\(E_a \approx 55330\text{ J mol}^{-1}\approx 55.34\text{ kJ mol}^{-1}\)

Final Answer:
The activation energy is about 55.34 kJ. \[ \boxed{\text{(A) }55.34\text{ kJ}} \]