For an elementary chemical reaction, the Arrhenius plot ( k vs 1/T) is given. If the activation energy is 6.64 kJ mol⁻¹ and R=8.3 J K⁻¹mol⁻¹, the temperature at which the rate constant becomes e² min⁻¹ is:

Solution & Explanation

Arrhenius equation: k = A·e^(−Ea/RT), so ln k = ln A − Ea/(RT). Plotting ln k against 1/T gives a straight line whose intercept (1/T → 0) equals ln A. From the given plot this intercept corresponds to A = e⁶, i.e. ln A = 6. We need the temperature where k = e² min⁻¹, so ln k = 2. Substitute: 2 = 6 − Ea/(RT) ⇒ Ea/(RT) = 4. Now Ea = 6.64 kJ/mol = 6640 J/mol and R = 8.3 J K⁻¹mol⁻¹. So 6640/(8.3 × T) = 4 ⇒ 800/T = 4 ⇒ T = 800/4 = 200 K. Trap: do not read the intercept as a k-value directly; it gives ln A. Answer: 200 K (option C).