Practice question

Given: The rate constant of a reaction is 1.2 × 10⁻⁴ s^{-1 at 320 K and 4.8 × 10⁻⁴ s^{-1 at 330 K. What is the activation energy in kJ mol^{-1 ? (R = 8.314 J/mol · K) These values define the system as per NCERT data. Formula: log k_2/k_1 = E_a/2.303R ( T_2 - T_1/T_1 T_2 ). This is the standard NCERT relation for this phenomenon. Substitution & Calculation: log frac4.8 × 10⁻⁴¹.2 × 10⁻⁴= log 4 = 0.602 . 0.602 = E_a/2.303 × 8.314 ( 10/320 × 330 ), E_a approx 64.7 kJ mol^{-1 . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.