Practice question

Given: A reaction has k = 1.0 × 10⁻³ s^{-1 at 27°C and E_a = 52 kJ mol^{-1 . What is the temperature (in °C) when k = 3.0 × 10⁻³ s^{-1 ? (R = 8.314 J/mol · K) These values define the system as per NCERT data. Formula: log frac3.0 × 10⁻³¹.0 × 10⁻³= 52000/2.303 × 8.314 ( T_2 - 300/300 T_2 ). This is the standard NCERT relation for this phenomenon. Substitution & Calculation: 0.477 = 2717.5 ( T_2 - 300/300 T_2 ), T_2 approx 316 K = 43 °C . 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.