Practice question

Given: The rate constant of a reaction is 1.0 × 10⁻² s^{-1 at 27°C. If the activation energy is 60 kJ/mol, what is the rate constant at 37°C? (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 frack_21.0 × 10⁻²= 60000/2.303 × 8.314 ( 10/300 × 310 ) = 0.336 . k_2 = 1.0 × 10⁻² × __10POW₀__.336 = 2.17 × 10⁻² s^{-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.