Activation Energy From Rate Constants

The activation energy can be extracted from two rate constants at two temperatures using the logarithmic Arrhenius form log(k_2/k_1) = (Ea/2.303R)((T_2 - T_1)/(T_1 T_2)). Here k_2/k_1 = 100, so log 100 = 2. The temperature factor is (T_2 - T_1)/(T_1 T_2) = (400 - 300)/(300 × 400) = 100/120000 = 8.33 × 10^-4 K^-1. Rearranging, Ea = (log(k_2/k_1) × 2.303 × R)/(temperature factor) = (2 × 2.303 × 8.314)/(8.33 × 10^-4) ≈ 38.29/8.33 × 10^-4 ≈ 45960 J/mol, about 46 kJ/mol. Option 23 kJ/mol halves the result through a logarithm error. Option 92 kJ/mol doubles it. Option 230 kJ/mol mishandles the temperature factor by an order of magnitude. This two-point determination of activation energy is a frequent JEE Advanced calculation. It is worth emphasising that this is not a special case but a representative example of how activation energy from rate constants operates throughout chemical kinetics. A common JEE pitfall is to ignore the role of activation energy calculation, yet it is exactly this factor that distinguishes the correct answer from the tempting alternatives. Plausibility check: a hundredfold rate increase over a wide 100 K span corresponds to a moderate barrier, and the value near 46 kJ/mol is physically reasonable for many reactions.