Consider the following plots of log of rate constant k(log k) vs 1/T for three different reactions. The correct order of activation energies of these reactions is :

Solution

Chemistry ma’am · 3 steps

On a log k vs 1/T plot, the steeper the line falls, the bigger the activation energy.

1. Arrhenius as a line. Take the log of k = A e^(-Eₐ/RT). You get a straight line in 1/T.

   log k = log A - Eₐ/2.303R(1/T)

2. Read the slope. The slope is -Eₐ/2.303R, so Eₐ is proportional to how steeply the line falls. The height of the line depends on log A and tells you nothing about Eₐ.

   Line	How steeply it falls	Activation energy
   2	almost vertical, steepest	largest
   1	moderate	middle
   3	nearly flat, gentlest	smallest

3. So. Ordering by steepness gives Eₐ₂ > Eₐ₁ > Eₐ₃, which is option (a).

The idea to keep. In any Arrhenius plot, compare slopes, not positions. A steep line means k is very sensitive to temperature, and that comes from a high Eₐ.

Answer from the NTA final answer key. Solution by PrepWiser.