JEE Main 2020 · 8 January · evening · Single correct · +4 / −1

Consider the following plots of rate constant versus 1/T for four different reactions. Which of the following orders is correct for the activation energies of these reactions?

JEE Main PYQ figure
  1. (a) E c > Eₐ > E d > E b
  2. (b) Eₐ > E c > E d > E b
  3. (c) E b > Eₐ > E d > E c
  4. (d) E b > E d > E c > Eₐ
Answer and solution

Answer (a) E c > Eₐ > E d > E b

Solution

Chemistry ma’am · 3 steps

On a log k versus 1/T plot only the steepness matters: the steeper the line, the bigger the Ea.

  1. Straight-line form. 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 a steeper downhill line means a larger Eₐ. How high the line sits depends on A and tells you nothing about Eₐ.
    LineSteepness in the plotEₐ rank
    csteepest, falls almost vertically1st (largest)
    afairly steep, but less than c2nd
    dgentle slope3rd
    balmost flat4th (smallest)
  3. So. Ranking the lines by steepness gives E_c > Eₐ > E_d > E_b, which is option (a).

The idea to keep. In Arrhenius plots, Eₐ lives in the slope, not in the position. Rank the lines by steepness and ignore which one sits higher or lower on the graph.

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

Study Chemical Kinetics

Asked about 150 times in JEE Main since 2020. This question is from 2020.