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JEE Main 8 January 2020, Shift 2, Chemistry Q27

Question 27 of 75 in this shift, Chemistry question 2 of 25, Section A.

Consider the following plots of rate constant versus

\frac{1}{T}

for four different reactions. Which of the following orders is correct for the activation energies of these reactions ?

The figure, in words

Graph with vertical axis labelled 'log k' (arrow pointing up) and horizontal axis labelled '1/T' (arrow pointing right). Four straight lines with negative slope are drawn, labelled a, b, c and d; line c is the steepest (it falls steeply from the middle of the plot to near the 1/T axis); line a, at the upper right, is less steep than c; line d slopes gently and crosses c and then b; line b is nearly flat and is the lowest line at the left.

(1) $\mathrm{E_b > E_d > E_c > E_a}$
(2) $\mathrm{E_a > E_c > E_d > E_b}$
(3) $\mathrm{E_c > E_a > E_d > E_b}$ Official answer
(4) $\mathrm{E_b > E_a > E_d > E_c}$

Official answer

Option 3

NTA final key (Jan 2020).

Chapter: Chemical Kinetics
Topic: Arrhenius Equation
Idea tested: Log Base 10 Arrhenius Plot Slope

JEE Main 8 January 2020, Shift 2, Chemistry Q27

Same idea in other shifts