JEE Main · 2020 · Shift-IIhardCK-106

Consider the following plots of rate constant versus 1T for four different reactions. Which of the following orders is…

Chemical Kinetics · Class 12 · JEE Main Previous Year Question

Question

Consider the following plots of rate constant versus 1T\frac{1}{T} for four different reactions. Which of the following orders is correct for the activation energies of these reactions? image

Options
  1. a

    Eb>Ea>Ed>EcE_b > E_a > E_d > E_c

  2. b

    Ea>Ec>Ed>EbE_a > E_c > E_d > E_b

  3. c

    Ec>Ea>Ed>EbE_c > E_a > E_d > E_b

  4. d

    Eb>Ed>Ec>EaE_b > E_d > E_c > E_a

Correct Answerc

Ec>Ea>Ed>EbE_c > E_a > E_d > E_b

Detailed Solution

For the Arrhenius equation, the slope of lnk\ln k vs 1T\frac{1}{T} is EaR-\frac{E_a}{R}.

A steeper negative slope corresponds to a higher activation energy.

From the graph, reading the slopes in order of steepness (most negative to least negative): c > a > d > b

Therefore: Ec>Ea>Ed>EbE_c > E_a > E_d > E_b

Answer: Option (3)

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Consider the following plots of rate constant versus 1T for four different reactions. Which of the… (JEE Main 2020) | Canvas Classes