JEE Main · 2019 · Shift-IeasyCK-034

If a reaction follows the Arrhenius equation, the plot k vs 1RT gives straight line with a gradient (-y) unit. The…

Chemical Kinetics · Class 12 · JEE Main Previous Year Question

Question

If a reaction follows the Arrhenius equation, the plot ln⁡k\ln k vs 1RT\frac{1}{RT} gives straight line with a gradient (−y)(-y) unit. The energy required to activate the reactant is:

Options
  1. a

    y/Ry/R unit

  2. b

    yy unit

    ✓
  3. c

    yRyR unit

  4. d

    −y-y unit

Correct Answerb

yy unit

Detailed Solution

Arrhenius equation: ln⁡k=ln⁡A−EaRT\ln k = \ln A - \frac{E_a}{RT}

Rewriting as a function of 1RT\frac{1}{RT}: ln⁡k=ln⁡A−Ea⋅1RT\ln k = \ln A - E_a \cdot \frac{1}{RT}

This is a straight line with slope =−Ea= -E_a when plotted as ln⁡k\ln k vs 1RT\frac{1}{RT}.

Given slope =−y= -y, therefore Ea=yE_a = y unit.

Answer: Option (2) — y unit

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