Q22 / 160 · AP EAMCET 2016 · Chemistry

The slope of the graph drawn between $\ln k$ and $\frac{1}{T}$ as per Arrhenius equation gives the value $(R=$ gas constant, $E_\alpha=$ Activation energy)

  • A
    $\frac{R}{E_a}$
  • B
    $\frac{E_a}{R}$
  • C
    $\frac{-E_a}{R}$
  • Correct answer
    D
    $\frac{-R}{E_a}$

Solution

Arrhenius equation, $k=A e^{-E_a / \mathrm{RT}}$ On taking log on both sides, we get $\ln k=\ln \mathrm{A}-\frac{E_a}{R T}$ A graph between in $k$ and $1 / \mathrm{T}$ is a straight line with $-\frac{E_a}{R}$ slope.

Hence, the correct answer is option: 'D'.

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