Chemistry | MHT CET 2018 | Discussion Page

The slope of the straight line obtained by plotting log₁₀k against 1/T represents what term?

$-E_{a}$

$-2.303E_{a}/R$

$-E_{a}/2.303R$

$-E_{a}/R$

Option C

According to Arrhenius equation,

$k= Ae^{-E_{a}/RT}$

By taking log on both sides

$\log k= \log_{e}A-\frac{E_{a}}{RT}$

$\log_{10} k= \log_{10}A-\frac{E_{a}}{2.303RT}$

Comparing it with y=mx+c

A plot between log₁₀ k and ( $\frac{1}{T}$ ) is a straight line and slope is given by

Slope = $-\frac{E_{a}}{2.303R}$

( slope has negative value)

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