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Slope of the straight line obtained by plotting $\log _{10} k$ against $\frac{1}{T}$ represents what term ? (A) $-E_{a}$ (B) $-2.303 E_{a} / R$ (C) $-E_{a} / 2.303 R$ (D) $-E_{a} / R$
Slope of the straight line obtained by plotting $\log _{10} k$ against $\frac{1}{T}$ represents what term ? (A) $-E_{a}$ (B) $-2.303 E_{a} / R$ (C) $-E_{a} / 2.303 R$ (D) $-E_{a} / R$
Chemistry
Slope of the straight line obtained by plotting $\log _{10} k$ against $\frac{1}{T}$ represents what term ? (A) $-E_{a}$ (B) $-2.303 E_{a} / R$ (C) $-E_{a} / 2.303 R$ (D) $-E_{a} / R$

Correct option is C ( $-\mathrm{E}_{\mathrm{a}} / 2.303 \mathrm{R}$)

Explanation:

The Arrhenius equation is:
$
\log _{10} \mathrm{k}=\log _{10} \mathrm{~A}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}
$
The y-intercept is $\log _{10} \mathrm{~A}$.
The slope is $\frac{-\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}$.
Hence, option $\mathrm{C}$ is correct.

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