A reaction rate constant is given by $k=1.2 \times 10^{14} e^{-25000 / \mathrm{RT}} \mathrm{sec}^{-1} .$ It…

A reaction rate constant is given by
$k=1.2 \times 10^{14} e^{-25000 / \mathrm{RT}} \mathrm{sec}^{-1} .$ It means
  1. $\log_e k$ versus $\log \mathrm{T}$ will give a straight line with a slope as $-25000$
  2. $\log_e k$ versus $\mathrm{T}$ will give a straight line with slope as 25000
  3. $\log_e k$ versus $1 / \mathrm{T}$ will give a straight line with slope as $-25000 / \mathrm{R}$
  4. None of these

Solution

$k=1.2 \times 10^{14} e^{-25000 / \mathrm{RT}} \sec ^{-1} \mathrm{or}$
$\log_e k=\log 1.2 \times 10^{14}-\frac{25000}{\mathrm{R}} \cdot \frac{1}{\mathrm{~T}}$

Equation of straight line
slope $=-\frac{25000}{R}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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