Search any question & find its solution
Question:
Answered & Verified by Expert
The graph obtained between $\ln \mathrm{k}(\mathrm{k}=$ Rate constant $)$ on $y$-axis $1 / T$ on $x$-axis is a straight line. The slope of it is $-4 \times 10^4 \mathrm{k}$. The activation energy of the reaction (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) is
$$
\left(\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)
$$
Options:
$$
\left(\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)
$$
Solution:
2632 Upvotes
Verified Answer
The correct answer is:
$332$
The equation for the corresponding graph is :-
$$
\ln \mathrm{K}=-\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}+\ln \mathrm{A}
$$
where $-\frac{E_a}{R}$ is the slope and $\ln A$ is the $y$-intercept.
$$
\begin{gathered}
\quad \mathrm{R} \\
\Rightarrow \quad \text { Slope }=-4 \times 10^4 \mathrm{~K}=-\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}} \\
\Rightarrow \quad \mathrm{E}_{\mathrm{a}}=4 \times 10^4 \times \mathrm{R}=4 \times 10^4 \times 8.3 \\
\quad=332000 \mathrm{~J}=332 \mathrm{~kJ} .
\end{gathered}
$$
$$
\ln \mathrm{K}=-\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}+\ln \mathrm{A}
$$
where $-\frac{E_a}{R}$ is the slope and $\ln A$ is the $y$-intercept.
$$
\begin{gathered}
\quad \mathrm{R} \\
\Rightarrow \quad \text { Slope }=-4 \times 10^4 \mathrm{~K}=-\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}} \\
\Rightarrow \quad \mathrm{E}_{\mathrm{a}}=4 \times 10^4 \times \mathrm{R}=4 \times 10^4 \times 8.3 \\
\quad=332000 \mathrm{~J}=332 \mathrm{~kJ} .
\end{gathered}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.