Question

For a reaction, consider the plot of $lnk\text{versus}1/T$ given in the figure where k is rate constant and T is temperature. If the rate constant of this reaction at $500K$ is ${10}^{-4}{s}^{-1}$, then the rate constant at $600K$ is:

Given :
$l{n}_e\left(10\right)\approx 2.303$

• A. $10s^{-1}$​
• B. ${10}^{-6}{s}^{-1}$​
• C. ${10}^{-2}{s}^{-1}$​
• D. ${10}^{-3}{s}^{-1}$​

Answer 1

The correct option is D ${10}^{-3}{s}^{-1}$​

According to Arrhenius equation,
$k=A{e}^{-E_a/RT}$
Taking ln on both sides,
​
$lnk=lnA-\frac{E_a}{RT}$
This equation is in the form of $y=mx+c$
The plot of $lnk\text{vs}\frac{1}{T}$ gives a strraight line with negative slope.
$\text{Slope}=-\frac{E_a}{R}$
$y-intercept=lnA$
From the graph we know,
$\text{Slope}=m=-\frac{E_a}{R}=-6909K$
Let $k_1$ and $k_2$ are the rate constant of the reaction at temperature $T_1$ and $T_2$

$\frac{k_1}{k_2}=\frac{A{e}^{-Ea/R{T}_1}}{A{e}^{-E_a/R{T}_2}}$​

$\frac{{10}^{-4}}{k_2}=e^{\frac{Ea}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)}$​

$\frac{{10}^{-4}}{k_2}=e^{\frac{Ea}{R}\left(\frac{1}{600}-\frac{1}{500}\right)}$

$\frac{{10}^{-4}}{k_2}=e^{\left(6909\right)(-1/3000)}$​

$\frac{{10}^{-4}}{k_2}\approx 0.1$​
$k_2={10}^{-3}{s}^{-1}$​