Animated Solution for Chemistry - Chemical Kinetics: For a reaction, consider the plot of $\ln k$ versus $1/T$ given in the figure. If the rate constant of this reaction at $400\text{ K}$ is $10^{-5}{\text{ s}}^{-1}$, then the rate constant at $500\text{ K}$ is

The Arrhenius equation is one of the most beautiful and profound relationships in chemical kinetics. It bridges the macroscopic world of reaction rates with the microscopic world of molecular collisions and activation energies. In this problem, we are given a graphical representation of this equation and asked to find the rate constant at a higher temperature. Let's embark on this journey to decode the graph and solve the mystery!

Imagine you are observing a chemical reaction. You notice that as you heat the reaction mixture, it proceeds much faster. Svante Arrhenius quantified this observation with his famous equation:

Here, $k$ is the rate constant, $A$ is the pre-exponential factor (related to collision frequency), $E_a$ is the activation energy, $R$ is the universal gas constant, and $T$ is the absolute temperature.

While this exponential form is powerful, it's often easier to work with linear relationships. By taking the natural logarithm ($\ln$) of both sides, we transform the equation into a straight line:

We can rewrite this to clearly see the linear form $y=mx+c$:

When we plot $\ln k$ on the y-axis and $\frac{1}{T}$ on the x-axis, we get a straight line. The slope of this line is $m=-\frac{E_a}{R}$, and the y-intercept is $c=\ln A$.

In our problem, the graph explicitly gives us the slope:

This immediately tells us that:

This is a crucial piece of information! We don't need to know the individual values of $E_a$ or $R$; their ratio is all we need to proceed.

We are given the rate constant at one temperature and asked to find it at another. This is a classic scenario for the two-point form of the Arrhenius equation. Let's set up our knowns and unknowns:

- Initial temperature, $T_1=400\text{ K}$ - Initial rate constant, $k_1=10^{-5}{\text{ s}}^{-1}$ - Final temperature, $T_2=500\text{ K}$ - Final rate constant, $k_2=?$

By writing the logarithmic Arrhenius equation for both temperatures and subtracting them, we eliminate the constant $\ln A$:

Using the property of logarithms, $\ln a-\ln b=\ln \left(\frac{a}{b}\right)$, we get:

This equation is our master key. It elegantly connects the ratio of rate constants to the difference in the reciprocals of temperatures.

Now, let's substitute our known values into the master equation. We know $\frac{E_a}{R}=4606$, $T_1=400$, $T_2=500$, and $k_1=10^{-5}$:

Let's focus on the temperature term inside the parentheses. Finding a common denominator makes the subtraction straightforward:

Substituting this back into our equation:

At this point, you might reach for a calculator to find the inverse natural logarithm ($e^{2.303}$). However, in competitive exams like JEE, numbers are rarely random. The value $2.303$ is a very special number in chemistry and mathematics!

Recall the conversion factor between the natural logarithm ($\ln$, base $e$) and the common logarithm (${\log }_{10}$, base 10):

If we set $x=10$, we get:

Since ${\log }_{10}10=1$, we have a beautiful identity:

This is the "aha!" moment of the problem. We can replace $2.303$ with $\ln 10$:

Because the natural logarithm function is one-to-one, if $\ln (A)=\ln (B)$, then $A=B$. Therefore, we can drop the logarithms entirely:

The rest is simple algebra. Multiply both sides by $10^{-5}$ to isolate $k_2$:

And there we have it! By understanding the graphical representation of the Arrhenius equation, setting up the two-point formula, and recognizing a clever mathematical identity, we've successfully found the rate constant at the higher temperature. The rate constant increased from $10^{-5}$ to $10^{-4}$, which makes perfect physical sense: as temperature increases, the rate of reaction increases.