Animated Solution for Chemistry - Chemical Kinetics: For the reaction, $aA+bB\to cC+dD$, the plot of $\log k$ vs $\frac{1}{T}$ is given below The temperature at which the rate constant of the reaction is $10^{-4}{\text{ s}}^{-1}$ is ......... K (Rounded off to the nearest integer). [Given : The rate constant of the reaction is $10^{-5}{\text{ s}}^{-1}$ at $500\text{ K}$]

Have you ever wondered why food spoils faster in the summer or why a refrigerator keeps your milk fresh? The answer lies in the microscopic world of molecular collisions, beautifully captured by the Arrhenius equation. This equation is the bridge between the macroscopic temperature we feel and the microscopic rate at which chemical reactions occur.

The Arrhenius equation is given by:

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

While the exponential form is powerful, it's often easier to work with straight lines. By taking the base-10 logarithm of both sides, we transform the Arrhenius equation into a linear format:

If you look closely, this resembles the classic equation of a straight line, $y=mx+c$.

- Our $y$-axis is $\log k$. - Our $x$-axis is $\frac{1}{T}$. - The $y$-intercept $c$ is $\log A$. - Most importantly, the slope $m$ is $-\frac{E_a}{2.303R}$.

In our problem, the graph explicitly gives us this slope: $-10000$ K. This is a massive shortcut! It means we don't need to calculate the activation energy $E_a$ separately. We already have the entire term we need.

When a problem gives you data at two different temperatures, the best tool in your arsenal is the two-point form of the Arrhenius equation. By subtracting the equation for state 1 from the equation for state 2, the intercept $\log A$ cancels out entirely:

Notice the term $\frac{E_a}{2.303R}$? That is exactly the negative of our slope! Since our slope is $-10000$, the term outside the bracket is simply $10000$.

Let's gather our knowns: - Initial rate constant, $k_1=10^{-5}{\text{ s}}^{-1}$ - Initial temperature, $T_1=500\text{ K}$ - Final rate constant, $k_2=10^{-4}{\text{ s}}^{-1}$ - The slope term, $\frac{E_a}{2.303R}=10000$

Substituting these into our master equation:

Now, let's simplify the left-hand side. The fraction $\frac{10^{-4}}{10^{-5}}$ simplifies beautifully to $10^1$, or just $10$.

Since we are working with base-10 logarithms, ${\log }_{10}(10)$ is exactly $1$. The equation suddenly looks much less intimidating:

Our goal is to isolate $T_2$. First, let's divide both sides by $10000$:

Next, rearrange the terms to get $\frac{1}{T_2}$ by itself on one side:

To subtract these fractions, we need a common denominator, which is $10000$. Multiply the numerator and denominator of the first fraction by $20$:

Finally, take the reciprocal of both sides to find $T_2$:

The question asks us to round off to the nearest integer, which gives us $526$.

Does this answer make physical sense? Absolutely. The rate constant increased from $10^{-5}$ to $10^{-4}$ (it became 10 times faster). For a reaction to speed up, the temperature must increase. Our calculated final temperature of $526\text{ K}$ is indeed higher than the initial $500\text{ K}$, confirming that our mathematical journey aligns perfectly with physical reality.