Animated Solution for Chemistry - Chemical Kinetics: For the reaction $A\to B$, the rate constant $k$ (in ${\text{s}}^{-1}$) is given by $${\log }_{10}k=20.35-\frac{(2.47\times 10^3)}{T}$$ The energy of activation in ${\text{kJ mol}}^{-1}$ is ......... (Nearest integer) [Given : $R=8.314{\text{ J K}}^{-1}{\text{ mol}}^{-1}$]

Have you ever wondered why milk spoils faster on a hot summer day than in the refrigerator? Or why we use a pressure cooker to cook food quickly? The answer lies in the profound relationship between temperature and the rate of a chemical reaction. This relationship is elegantly captured by the Arrhenius equation, a cornerstone of chemical kinetics. In this problem, we are given an empirical equation for the rate constant of a reaction and asked to find its activation energy. Let's embark on this journey to decode the math and uncover the physical reality it represents.

Arrhenius Theory

The Arrhenius equation is typically written as $k=A{e}^{-E_a/RT}$. Here, $k$ is the rate constant, $A$ is the pre-exponential factor (related to the frequency of collisions), $E_a$ is the activation energy, $R$ is the universal gas constant, and $T$ is the absolute temperature in Kelvin.

To make this equation easier to work with, especially when plotting data, we often take the natural logarithm ($\ln$) or the base-10 logarithm (${\log }_{10}$) of both sides. Taking the base-10 logarithm gives us:

This form is incredibly powerful because it resembles the equation of a straight line, $y=mx+c$.

The Graphical Interpretation

Let's map our logarithmic Arrhenius equation to the straight-line equation: - Our $y$-variable is ${\log }_{10}k$. - Our $x$-variable is $\frac{1}{T}$. - The $y$-intercept ($c$) is ${\log }_{10}A$. - The slope ($m$) is $-\frac{E_a}{2.303R}$.

If we plot ${\log }_{10}k$ against $\frac{1}{T}$, we get a straight line with a negative slope. The steepness of this slope is directly proportional to the activation energy. A steeper slope means a higher activation energy, indicating that the reaction rate is highly sensitive to temperature changes.

The Logic Bridge

Now, let's look at the equation provided in the problem:

By comparing this given equation with our standard straight-line Arrhenius equation, we can immediately identify the corresponding terms. The constant term $20.35$ corresponds to the intercept, ${\log }_{10}A$. The coefficient of $\frac{1}{T}$ corresponds to the slope.

Therefore, we can set up the following equality:

The negative signs cancel out, leaving us with a straightforward algebraic equation to solve for the activation energy, $E_a$.

Watch Your Units!

Let's rearrange the equation to isolate $E_a$:

We are given the value of the universal gas constant, $R=8.314{\text{ J K}}^{-1}{\text{ mol}}^{-1}$. Substituting this value into our equation:

Now, it's time for the atomic compute. Multiplying these numbers together:

Crucial Trap Warning: The value we just calculated is in Joules per mole (${\text{J mol}}^{-1}$). However, the question specifically asks for the activation energy in kilojoules per mole (${\text{kJ mol}}^{-1}$). This is a classic trap where many students lose marks!

To convert Joules to kilojoules, we divide by $1000$:

Finally, the question asks us to round off to the nearest integer. Rounding $47.2898$ gives us our final answer:

We've successfully found the activation energy, but what else could this problem have asked? As we noted earlier, the intercept of the given equation is $20.35$. This means ${\log }_{10}A=20.35$. If the question had asked for the pre-exponential factor $A$, we could easily calculate it as $A=10^{20.35}{\text{ s}}^{-1}$. Always try to extract as much information as possible from a given equation; it builds a deeper intuition for the subject!