Animated Solution for Chemistry - Chemical Kinetics: A reactant (A) forms two products $A\stackrel{k_1}{\to }B$, Activation energy $E_{a_1}$ $A\stackrel{k_2}{\to }C$, Activation energy $E_{a_2}$ If $E_{a_2}=2E_{a_1}$, then $k_1$ and $k_2$ are related as

Imagine you are standing at a crossroads. You are the reactant molecule, A, and you have a choice to make. You can either take the scenic, slightly bumpy road to become product B, or you can take the steep, mountainous hike to become product C.

In the world of chemical kinetics, these "bumps" and "mountains" are what we call Activation Energy ($E_a$). The problem tells us that the mountain to reach C is exactly twice as high as the bump to reach B. Mathematically, this is written as:

Our goal is to figure out how the speed of taking the first path ($k_1$) compares to the speed of taking the second path ($k_2$).

Arrhenius Equation

To connect the speed of a reaction (the rate constant, $k$) to the height of its energy barrier (the activation energy, $E_a$), we use the legendary Arrhenius Equation:

Here, $A$ is the pre-exponential factor (think of it as the number of times molecules attempt the journey), $R$ is the universal gas constant, and $T$ is the absolute temperature.

Since both paths start from the exact same reactant A, it is a very safe and standard assumption in kinetics that the attempt frequency, $A$, is the same for both reactions. Let's call it $A^{\prime }$.

Let's write down the Arrhenius equation for both of our possible journeys.

For the path to product B:

For the path to product C:

Now, we bring in our crucial piece of intel: $E_{a_2}=2E_{a_1}$. Let's substitute this into the equation for $k_2$:

We want to find a relationship between $k_1$ and $k_2$. The easiest way to compare two multiplicative equations and get rid of pesky constants (like $A^{\prime }$) is to divide them! Let's divide $k_1$ by $k_2$:

The $A^{\prime }$ terms beautifully cancel out, leaving us with:

Now, we use the basic rule of exponents: $\frac{x^a}{x^b}=x^{a-b}$.

To match the options provided in the question, we simply rearrange our final expression by multiplying both sides by $k_2$:

And there we have it! This perfectly matches option (b).

Physical Insight: Because the exponent $\frac{E_{a_1}}{RT}$ is a positive number, $e^{\frac{E_{a_1}}{RT}}$ is greater than 1. This mathematically proves what our intuition tells us: $k_1$ is significantly larger than $k_2$. The reaction will overwhelmingly prefer the easier path to form product B!