Animated Solution for Chemistry - Chemical Kinetics: For following reactions : $A\stackrel{700\text{K}}{\to }\text{Product}$; $A\underset{\text{Catalyst}}{\overset{500\text{K}}{\to }}\text{Product}$ It was found that the $E_a$ is decreased by $30\text{ kJ/mol}$ in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same):

Imagine you are standing at the base of a massive mountain, and you need to get to the other side. You could hike all the way over the peak, which requires a tremendous amount of energy. Or, you could find a secret tunnel that cuts straight through the mountain. A catalyst is exactly like that secret tunnel. It provides an alternative reaction pathway with a lower activation energy barrier, allowing the reaction to proceed much faster at a given temperature.

In our problem, we are given two scenarios for the same reaction: one without a catalyst at $700\text{ K}$, and one with a catalyst at $500\text{ K}$. We are told that the catalyst lowers the activation energy by $30\text{ kJ/mol}$. If we let the activation energy of the catalysed reaction be $x$, then the activation energy of the uncatalysed reaction must be $x+30$.

Our Master Key

To connect activation energy, temperature, and the rate of a reaction, we rely on one of the most beautiful relationships in physical chemistry: the Arrhenius equation.

Here, $k$ is the rate constant, $A$ is the pre-exponential factor (which relates 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.

The problem gives us a crucial piece of information: the rate of the reaction remains unchanged in both scenarios. Since the pre-exponential factor $A$ is assumed to be the same, this means the rate constants for both scenarios must be exactly equal.

Let's substitute our specific values into the Arrhenius equation for both cases:

Now, we get to witness the beauty of algebra as it simplifies complex physical realities. First, we can divide both sides by $A$, completely eliminating the pre-exponential factor. Next, we take the natural logarithm ($\ln$) of both sides. This allows the exponents to drop down, giving us a much cleaner linear equation:

Notice how the negative signs and the gas constant $R$ appear in the denominators on both sides? They cancel out perfectly! We can also simplify the denominators by dividing both by $100$:

We are now left with a simple linear equation. Let's cross-multiply to solve for $x$:

Expanding the bracket gives:

Subtracting $5x$ from both sides leaves us with:

Finally, dividing by $2$, we find our answer:

This is the activation energy for our catalysed reaction. By carefully setting up the Arrhenius equation and trusting the algebra, we've successfully unlocked the physical reality of the reaction's energy profile.