Animated Solution for Chemistry - Chemical Kinetics: Consider the given plot of enthalpy of the following reaction between $A$ and $B$. $A+B⟶C+D$ Identify the incorrect statement.

Kinetic vs Thermodynamic Control

When analyzing a chemical reaction, a potential energy surface (or enthalpy plot) is like a topographic map of a mountain range. It tells us not only where the reaction starts and ends but also how difficult the journey is. In this problem, we are presented with a fascinating scenario where reactants $A$ and $B$ can take two different paths to form either product $C$ or product $D$.

Let's orient ourselves with the graph. On the y-axis, we have the enthalpy in ${\text{kJ mol}}^{-1}$, and on the x-axis, the reaction coordinate, which tracks the progress of the reaction. Right in the middle, our reactants $A+B$ are sitting comfortably at an enthalpy level of $15{\text{ kJ mol}}^{-1}$.

From this starting point, the reaction can proceed in two directions. To the left, it climbs a small hill to form product $D$. To the right, it climbs a much steeper mountain to form product $C$.

Let's look at the pathway to form product $D$. The reactants start at $15{\text{ kJ mol}}^{-1}$, and the peak of the curve (the transition state) is exactly at the $20{\text{ kJ mol}}^{-1}$ mark.

The activation enthalpy ($E_a$) is the energy difference between the peak and the reactants.

Because this energy barrier is relatively small, the reaction to form $D$ will happen quickly. This makes $D$ the kinetically favored product.

Now, let's trace the pathway to form product $C$ on the right. Notice how high this peak is! It goes way above the $20{\text{ kJ mol}}^{-1}$ mark. While the exact number isn't explicitly labeled on the y-axis, it is visually obvious that the activation enthalpy to form $C$ is significantly greater than $5{\text{ kJ mol}}^{-1}$.

However, look at where product $C$ ends up. It settles at an enthalpy of $5{\text{ kJ mol}}^{-1}$, which is the lowest energy state on the entire graph. Because systems naturally prefer lower energy states, $C$ is the thermodynamically stable product.

With this understanding, let's evaluate the given options to find the incorrect one.

Option (a) states that $D$ is the kinetically stable product. As we calculated, $D$ has the lowest activation barrier, meaning it forms the fastest. This statement is correct.

Option (b) claims that the formation of $A$ and $B$ from $C$ has the highest enthalpy of activation. To go from $C$ back to $A+B$, you must climb from the lowest valley ($5{\text{ kJ mol}}^{-1}$) to the highest peak on the graph. This is indeed the largest energy difference, making this statement correct.

Option (c) says $C$ is the thermodynamically stable product. Since $C$ rests at the lowest enthalpy level ($5{\text{ kJ mol}}^{-1}$), it is the most stable. This statement is correct.

Finally, option (d) claims that the activation enthalpy to form $C$ is $5{\text{ kJ mol}}^{-1}$ less than that to form $D$. If $E_{a,D}$ is $5{\text{ kJ mol}}^{-1}$, this would mean $E_{a,C}$ is $0{\text{ kJ mol}}^{-1}$. But we clearly saw that the peak for $C$ is a massive mountain, much higher than the peak for $D$. Therefore, this statement is completely false.

Option (d) is the incorrect statement and the correct answer to our problem.