Animated Solution for Chemistry - Chemical Equilibrium: For the reaction, $2{\text{SO}}_2(g)+{\text{O}}_2(g)⟶2{\text{SO}}_3(g)$, $\Delta H=-57.2{\text{ kJ mol}}^{-1}$ and $K_c=1.7\times 10^{16}$. Which of the following statement is incorrect?

The Tale of Thermodynamics vs. Kinetics: Le Chatelier's Principle in Action

Have you ever wondered why some chemical reactions, despite being highly favorable, just sit there doing nothing? It’s like having a massive boulder at the top of a hill—it wants to roll down, but a tiny pebble is blocking its path. This classic JEE problem perfectly captures the beautiful, yet often misunderstood, relationship between thermodynamics and kinetics. Let's dive into the Contact process and dissect each statement to uncover the truth!

We are given the reaction for the oxidation of sulfur dioxide:

The problem provides two crucial pieces of information. First, the enthalpy change is $\Delta H=-57.2{\text{ kJ mol}}^{-1}$. The negative sign screams exothermic—this reaction releases heat into its surroundings. Second, the equilibrium constant is astronomically large, $K_c=1.7\times 10^{16}$. This tells us that at equilibrium, the reaction vessel is practically swimming in ${\text{SO}}_3$ with barely any reactants left.

What happens if we turn up the heat? According to Le Chatelier's Principle, a system at equilibrium will always fight back against any change you impose. If you increase the temperature, the system will try to cool itself down by favoring the endothermic (heat-absorbing) direction.

Since our forward reaction is exothermic, the backward reaction must be endothermic. Therefore, increasing the temperature shifts the equilibrium backward, causing the equilibrium constant $K_c$ to decrease. Statement (a) is absolutely correct!

Imagine a crowded room (our reaction vessel) where people (molecules) are mingling. If you suddenly throw in a bunch of mannequins (inert gas), the room gets more cramped (higher total pressure).

However, the actual people can still mingle just as easily because the size of the room hasn't changed, and the mannequins don't interact with them. In chemistry terms, the partial pressures and concentrations of the reacting gases remain completely unchanged. Thus, the equilibrium is undisturbed. Statement (b) is correct.

What about increasing the pressure of the system? To understand this, we need to look at the change in the number of gaseous moles, $\Delta n_g$.

Because the reaction produces fewer moles of gas, it naturally reduces the pressure of the system. If we externally increase the pressure, Le Chatelier's Principle dictates that the system will shift in the direction that reduces pressure—the forward direction. Statement (c) is correct.

Statement (d) claims that because $K_c$ is so large, the reaction goes to completion and no catalyst is required. This is where mistakes happen!

Thermodynamics (represented by $K_c$) only tells us the destination. It says, "If you wait long enough, you will get a lot of product." But it says absolutely nothing about the journey or how long it will take. That is the domain of kinetics.

In reality, the oxidation of ${\text{SO}}_2$ at normal temperatures is agonizingly slow. To make this reaction industrially viable in the Contact process, we absolutely must use a catalyst, typically Vanadium Pentoxide (${\text{V}}_2{\text{O}}_5$), to speed things up. A catalyst lowers the activation energy, allowing the system to reach that highly favorable equilibrium much faster. Therefore, statement (d) is fundamentally flawed.

The final answer is option (d). Always remember: a large equilibrium constant guarantees a high yield, but it never guarantees a fast reaction!