Animated Solution for Chemistry - Chemical Equilibrium: Comprehension Passage

This comprehension passage from JEE Advanced is a beautiful amalgamation of chemical equilibrium, stoichiometry, and thermodynamics. It tests not just your ability to manipulate algebraic expressions for equilibrium constants, but also your deep conceptual understanding of how Gibbs free energy dictates the position of equilibrium. Let's embark on a detailed journey to unravel both questions.

The reaction given is the thermal decomposition of a diatomic gas $X_2$ into its constituent atoms:

We are told that initially, we have exactly $1$ mole of $X_2$ and $0$ moles of $X$. As the reaction proceeds towards equilibrium, some of the $X_2$ molecules dissociate. The problem defines a specific variable, ${\beta }_{\text{equilibrium}}$, as the number of moles of $X$ formed at equilibrium.

This is where many students make a critical error. You must look at the stoichiometry of the balanced equation. For every $2$ moles of $X$ produced, exactly $1$ mole of $X_2$ must have been consumed. Therefore, if ${\beta }_{\text{equilibrium}}$ moles of $X$ are formed, the amount of $X_2$ that reacted is exactly half of that, which is $\frac{{\beta }_{\text{equilibrium}}}{2}$.

Let's formalize this logic using an ICE (Initial, Change, Equilibrium) table. To keep the notation clean, let's denote ${\beta }_{\text{equilibrium}}$ simply as $\beta$.

Initial Moles: $X_2=1$ $X=0$

Change in Moles: $X_2=-\frac{\beta }{2}$ $X=+\beta$

Equilibrium Moles: $X_2=1-\frac{\beta }{2}$ $X=\beta$

To find the partial pressures, we first need the total number of moles at equilibrium. We simply sum the equilibrium moles of all gaseous species:

The equilibrium constant in terms of partial pressures, $K_P$, is defined as the product of the partial pressures of the products raised to their stoichiometric coefficients, divided by that of the reactants.

The partial pressure of any gas is its mole fraction multiplied by the total pressure, $P_T$. Let's write out the partial pressures:

Now, we substitute these into our $K_P$ expression:

Notice how one of the $P_T$ terms cancels out. We can simplify the complex fraction:

Using the algebraic identity $(a+b)(a-b)=a^2-b^2$ in the denominator, we get:

To clear the fraction in the denominator, multiply the numerator and denominator by $4$:

The problem explicitly states that the reaction is carried out at a constant total pressure of $2$ bar ($P_T=2$). Substituting this value yields our final expression for the first question:

This perfectly matches option (B) for Question 4.

Now, let's tackle Question 5, which asks us to identify the INCORRECT statement among four conceptual claims. This requires a deep dive into Le Chatelier's principle and chemical thermodynamics.

Analyzing Statement (A): "Decrease in the total pressure will result in formation of more moles of gaseous X." According to Le Chatelier's principle, if a system at equilibrium experiences a decrease in pressure, it will shift in the direction that produces more moles of gas to counteract the change. In our reaction, $\Delta n_g=2-1=1>0$. The forward reaction produces more gas molecules. Therefore, decreasing the pressure shifts the equilibrium forward, forming more $X$. Statement (A) is correct.

Analyzing Statement (B): "At the start of the reaction, dissociation of gaseous $X_2$ takes place spontaneously." Spontaneity is governed by the actual Gibbs free energy change, $\Delta G$, not the standard Gibbs free energy change, $\Delta G^{\circ }$. The relationship is given by the isotherm equation:

At the very start of the reaction ($t=0$), there is absolutely no product $X$. Therefore, the reaction quotient $Q=0$. As $Q$ approaches $0$, $\ln Q$ approaches $-\infty$. This makes the $RT\ln Q$ term infinitely negative, completely overpowering any positive $\Delta G^{\circ }$. Thus, $\Delta G$ is highly negative, meaning the forward reaction is highly spontaneous initially. Statement (B) is correct.

Analyzing Statement (C): "${\beta }_{\text{equilibrium}}=0.7$" Let's test this hypothesis. If $\beta =0.7$, what would be the value of $K_P$?

Since $3.92>3.51$, this would mean $K_P>1$.

However, the problem explicitly states that the standard reaction Gibbs energy, ${\Delta }_r{G}^{\circ }$, is positive. The fundamental link between thermodynamics and equilibrium is:

If ${\Delta }_r{G}^{\circ }>0$, then $-RT\ln K_P$ must be positive. Since $R$ and $T$ are positive constants, $\ln K_P$ must be negative. The natural logarithm of a number is negative if and only if that number is strictly between $0$ and $1$. Therefore, thermodynamics dictates that $K_P<1$.

Our calculation showed that if $\beta =0.7$, $K_P$ would be greater than $1$, which directly contradicts the thermodynamic constraint ${\Delta }_r{G}^{\circ }>0$. Therefore, $\beta$ cannot possibly be $0.7$. Statement (C) is mathematically and thermodynamically impossible, making it the INCORRECT statement.

Analyzing Statement (D): "$K_C<1$" We know the relationship between $K_P$ and $K_C$:

For this reaction, $\Delta n_g=1$, so $K_P=K_C(RT)$. Rearranging for $K_C$ gives:

We have already established that $K_P<1$. Let's look at the denominator, $RT$. At $298$ K, $RT=0.083\times 298\approx 24.7$.

We are dividing a number that is already less than $1$ by a number that is approximately $24.7$. The result must unequivocally be less than $1$. Therefore, $K_C<1$ is a true statement. Statement (D) is correct.

By systematically applying the principles of stoichiometry, Le Chatelier's principle, and the thermodynamic definitions of equilibrium, we have rigorously proven that the equilibrium constant expression is $\frac{8{\beta }^2}{4-{\beta }^2}$, and that the claim ${\beta }_{\text{equilibrium}}=0.7$ violates the fundamental thermodynamic constraints of the system.