Q.At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is 8.3 × 10⁻³. If decomposition is depicted as, PCl5

Concept: Equilibrium constant expressions, reverse reactions, and Le Chatelier's principle applied to $K_c$.

(a) Expression for $K_c$:

For the decomposition $\ce{PCl5(g) \rightleftharpoons PCl3(g) + Cl2(g)}$, the equilibrium constant is the ratio of product concentrations to reactant concentration:

(b) $K_c$ for the reverse reaction:

The reverse reaction is $\ce{PCl3(g) + Cl2(g) \rightleftharpoons PCl5(g)}$. Its equilibrium constant is the reciprocal:

(c) Effect on $K_c$:

(i) Adding more $\ce{PCl5}$: $K_c$ is unchanged. The equilibrium shifts right to consume the added reactant, but the ratio of concentrations at the new equilibrium remains $8.3 \times 10^{-3}$ at 473 K. …

The equilibrium constant $K_c$ is a ratio of product concentrations to reactant concentrations at equilibrium, each raised to its stoichiometric coefficient. It captures the position of equilibrium—how far a reaction proceeds before the forward and reverse rates balance. Crucially, $K_c$ depends only on temperature for a given reaction; adding more substance or changing pressure shifts the equilibrium position but leaves $K_c$ unchanged.

Let me walk through each part systematically.

(a) Expression for $K_c$

For the reaction

the equilibrium constant is the ratio of the concentrations of products to reactants, each raised to the power of its coefficient (all coefficients here are 1):

The square brackets denote molar concentrations at equilibrium. This expression tells us that at 473 K, when equilibrium is reached, the product of the concentrations of PCl₃ and Cl₂ divided by the concentration of PCl₅ will always equal $8.3 \times 10^{-3}$.

(b) $K_c$ for the reverse reaction

When we reverse a reaction, the products become reactants and vice versa. The equilibrium constant for the reverse reaction is simply the reciprocal of the forward constant.

For the reverse reaction:

we have

Substituting the given value:

So $K_c$ for the reverse reaction is approximately $1.2 \times 10^{2}$ or 120 at the same temperature.

(c) Effect on $K_c$ under different conditions

This is where understanding the nature of $K_c$ becomes essential. The equilibrium constant is a function of temperature alone for a given reaction. Changes in concentration or pressure shift the equilibrium position (the actual concentrations change) but do not alter $K_c$ itself.

(i) Adding more PCl₅

When you add more PCl₅, you increase its concentration. The system responds by shifting the equilibrium to the right (toward products) to consume some of the added PCl₅, according to Le Chatelier's principle. The concentrations of PCl₃ and Cl₂ increase, and the concentration of PCl₅ decreases from its new higher value until the ratio $\frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}$ once again equals $8.3 \times 10^{-3}$.

Effect on $K_c$: No change. The value remains $8.3 \times 10^{-3}$ because $K_c$ depends only on temperature.

(ii) Increasing pressure

Increasing the total pressure (say, by decreasing volume) affects the equilibrium position for reactions involving gases where the number of moles changes. Here, 1 mole of PCl₅ produces 2 moles of gas (PCl₃ + Cl₂). The system shifts toward the side with fewer moles—toward the left (reactants)—to reduce pressure.

However, $K_c$ is defined in terms of concentrations, not partial pressures. While the equilibrium shifts and individual concentrations change, the ratio $\frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}$ adjusts to maintain the same value.

Effect on $K_c$: No change. Again, $K_c = 8.3 \times 10^{-3}$ because temperature is constant.