Write the expression for the equilibrium constant, Kc for each of the following reactions: (i) 2NOCl (g) ⇌ 2NO

Exercises · 6.4

Q. Write the expression for the equilibrium constant, Kc for each of the following reactions:

(i) 2NOCl

(g) ⇌ 2NO

(g) + Cl2

(g)

(ii) 2Cu(NO3)2 (s) ⇌ 2CuO (s) + 4NO2

(g) + O2

(g)

(iii) CH3COOC2H5(aq) + H2O(l) ⇌ CH3COOH (aq) + C2H5OH (aq)

(iv) Fe3+ (aq) + 3OH– (aq) ⇌ Fe(OH)3 (s)

(v) I2 (s) + 5F2 ⇌ 2IF5

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Concept understanding — Equilibrium Constant Calculation

Equilibrium Constant Calculation: From Intuition to Precision

Imagine you're at a party where people can move between two rooms. Some people prefer the kitchen (more snacks), others prefer the living room (better music). After a while, the number of people in each room stops changing — not because everyone froze, but because the rate of people leaving the kitchen equals the rate of people entering it. The system is in dynamic equilibrium.

Chemical reactions work the same way. A reversible reaction like

$A + B \rightleftharpoons C + D$ doesn't stop when it reaches equilibrium. Instead, the forward reaction (making C and D) and the reverse reaction (making A and B) happen at the same rate. The concentrations of A, B, C, and D become constant — not equal, but constant.

The equilibrium constant $K$ is a number that tells you where this balance lies. It answers the question: At equilibrium, which side of the reaction is favoured?

The Intuitive Idea

Think of a seesaw. If $K$ is very large (say $10^6$ ), the equilibrium sits heavily on the product side — almost all A and B have turned into C and D. If $K$ is very small (say $10^{-6}$ ), the opposite is true: hardly any product forms. If $K$ is around 1, both sides have comparable amounts.

So $K$ is a ratio — a comparison of product concentrations to reactant concentrations at equilibrium, each raised to the power of their stoichiometric coefficients.

The Precise Statement

For a general reversible reaction at a given temperature:

aA + bB \rightleftharpoons cC + dD

the equilibrium constant $K_c$ (for concentrations in mol/L) is:

K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}

where $[X]$ means the equilibrium concentration of species X in moles per litre.

Important

The exponents come directly from the balanced chemical equation. If the coefficient of A is 2, you square its concentration. This is not optional — it's built into the definition.

What $K$ Actually Depends On

$K$ is constant at a given temperature. Change the temperature, and $K$ changes. But $K$ does not depend on:

Initial concentrations
Presence of a catalyst (catalysts speed up both directions equally)
Pressure or volume changes (for $K_c$ ; $K_p$ for gases has its own rules)

This is a critical exam point: if a problem gives you initial concentrations and asks for $K$ , you must first find equilibrium concentrations — you cannot plug in initial values.

A Worked Example

Problem:

$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$

At 500°C, equilibrium concentrations are:

$[N_2] = 0.50\ \text{M}$ , $[H_2] = 1.50\ \text{M}$ , $[NH_3] = 0.20\ \text{M}$

Calculate $K_c$ .

Solution:

Write the expression:

K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}

Substitute:

K_c = \frac{(0.20)^2}{(0.50)(1.50)^3} = \frac{0.04}{0.50 \times 3.375} = \frac{0.04}{1.6875} \approx 0.0237

Note

The units cancel because the numerator and denominator both have units of (mol/L) $^2$ and (mol/L) $^4$ respectively — but by convention, $K_c$ is reported without units. The numerical value is what matters.

Common Pitfalls

Forgetting the exponents: A coefficient of 2 means square the concentration, not double it.
Using initial concentrations: You must use equilibrium concentrations only. …

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The equilibrium constant $K_c$ is written using active masses (molar concentrations) of only those species whose concentration can change measurably — gases and aqueous solutes. Pure solids and pure liquids are omitted because their activity is taken as 1. For each reaction, we identify the phases and write $K_c$ accordingly.

The core idea is simple: $K_c$ is the ratio of product concentrations (raised to their stoichiometric coefficients) to reactant concentrations (raised to their coefficients), but only for species that are in homogeneous solution or gas phase. Solids and pure liquids have constant density (and hence constant "concentration") during the reaction, so they don't appear in the expression — they are absorbed into the constant.

Let's apply this to each case.

Reaction (i): $2\text{NOCl}(g) \rightleftharpoons 2\text{NO}(g) + \text{Cl}_2(g)$ All species are gases. Every gas has a concentration that changes as the reaction proceeds. So we include all of them:

K_c = \frac{[\text{NO}]^2 [\text{Cl}_2]}{[\text{NOCl}]^2}

Tip

The coefficient becomes the exponent. For $2\text{NO}$ , it's $[\text{NO}]^2$ ; for $2\text{NOCl}$ , it's $[\text{NOCl}]^2$ .

Reaction (ii): $2\text{Cu(NO}_3)_2(s) \rightleftharpoons 2\text{CuO}(s) + 4\text{NO}_2(g) + \text{O}_2(g)$ $\text{Cu(NO}_3)_2$ and $\text{CuO}$ are solids — their "concentrations" are constant and do not appear. Only the gases $\text{NO}_2$ and $\text{O}_2$ appear:

K_c = [\text{NO}_2]^4 [\text{O}_2]

Watch out

A common mistake is to write the solids in the denominator or numerator. Remember: solids are omitted entirely, not set to 1 in the expression — they are simply not written.

Reaction (iii): $\text{CH}_3\text{COOC}_2\text{H}_5(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{CH}_3\text{COOH}(aq) + \text{C}_2\text{H}_5\text{OH}(aq)$ Water is a pure liquid. In dilute aqueous solutions, the concentration of water is essentially constant (about 55.5 M) and does not change measurably, so it is omitted. The ester, acid, and alcohol are all aqueous solutes:

K_c = \frac{[\text{CH}_3\text{COOH}][\text{C}_2\text{H}_5\text{OH}]}{[\text{CH}_3\text{COOC}_2\text{H}_5]}

Note

If water were the solvent in a non-aqueous reaction, or if the reaction were not in dilute solution, water might be included. But in standard aqueous organic reactions, it is treated as a pure liquid and omitted.

Reaction (iv): $\text{Fe}^{3+}(aq) + 3\text{OH}^-(aq) \rightleftharpoons \text{Fe(OH)}_3(s)$

The product $\text{Fe(OH)}_3$ is a solid — omitted. Only the aqueous ions appear:

$K_c = \frac{1}{[\text{Fe}^{3+}][\text{OH}^-]^3}$ …

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Q.Write the expression for the equilibrium constant, Kc for each of the following reactions:

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