What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentrat

Exercises · 6.16

Q. What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ? 2ICl

(g) ⇌ I2

(g) + Cl2 (g); Kc = 0.14

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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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For the reaction $2\text{ICl} \rightleftharpoons \text{I}_2 + \text{Cl}_2$ with $K_c = 0.14$ and initial $[\text{ICl}] = 0.78\ \text{M}$ , the equilibrium concentrations are $[\text{ICl}] = 0.48\ \text{M}$ , $[\text{I}_2] = [\text{Cl}_2] = 0.15\ \text{M}$ .

The key idea here is that the equilibrium constant tells us the ratio of products to reactants at equilibrium. Since the reaction starts with only ICl, the products must form in equal amounts (one mole of I₂ and one mole of Cl₂ per two moles of ICl consumed). This symmetry simplifies the algebra.

Let’s walk through it step by step.

Write the balanced equation and the expression for $K_c$ . The reaction is:

2\text{ICl}(g) \rightleftharpoons \text{I}_2(g) + \text{Cl}_2(g)

The equilibrium constant expression is:

K_c = \frac{[\text{I}_2][\text{Cl}_2]}{[\text{ICl}]^2} = 0.14

Set up the change in concentrations using a variable. Let $x$ be the concentration of I₂ that forms at equilibrium. Because the stoichiometry is 1:1 for I₂ and Cl₂, the same $x$ applies to Cl₂:

[\text{I}_2] = x \quad \text{and} \quad [\text{Cl}_2] = x

For every mole of I₂ formed, 2 moles of ICl are consumed. So the change in ICl is $-2x$ , and its equilibrium concentration is:

[\text{ICl}] = 0.78 - 2x

Substitute into the $K_c$ expression.

0.14 = \frac{(x)(x)}{(0.78 - 2x)^2} = \frac{x^2}{(0.78 - 2x)^2}

Solve for $x$ by taking the square root of both sides. This is a neat shortcut — because both numerator and denominator are perfect squares, we avoid solving a quadratic:

\sqrt{0.14} = \frac{x}{0.78 - 2x}

Calculate $\sqrt{0.14}$ :

\sqrt{0.14} \approx 0.3742

So:

0.3742 = \frac{x}{0.78 - 2x}

Solve the linear equation. Multiply both sides by $(0.78 - 2x)$ :

0.3742(0.78 - 2x) = x

0.2919 - 0.7484x = x

0.2919 = x + 0.7484x = 1.7484x

x = \frac{0.2919}{1.7484} \approx 0.167\ \text{M}

Watch out

A common mistake is to forget that the denominator is squared. If you mistakenly write $K_c = \frac{x^2}{0.78 - 2x}$ , you’ll get a different (wrong) answer. Always check the exponent in the $K_c$ expression.

Check if the approximation is valid.

The change $2x = 0.334\ \text{M}$ is less than the initial 0.78 M, so the equilibrium concentration of ICl is positive: $0.78 - 0.334 = 0.446\ \text{M}$ . This is fine — no negative concentrations.
Compute the equilibrium concentrations precisely.

Let’s use more exact arithmetic to avoid rounding errors.

x = \frac{0.78 \cdot \sqrt{0.14}}{1 + 2\sqrt{0.14}}

With $\sqrt{0.14} = \sqrt{\frac{14}{100}} = \frac{\sqrt{14}}{10} \approx \frac{3.7417}{10} = 0.37417$ : …

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Q.What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ? 2ICl

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