At a certain temperature and total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms: I2 (g)

Exercises · 6.3

Q. At a certain temperature and total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms: I2

(g) ⇌ 2I (g). Calculate Kp for the equilibrium.

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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.
Ignoring pure solids and liquids: Their concentrations are constant and are absorbed into $K$ — they do not appear in the expression. For example, in $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$ , $K_c = [CO_2]$ .
Confusing $K_c$ and $K_p$ : $K_p$ uses partial pressures (in atm or bar) instead of concentrations. The form is identical, but the numerical value differs unless $\Delta n = 0$ .

The Big Picture

$K$ is a thermodynamic fingerprint of a reaction at a given temperature. It tells you:

Direction: Compare $Q$ (the reaction quotient, same formula but with any concentrations) to $K$ . If $Q < K$ , the reaction moves forward. If $Q > K$ , it moves backward.
Extent: Large $K$ → products favoured; small $K$ → reactants favoured.
Temperature dependence: Use Le Chatelier's principle or the van't Hoff equation (for advanced problems).

Once you see $K$ as a ratio of "what's made" to "what's left" at equilibrium, the calculations become straightforward — just careful algebra with the right numbers.

Many students find this page while searching "Equilibrium Constant Calculation formula chemistry" or "Equilibrium Constant Calculation important questions and answers"; the concept sits firmly within the Class 11 Chemistry NCERT/CBSE syllabus. It's also a frequent building block for numericals in JEE Main, NEET and state CET Chemistry papers, so treating it as a one-time memorisation task rather than an understood idea tends to backfire later.

The equilibrium constant $K_p$ is calculated from the partial pressures of the species at equilibrium. Given 40% dissociation by volume, the mole fractions give $P_I = 0.4 \times 10^5$ Pa and $P_{I_2} = 0.6 \times 10^5$ Pa, so $K_p = \frac{(0.4 \times 10^5)^2}{0.6 \times 10^5} = 2.67 \times 10^4$ Pa.

The key idea here is that for a gaseous equilibrium, $K_p$ is expressed in terms of partial pressures. The phrase "40% by volume of I atoms" is crucial — in a gas mixture, volume percent equals mole percent (Avogadro's law). So we can directly find the mole fractions of I and I₂ at equilibrium, then convert to partial pressures using the total pressure.

Let’s walk through it step by step.

Interpret the given data. The equilibrium is:

I_2(g) \rightleftharpoons 2I(g)

Total pressure $P_{\text{total}} = 10^5$ Pa.

"40% by volume of I atoms" means that out of every 100 volumes of the equilibrium mixture, 40 volumes are contributed by I atoms. Since volume fraction = mole fraction, the mole fraction of I atoms is 0.4. But careful: this is the mole fraction of atoms, not of I₂ molecules. The mixture contains I₂ molecules and I atoms. So if the mole fraction of I atoms is 0.4, then the mole fraction of I₂ molecules must be 0.6 (since total mole fraction = 1).

Relate mole fractions to partial pressures. For an ideal gas mixture, partial pressure of a component = (mole fraction) × (total pressure). So:

P_I = 0.4 \times 10^5 \text{ Pa}

P_{I_2} = 0.6 \times 10^5 \text{ Pa}

Write the expression for $K_p$ . For the reaction $I_2 \rightleftharpoons 2I$ ,

K_p = \frac{(P_I)^2}{P_{I_2}}

Notice that the stoichiometric coefficient of I is 2, so its partial pressure is squared.

Substitute the values.

K_p = \frac{(0.4 \times 10^5)^2}{0.6 \times 10^5}

Simplify step by step:

(0.4 \times 10^5)^2 = 0.16 \times 10^{10}

Divide by $0.6 \times 10^5$ :

K_p = \frac{0.16 \times 10^{10}}{0.6 \times 10^5} = \frac{0.16}{0.6} \times 10^{5}

\frac{0.16}{0.6} = \frac{16}{60} = \frac{4}{15} \approx 0.2667

So:

K_p = 0.2667 \times 10^5 = 2.667 \times 10^4 \text{ Pa}

Watch out

A common mistake is to treat "40% by volume of I atoms" as the degree of dissociation or as the mole fraction of I₂. Remember: it directly gives the mole fraction of atomic iodine in the equilibrium mixture. Also, don't forget to square the partial pressure of I — the exponent comes from the stoichiometric coefficient.

Tip

If you prefer working with a degree of dissociation $\alpha$ , you can set up: let initial moles of I₂ = 1, then at equilibrium, moles of I₂ = $1 - \alpha$ , moles of I = $2\alpha$ , total moles = $1 + \alpha$ . The mole fraction of I atoms is $\frac{2\alpha}{1+\alpha} = 0.4$ , solve to get $\alpha = 0.25$ , then partial pressures follow. Both methods give the same $K_p$ .

✓Final answer

K_p

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Q.At a certain temperature and total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms: I2

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