Chemical equilibrium and Kc in Year 4

End of September 2026: in Year 4 at the liceo scientifico (the Italian science-focused high school), after the block on solutions and concentration and the one on thermochemistry, natural sciences reaches kinetics and equilibrium — 3 hours a week (5 in the applied-sciences track). The test on equilibrium and $K_{c}$ typically falls six to nine weeks from now, so between the beginning of November and the beginning of December. It is not an isolated block: straight afterwards come acids, bases, pH and hydrolysis, which are built entirely on what you understand now — if the idea of equilibrium stays vague, pH becomes a list of meaningless formulas instead of a special case of the same logic.

What “dynamic equilibrium” really means

The first misunderstanding to clear up right away: equilibrium does not mean that the reaction has stopped. In a closed system, when a reaction reaches equilibrium, the forward and reverse reactions both continue to proceed, but at the same rate. The net result is that the concentrations of reactants and products stay constant over time, not because nothing happens any more, but because what is consumed on one side is re-formed at exactly the same rate on the other. That is why we speak of dynamic, not static, equilibrium: if you could “label” individual molecules, you would see them continue to turn back and forth even when the concentrations measured in the lab no longer change.

The equilibrium constant Kc

For a generic reaction at equilibrium

a A + b B ⇌ c C + d D

the equilibrium constant is written as the ratio between the concentrations of the products and those of the reactants, each raised to its own stoichiometric coefficient:

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

At a given temperature, $K_{c}$ is a fixed number for that reaction: it changes only if the temperature changes, not if the starting concentrations change. One point that causes systematic mistakes: pure solids and pure solvents do not appear in the expression for $K_{c}$ . Their concentration, expressed as moles per volume of the solid itself, is an intrinsic constant of the material and does not vary during the reaction, so it is absorbed into the constant instead of appearing as a variable. For the decomposition of calcium carbonate,

CaC O_{3} (s) ⇌ CaO (s) + C O_{2} (g)

the correct expression is $K_{c} = [C O_{2}]$ : the two solids do not appear, only the gas remains.

The reaction quotient Q: predicting the direction before reaching equilibrium

$Q$ is calculated with the exact same expression as $K_{c}$ , but using the concentrations at any instant, not necessarily the equilibrium ones. Comparing $Q$ with $K_{c}$ tells you in which direction the system will move to reach equilibrium:

if $Q < K_{c}$ , there are relatively too many reactants compared with products: the reaction proceeds forwards, towards the products, until $Q$ rises to equal $K_{c}$ ;
if $Q > K_{c}$ , it is the opposite: the reaction proceeds backwards, towards the reactants, until $Q$ falls to $K_{c}$ ;
if $Q = K_{c}$ , the system is already at equilibrium and does not shift.

Le Châtelier’s principle

When a system at equilibrium undergoes an external disturbance, the system responds by shifting in the direction that tends to minimise the effect of the disturbance, until it reaches a new equilibrium.

Concentration. Adding a reactant (or removing a product) shifts the equilibrium towards the products, because the system “consumes” the added excess by producing more of the species on the other side. The exact mechanism is that of $Q$ and $K_{c}$ : adding a reactant instantly lowers $Q$ below $K_{c}$ , and the reaction proceeds forwards until $Q$ returns to equal $K_{c}$ .

Pressure and volume. Reducing the volume of a gaseous system (and so increasing the pressure), the equilibrium shifts towards the side with fewer total moles of gas, because that is the direction that reduces the overall pressure of the system. If the two sides of the reaction have the same number of gaseous moles, a change in volume does not shift the equilibrium at all. This disturbance has no effect if there are no gaseous species in the reaction, or if they appear in equal numbers on both sides.

Temperature. It is the only disturbance that changes the value of $K_{c}$ itself, not just the position of the equilibrium. It helps to treat heat as if it were a reactant (in an endothermic reaction) or a product (in an exothermic reaction): raising the temperature of an exothermic reaction shifts the equilibrium towards the reactants (as if an excess “product” had been added), whereas in an endothermic reaction it shifts it towards the products.

A catalyst does not shift the equilibrium. A catalyst lowers the activation energy of both the forward and the reverse reaction, to the same extent: it makes equilibrium be reached faster, but changes neither the position of that equilibrium nor the value of $K_{c}$ . It is a very common mistake to think that a catalyst increases the yield of the reaction: it only increases the speed with which you get there.

The typical mistake to avoid

The most frequent of all is including pure solids or solvents in the expression for $K_{c}$ : every time there is an $(s)$ or the solvent of the reaction, it has to be left out. The second is saying generically “the equilibrium shifts to the right” without being able to link that statement to the mechanism — a comparison between $Q$ and $K_{c}$ , or the direction that minimises the disturbance — because in a test a correct statement without the explanation of why is usually worth only half a mark. The third is confusing $Q$ with $K_{c}$ : they are the same formula, but $K_{c}$ is a fixed value at a given temperature, whereas $Q$ changes at every instant until the system is at equilibrium.

Four worked examples

1. Writing Kc and calculating it from experimental data

Write the expression for $K_{c}$ for each of these reactions:

CaC O_{3} (s) ⇌ CaO (s) + C O_{2} (g) K_{c} = [C O_{2}]

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g) K_{c} = \frac{[ N H _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}}

In the first, the two solids do not appear. In the second, all the species are gaseous and all appear, each with its own stoichiometric coefficient as the exponent.

Now calculate $K_{c}$ for the synthesis of ammonia, knowing that at equilibrium $[N_{2}] = 0.50$ M, $[H_{2}] = 0.30$ M, $[N H_{3}] = 0.20$ M:

K_{c} = \frac{( 0.20 ) ^{2}}{( 0.50 ) ( 0.30 ) ^{3}} = \frac{0.0400}{0.50 \times 0.0270} = \frac{0.0400}{0.0135} \approx 2.96

The ready-made tests, for maths and physics, are in the practice-test category for high school: there is no dedicated science quiz in the catalogue, so there you train on the other subjects — the correct answer is free, the guided step-by-step solution unlocks with the first lesson.

2. ICE table with a quadratic equation

In the reactor, $H_{2}$ and $I_{2}$ are made to react according to

H_{2} (g) + I_{2} (g) ⇌ 2 HI (g) K_{c} = 54.0

The initial concentrations are $[H_{2}]_{0} = 0.30$ M and $[I_{2}]_{0} = 0.20$ M, with no HI at the start. Find the concentrations at equilibrium.

ICE table (initial — change — equilibrium), with $x$ the mol/L of $H_{2}$ and $I_{2}$ consumed:

	$H_{2}$	$I_{2}$	$HI$
Initial	$0.30$	$0.20$	$0$
Change	$- x$	$- x$	$+ 2 x$
Equilibrium	$0.30 - x$	$0.20 - x$	$2 x$

Substituting into the expression for $K_{c}$ :

K_{c} = \frac{( 2 x ) ^{2}}{( 0.30 - x ) ( 0.20 - x )} = 54.0

I expand the denominator and bring everything to one side:

4 x^{2} = 54.0 \times (0.06 - 0.50 x + x^{2})

4 x^{2} = 3.24 - 27 x + 54 x^{2}

50 x^{2} - 27 x + 3.24 = 0

Discriminant:

Δ = 2 7^{2} - 4 \times 50 \times 3.24 = 729 - 648 = 81

x = \frac{27 \pm 81}{100} = \frac{27 \pm 9}{100}

The two solutions are $x = 0.36$ and $x = 0.18$ . The first has to be discarded: it would exceed the initial concentration of $I_{2}$ ( $0.20$ M), which is physically impossible — you cannot consume more reactant than there is. That leaves $x = 0.18$ .

Concentrations at equilibrium: $[H_{2}] = 0.30 - 0.18 = 0.12$ M, $[I_{2}] = 0.20 - 0.18 = 0.02$ M, $[HI] = 2 \times 0.18 = 0.36$ M.

Check: $K_{c} = \frac{( 0.36 ) ^{2}}{( 0.12 ) ( 0.02 )} = \frac{0.1296}{0.0024} = 54.0$ — it matches exactly.

3. Comparing Q and Kc to predict the direction

For the reaction $A (g) + B (g) ⇌ C (g)$ , $K_{c} = 4.00$ . At a given instant the concentrations are $[A] = 1.0$ M, $[B] = 2.0$ M, $[C] = 3.0$ M. In which direction does the system evolve?

Q = \frac{[ C ]}{[ A ] [ B ]} = \frac{3.0}{1.0 \times 2.0} = 1.5

$Q = 1.5$ is less than $K_{c} = 4.00$ : there are, proportionally, too many reactants and too little product compared with equilibrium. The reaction proceeds forwards, consuming A and B and forming more C, until $Q$ rises to reach $4.00$ .

4. Three Le Châtelier disturbances

The synthesis of sulfur trioxide is an exothermic reaction:

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g) Δ H < 0

Disturbance (a): I add $O_{2}$ . Increasing the concentration of a reactant lowers $Q$ below $K_{c}$ : the equilibrium shifts towards the products, that is, more $S O_{3}$ forms.

Disturbance (b): I reduce the volume of the reactor (I increase the pressure). I count the gaseous moles on the two sides: 3 moles in total on the left ( $2 + 1$ ), 2 moles on the right. The equilibrium shifts towards the side with fewer moles of gas, so towards the right, forming more $S O_{3}$ — it is the direction that reduces the overall pressure of the system.

Disturbance (c): I raise the temperature. The forward reaction is exothermic: treating heat as a product in excess, raising the temperature shifts the equilibrium towards the reactants, so towards the left — less $S O_{3}$ forms, not more. This is the case where intuition (“hotter, faster, so more product”) leads you astray: the rate increases in both directions, but the position of the equilibrium shifts the opposite way from what you would instinctively expect.

Before the test

The method I use in lessons is always the same, however complex the reaction looks: first the expression for $K_{c}$ (checking which solids and solvents to leave out), then the ICE table if you need the equilibrium concentrations, then the $Q$ against $K_{c}$ comparison if the question asks only for the direction. For the part on solutions and concentration that precedes this block in the Year 4 syllabus, you will find the step-by-step method in this post on molarity and dilutions; for the conversions between mass and moles that every line of these calculations needs, there is this post on the mole and stoichiometry.

The full picture of the subject is on the science page, while the ready-made maths and physics tests are in the practice-test category for high school. If you want to revise equilibrium before the November test, or already have a test to go over together, write to me via contact. The next step is Acids, bases and pH in Year 4: how to calculate pH, which uses the same logic as $K_{c}$ ; for the temperature disturbance and the sign of $Δ H$ see Thermochemistry: enthalpy and Hess’s law with exercises.

Chemical equilibrium and Le Châtelier: Kc made clear