Calculating Equilibrium Concentrations: Study Guide (Chemistry 7.7)

The question this topic answers

In Topic 7.4 you were given the concentrations at equilibrium and you calculated K. Now you work the other way round. You know:

• the balanced equation,
• the initial concentrations (or partial pressures), and
• the value of K at the temperature of the experiment.

You must predict the concentrations, or partial pressures, of every species once equilibrium is reached. Every problem of this type follows the same two steps:

1. Compare Q with K to find which way the reaction will go.
2. Use an ICE table to track how far it goes, then solve for the unknown.

Step 1: compare Q with K

The reaction quotient Q has exactly the same form as K, but you fill it with the concentrations you have now, not the equilibrium ones (Topic 7.3). A mixture always moves so that Q gets closer to K.

Comparison	What happens next	Why
Q < K	Net forward reaction: reactants are used up, products form	The ratio of products to reactants is too small, so products must increase (and reactants decrease) until Q = K
Q > K	Net reverse reaction: products are used up, reactants form	The ratio of products to reactants is too large, so products must decrease (and reactants increase) until Q = K
Q = K	No net change: the mixture is already at equilibrium	Forward and reverse rates are already equal

Two special cases come up often:

• If there are no products at the start, Q = 0, so Q < K and the reaction must go forwards.
• If there are no reactants at the start, Q cannot be calculated (division by zero) and you can treat it as infinitely large. The reaction must go backwards.

What “at equilibrium” really means

When Q = K, the reaction has not stopped. Particles still collide and react in both directions. The forward rate equals the reverse rate, so every species is made as fast as it is used up. The result is that concentrations, and therefore the proportion of reactants to products, stay constant. This is dynamic equilibrium. A flat line on a concentration–time graph means “no net change”, not “no reaction”.

Step 2: the ICE table

An ICE table has three rows under the balanced equation:

• I – Initial: the concentrations you start with (in mol L⁻¹ for Kc, atm for Kp).
• C – Change: how each one changes, written in terms of one unknown, x.
• E – Equilibrium: Initial + Change.

The Change row follows the coefficients. For a reaction 2 G + H ⇌ 3 J going forwards, if H falls by x then G falls by 2x and J rises by 3x. Use a minus sign for the side being used up and a plus sign for the side being formed. Which side is which comes from Step 1.

Three rules keep the table correct:

• Concentrations, not moles. If you are given moles and a volume, divide first: c = n / V.
• Leave out pure solids and liquids. They do not appear in K, so they do not need a row of concentrations.
• Kc with concentrations, Kp with partial pressures. Do not mix them. Converting between Kc and Kp is not assessed in this course, but you must use whichever one the question gives.

Then substitute the Equilibrium row into the K expression, solve for x and work out each equilibrium value. Finally, check: put your answers back into the K expression. You should get K back, to within rounding.

Representing the approach to equilibrium

A concentration–time graph shows the same story as the ICE table. Figure 1 shows the reaction in Worked example 1 below.

When you sketch a graph like this, check four things:

• The vertical axis is labelled with the quantity and its units (mol L⁻¹ for concentration, atm for pressure) and has an even scale.
• Each curve starts at its initial value and ends at its equilibrium value.
• The sizes of the changes are in the ratio of the coefficients.
• All the curves become flat at the same time, because equilibrium is reached by the whole mixture at once.

Worked example 1: a perfect square

Question. The fictional gases X₂ and Y₂ react: X₂(g) + Y₂(g) ⇌ 2 XY(g), with Kc = 25.0 at a fixed temperature. A flask starts with [X₂] = [Y₂] = 0.200 mol L⁻¹ and no XY. Calculate the equilibrium concentrations.

1. Direction. There is no XY, so Q = 0 < K. The reaction goes forwards.
2. ICE table.

	X₂	Y₂	2 XY
I (mol L⁻¹)	0.200	0.200	0
C (mol L⁻¹)	−x	−x	+2x
E (mol L⁻¹)	0.200 − x	0.200 − x	2x

3. Substitute. Kc = [XY]² / ([X₂][Y₂]) = (2x)² / (0.200 − x)² = 25.0
4. Solve. Both sides are perfect squares, so take the square root: 2x / (0.200 − x) = 5.00. Then 2x = 1.00 − 5x, so 7x = 1.00 and x = 0.1429 mol L⁻¹.
5. Equilibrium values. [XY] = 2x = 0.286 mol L⁻¹; [X₂] = [Y₂] = 0.200 − 0.1429 = 0.0571 mol L⁻¹.

Check. (0.2857)² / (0.05714)² = 25.0. ✓ The square root step is only allowed because the top and bottom are both squares. Use the positive root: the negative one would give a negative concentration.

Worked example 2: a very small K

Question. The fictional gas A₂ splits into atoms: A₂(g) ⇌ 2 A(g), with Kc = 4.0 × 10⁻⁶. A 0.500 L flask starts with 0.250 mol of A₂ and no A. Calculate [A] at equilibrium.

1. Concentration first. [A₂] = 0.250 mol ÷ 0.500 L = 0.500 mol L⁻¹.
2. Direction. No A, so Q = 0 < K: forwards.
3. ICE table.

	A₂	2 A
I (mol L⁻¹)	0.500	0
C (mol L⁻¹)	−x	+2x
E (mol L⁻¹)	0.500 − x	2x

4. Substitute. Kc = (2x)² / (0.500 − x) = 4.0 × 10⁻⁶.
5. Small-x shortcut. K is tiny, so very little A₂ splits (Topic 7.5). Assume x is so small that 0.500 − x ≈ 0.500. Then 4x² = 4.0 × 10⁻⁶ × 0.500 = 2.0 × 10⁻⁶, so x² = 5.0 × 10⁻⁷ and x = 7.07 × 10⁻⁴ mol L⁻¹.
6. Answer. [A] = 2x = 1.4 × 10⁻³ mol L⁻¹ (2 significant figures, matching K). [A₂] = 0.500 − 0.000707 = 0.499 mol L⁻¹, almost unchanged.

Check the shortcut. x is 7.07 × 10⁻⁴ ÷ 0.500 × 100 = 0.14% of the starting value, far below 5%. Solving the full quadratic gives x = 7.066 × 10⁻⁴, a difference of under 0.1%. The shortcut is safe here. If x had come out larger than about 5% of the starting value, you would need to solve the quadratic instead.

Worked example 3: Q > K, using partial pressures

Question. Two fictional isomers interconvert in the gas phase: E(g) ⇌ F(g), with Kp = 3.0. A container starts with P_E = 0.20 atm and P_F = 1.00 atm. Find the equilibrium partial pressures.

1. Direction. Qp = P_F / P_E = 1.00 / 0.20 = 5.0. Since Q > K, the reaction goes backwards: F is used up and E forms.
2. ICE table (signs set by step 1).

	E	F
I (atm)	0.20	1.00
C (atm)	+x	−x
E (atm)	0.20 + x	1.00 − x

3. Substitute and solve. (1.00 − x) / (0.20 + x) = 3.0, so 1.00 − x = 0.60 + 3x, 4x = 0.40 and x = 0.10 atm.
4. Answer. P_E = 0.30 atm, P_F = 0.90 atm.

Check. 0.90 / 0.30 = 3.0. ✓ The total pressure stays at 1.20 atm, which makes sense: one molecule of E becomes one molecule of F, so the number of gas molecules does not change.

What if you set the signs the wrong way? If you assume a forward reaction (−x for E, +x for F), the algebra still works and you get x = −0.10. A negative x is the algebra telling you the reaction went the other way. Checking Q first avoids the confusion.

Choosing a method

Situation	Method
Both sides of the K expression are perfect squares	Take the square root (Worked example 1)
K is very small compared with the starting concentration c	Assume c − x ≈ c, then check x is under 5% of c (Worked example 2)
Expression is linear in x	Rearrange directly (Worked example 3)
K is very large	Let the reaction go to completion on paper, then let a small amount come back (see Practice Q7)
None of these	Solve the quadratic

Common misconceptions

• “Put the initial concentrations into K.” Initial values give Q, not K. Only equilibrium values satisfy K.
• “Every species changes by x.” The changes follow the coefficients. For 2 XY the change is 2x, and it is squared in the K expression as (2x)².
• “Equilibrium means equal concentrations.” It means constant concentrations. In Worked example 1 there is five times as much XY as X₂ at equilibrium.
• “At equilibrium the reactions stop.” Both reactions continue at equal rates. Nothing changes overall, but particles still react.
• Using moles instead of concentrations in Kc. Divide by the volume first unless the question says the volume is 1.00 L.
• Using the small-x shortcut when K is not small. If x comes out bigger than about 5% of the starting value, or bigger than the starting value itself, the shortcut has failed.
• Assuming the reaction always goes forwards. Always compare Q with K first. Starting with products can make the reaction run backwards.
• Including solids or liquids in the ICE table. Pure solids and liquids do not appear in K, so their amounts do not enter the calculation.

Where this leads

Next, Topic 7.8 shows the same calculations with particle diagrams: you count particles in a box instead of reading concentrations from a table. Q-versus-K reasoning returns in Topics 7.9 and 7.10 on Le Châtelier’s principle, and ICE tables are the main tool for weak acids, weak bases and buffers in Unit 8. Try the practice questions now, then use the revision notes and the checklist to consolidate.