Properties of the Equilibrium Constant: Study Guide (Chemistry 7.6)

One reaction, many equations

The same chemical change can be written in several ways. You can write it forwards or backwards, with whole-number coefficients or with halves. Each way gives a different K expression, so each has its own value of K.

This matters for two reasons:

1. A data table may give K for one version of an equation when the question asks about another version.
2. Many processes happen in steps. If you know K for each step, you can find K for the overall process without doing any new experiments.

All the rules below come from one idea: write out the expressions and see what happens to them. If you can do that, you never need to memorise the rules blindly. In this guide the K values are invented for practice.

Rule 1: reversing an equation inverts K

Take 2 NO(g) + O₂(g) ⇌ 2 NO₂(g), with

K₁ = [NO₂]² / ([NO]²[O₂])

Written the other way, 2 NO₂(g) ⇌ 2 NO(g) + O₂(g), the products and reactants swap places:

K₂ = [NO]²[O₂] / [NO₂]² = 1 / K₁

So K_reverse = 1 / K_forward. This fits Topic 7.5: if the forward reaction is strongly product-favoured (large K), the reverse reaction is strongly reactant-favoured (small K).

Rule 2: multiplying the coefficients by c raises K to the power c

Coefficients become powers in the expression. Multiply every coefficient by c and every power is multiplied by c, so the whole expression is raised to the power c.

Halve the equation above: NO(g) + ½ O₂(g) ⇌ NO₂(g).

K₃ = [NO₂] / ([NO][O₂]^½) = (K₁)^½ = √K₁

Doubling an equation squares K; it does not double it. Tripling cubes K. Halving takes the square root.

Rule 3: adding equations multiplies K values

When you add two equations, you multiply their expressions. Anything that appears as a product of the first step and a reactant of the second step cancels, just as it cancels from the overall equation.

A two-step pathway that could make NO₂ (values invented):

Step	Equation	K
1	2 NO(g) ⇌ N₂O₂(g)	K_a = 80
2	N₂O₂(g) + O₂(g) ⇌ 2 NO₂(g)	K_b = 8.0 × 10⁷
Overall	2 NO(g) + O₂(g) ⇌ 2 NO₂(g)	K = K_a × K_b = 6.4 × 10⁹

Check with the expressions:

K_a × K_b = [N₂O₂]/[NO]² × [NO₂]²/([N₂O₂][O₂]) = [NO₂]² / ([NO]²[O₂])

The intermediate N₂O₂ cancels, and what is left is the expression for the overall equation. This is how you represent a multistep process with one overall expression: add the steps, multiply the K values, and check that the intermediates cancel.

Summary of the rules

What you do to the equation	What happens to K	What happens to ΔH (Topic 6.9)
Reverse it	K → 1/K	ΔH → −ΔH
Multiply all coefficients by c	K → Kᶜ	ΔH → c × ΔH
Add two equations	K = K₁ × K₂	ΔH = ΔH₁ + ΔH₂

A neat way to remember the link: whatever ΔH does by adding or multiplying, K does one level up, by multiplying or raising to a power.

The rules are the same for Kp. Partial pressures take the place of concentrations, but coefficients still become powers, so reversing, rescaling and adding work in exactly the same way. Just keep every K in one calculation the same type.

A strategy for any target equation

Questions on this topic usually give you two or three equations with their K values and ask for K for a new “target” equation. Before calculating anything, identify what you need from the information given.

1. Find each species in the target. Pick a substance that appears in only one of the given equations. Note which side of the target it is on and its coefficient.
2. Match that equation to the target. If the substance is on the wrong side, reverse the equation (K → 1/K). If its coefficient is wrong, multiply the whole equation (K → Kᶜ).
3. Repeat for the other given equations.
4. Add the adjusted equations and cancel anything that appears on both sides. The result must be exactly the target, with the same coefficients. If it is not, go back and check steps 1 to 3.
5. Multiply the adjusted K values. Only now do you use the calculator.
6. Sense-check the size. A reversed product-favoured reaction should give a small K, and squaring a K below 1 should make it smaller still.

Keep a short record of what you did to each equation, such as “(1) × 2, (2) unchanged, then reverse”. It makes the calculation easy to follow, and it shows your reasoning if the question asks for a justification.

Worked example 1: one equation, three versions

Question. At a certain temperature, 2 NO(g) + O₂(g) ⇌ 2 NO₂(g) has Kc = 6.4 × 10⁹ (invented value). Find Kc at the same temperature for: (a) 2 NO₂(g) ⇌ 2 NO(g) + O₂(g) (b) NO(g) + ½ O₂(g) ⇌ NO₂(g) (c) NO₂(g) ⇌ NO(g) + ½ O₂(g)

(a) Reversed. Kc = 1 / (6.4 × 10⁹) = 1.6 × 10⁻¹⁰ (1.5625 × 10⁻¹⁰ before rounding).

(b) Halved. Every coefficient × ½, so Kc = (6.4 × 10⁹)^½ = √(6.4 × 10⁹) = 8.0 × 10⁴.

(c) Halved and reversed. Apply both changes: Kc = 1 / (8.0 × 10⁴) = 1.3 × 10⁻⁵ (1.25 × 10⁻⁵). The order does not matter: √(1.5625 × 10⁻¹⁰) gives the same answer.

Check. Write (c) out: [NO][O₂]^½ / [NO₂]. This is 1 / √([NO₂]²/([NO]²[O₂])), which is 1 / √K. ✓

Common wrong answers. 3.2 × 10⁹ for (b) halves K instead of taking the square root. −6.4 × 10⁹ for (a) treats K like ΔH. A K value can never be negative.

Worked example 2: building K for a target equation

Question. Two invented values at the same temperature:

• Equation (1): S(s) + O₂(g) ⇌ SO₂(g), K₁ = 5.0 × 10¹²
• Equation (2): 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g), K₂ = 9.0 × 10⁴

Find K for (i) 2 S(s) + 3 O₂(g) ⇌ 2 SO₃(g) and (ii) SO₃(g) ⇌ S(s) + 3/2 O₂(g).

Plan. The target (i) has 2 S on the left, so double equation (1). Then add equation (2) so that the 2 SO₂ cancels.

1. Double (1): 2 S(s) + 2 O₂(g) ⇌ 2 SO₂(g). K = K₁² = (5.0 × 10¹²)² = 2.5 × 10²⁵.
2. Add (2): 2 S(s) + 2 O₂(g) + 2 SO₂(g) + O₂(g) ⇌ 2 SO₂(g) + 2 SO₃(g). Cancel 2 SO₂ and collect O₂: 2 S(s) + 3 O₂(g) ⇌ 2 SO₃(g). ✓
3. Multiply the K values: K(i) = K₁² × K₂ = 2.5 × 10²⁵ × 9.0 × 10⁴ = 2.3 × 10³⁰ (2.25 × 10³⁰).
4. For (ii), the target is (i) reversed and halved. Reverse: 1/K. Halve: square root. K(ii) = 1 / √(2.25 × 10³⁰) = 1 / (1.5 × 10¹⁵) = 6.7 × 10⁻¹⁶.

Check with the expressions. Sulfur is a solid, so it does not appear (Topic 7.3). K₁ = [SO₂]/[O₂] and K₂ = [SO₃]²/([SO₂]²[O₂]). K₁² × K₂ = [SO₂]²/[O₂]² × [SO₃]²/([SO₂]²[O₂]) = [SO₃]² / [O₂]³, which is the expression for (i). ✓

The same rules apply to Q

The reaction quotient Q has exactly the same form as K; the only difference is that Q uses the concentrations at any moment, not only at equilibrium (Topic 7.3). So every manipulation above works for Q too.

Example. A flask holds [NO] = 0.20 M, [O₂] = 0.10 M and [NO₂] = 0.40 M.

• For 2 NO + O₂ ⇌ 2 NO₂: Q = (0.40)² / ((0.20)² × 0.10) = 0.16 / 0.0040 = 40.
• For the reverse equation: Q = 1/40 = 0.025. Working it directly gives (0.20)² × 0.10 / (0.40)² = 0.025 ✓.
• For the halved equation NO + ½ O₂ ⇌ NO₂: Q = √40 = 6.3. Directly: 0.40 / (0.20 × √0.10) = 6.3 ✓.

Because Q and K change in the same way, comparing them always gives the same conclusion however you write the equation. With the forward equation, Q = 40 is far below K = 6.4 × 10⁹. With the reverse equation, Q = 0.025 is far above K = 1.6 × 10⁻¹⁰. Both say the mixture still has to make more NO₂ to reach equilibrium. You will use Q and K this way in Topic 7.7.

Common misconceptions

• “Doubling the equation doubles K.” It squares K. Coefficients are powers in the expression, so doubling every coefficient doubles every power, which squares the whole expression.
• “Adding equations adds the K values.” It multiplies them. Adding is the rule for ΔH, not for K.
• “Reversing makes K negative.” It makes K the reciprocal. K is always positive.
• “Halving the equation halves K.” It takes the square root.
• Forgetting to apply the power to the whole K. If you double equation (1) before adding, you must use K₁², not K₁.
• Mixing Kc and Kp. Every K in one calculation must be the same type.
• Including solids or pure liquids when checking. They still drop out of every expression, so they do not affect any of the rules.
• “Changing the equation changes the mixture.” It does not. The equilibrium concentrations stay the same; only the number that matches your chosen equation changes.

Where this leads

Now that you can match a value of K to any version of an equation, Topic 7.7 uses K to calculate the concentrations or partial pressures at equilibrium from the starting conditions. The same rules return in Unit 8, where the K values for acids and their conjugate bases are linked by multiplying equations. Try the practice questions now, then use the revision notes and the checklist to consolidate.