The equilibrium constant Kc — Chemistry

Before this

This leads to

Kp — partial pressures
Acid-base equilibria — Ka, Kb and pH · coming soon
Gibbs free energy

Different mixtures, the same number

In the 1890s, a young German chemist, Max Bodenstein, sealed hydrogen and iodine in glass tubes and held them at a high, steady temperature. The reaction

H_{2} (g) + I_{2} (g) ⇌ 2 HI(g)

reached equilibrium inside each tube. He then cooled the tubes suddenly, which froze the mixture, and measured how much of each gas was there. He also started some tubes from hydrogen iodide alone, and some from different mixtures.

The amounts at equilibrium were different in every tube. A tube with more iodine at the start ended with more iodine. So the equilibrium mixture is not a fixed recipe. And yet one particular combination of the three concentrations came out the same, tube after tube, at the same temperature.

The last lecture promised to put a number on the balance point. This is that number. It is called the equilibrium constant. By the end of this lecture, you will be able to write it for any reaction and calculate it. You will also use it to predict what an equilibrium mixture contains.

Finding the combination that stays the same

Which combination of concentrations stays constant? You could guess: perhaps [HI] divided by [H₂] × [I₂], or [HI] divided by [H₂] + [I₂]. The square brackets mean "concentration of", in mol/dm³. Test the guesses yourself below. The model uses the value that fits Bodenstein's measurements at 443 °C.

Explore — which ratio stays constant?

Set the starting concentrations of H₂, I₂ and HI. The model lets the mixture reach equilibrium at 443 °C and shows the equilibrium concentrations. Then look at the three ratios. Change the start several times. Which ratio never changes?

Start [H₂]: 1.0 mol/dm³

Start [I₂]: 1.0 mol/dm³

Start [HI]: 0.0 mol/dm³

Ratio at equilibrium	Value
[HI] ÷ ([H₂] × [I₂])
[HI] ÷ ([H₂] + [I₂])
[HI]² ÷ ([H₂] × [I₂])

This table shows four different starting mixtures at 443 °C, the equilibrium concentrations they reach (in mol/dm³), and three ratios. Only the last ratio is the same every time.

Start: H₂, I₂, HI	Equilibrium: H₂, I₂, HI	[HI] ÷ ([H₂][I₂])	[HI] ÷ ([H₂] + [I₂])	[HI]² ÷ ([H₂][I₂])
1.0, 1.0, 0	0.220, 0.220, 1.559	32.1	3.54	50.0
0, 0, 2.0	0.220, 0.220, 1.559	32.1	3.54	50.0
1.0, 2.0, 0	0.066, 1.066, 1.869	26.8	1.65	50.0
2.0, 2.0, 0	0.441, 0.441, 3.118	16.0	3.54	50.0

The first two ratios change every time you change the start. The third one does not. It is 50 every time, whether you start from hydrogen and iodine, from hydrogen iodide alone, or from a lopsided mixture. Notice that [HI] is squared. The equation makes two molecules of HI, and that 2 becomes a power.

The Norwegian chemists Cato Guldberg and Peter Waage proposed this rule in the 1860s, and Bodenstein's tubes were among the first careful tests of it. It is now called the equilibrium law.

The equilibrium constant

For a general reaction with coefficients a, b, c and d,

a A + b B ⇌ c C + d D

the equilibrium law says that, at equilibrium,

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

The equilibrium constant, K_c

Products go on top and reactants go underneath. Each equilibrium concentration, in mol/dm³, is raised to the power of its coefficient in the balanced equation.

At a fixed temperature, the value of K_c for a reaction is the same for every equilibrium mixture. The subscript c means "written with concentrations".

Two parts of this need care. The first is that only equilibrium concentrations go into K_c. Starting amounts do not. The second is which substances go in at all.

Solids are left out. The concentration of a pure solid is fixed by its density, however much of it there is. A lump of calcium carbonate twice as big is not more concentrated. Since its concentration cannot change, it is not written in the expression. Its constant value is absorbed into K_c.

Water as the solvent is left out. In a dilute solution, there is so much water that the reaction hardly changes its concentration. So water is left out when it is the solvent. But when the reacting substances are liquids mixed together, as in Berthelot's ester flasks, every one of them changes in amount. Then all of them, water included, go into the expression.

That is it. That is the whole idea. Everything whose concentration can change at equilibrium goes in, raised to its coefficient. Everything whose concentration is fixed stays out.

Stop and think

A mixture of H₂, I₂ and HI is at equilibrium at 443 °C. You inject extra hydrogen. When equilibrium is reached again, is K_c bigger, smaller or the same?

Show the answer

The same, about 50. The extra hydrogen makes the top of the fraction too small compared with the bottom, so the mixture is no longer at equilibrium. The reaction then runs to the right, using up some H₂ and I₂ and making HI, until the fraction is back to 50. This is Le Chatelier's "shift to the right", seen as a number. The mixture changes. K_c does not.

What the size of K_c tells you

K_c is a fraction with the products on top. So a very large K_c means the equilibrium mixture is mostly products: the reaction goes nearly to completion. A very small K_c means it is mostly reactants: the reaction hardly goes at all. A value somewhere near 1 means that both are present in useful amounts.

Treat this as a rough guide. "Large" and "small" mean many powers of ten, such as 10¹⁰ or 10⁻¹⁰, not 3 or 0.3. The exact amounts also depend on the starting concentrations, as the explorer showed. For hydrogen iodide, K_c ≈ 50 means that, from equal amounts of hydrogen and iodine, about 78% of each is converted to HI.

Only temperature changes K_c

The last lecture promised to show that temperature is the only one of Le Chatelier's changes that alters the constant. Here is why. A change in concentration, or in pressure, changes the mixture. For a moment the fraction is no longer equal to K_c. The reaction then runs one way until the fraction is back to the same K_c. A catalyst speeds up both directions equally, so it changes nothing at equilibrium.

Temperature is different. It changes the forward and reverse rates by different amounts, because the two directions have different activation energies. So the balance point itself moves, and K_c takes a new value. Bodenstein measured this too, at eleven temperatures.

K_c for H₂ + I₂ ⇌ 2HI, calculated from the fraction of HI that Bodenstein found decomposed at each temperature (published 1899). K_c falls as the temperature rises.

Forming HI from the gases releases a little heat, so it is exothermic. Le Chatelier says that heating shifts an exothermic equilibrium to the left. The graph shows the same thing as a number: K_c falls from about 85 at 283 °C to about 40 at 508 °C. Every value of K_c must therefore be quoted with its temperature.

The reaction quotient makes these ideas precise. For any mixture, equilibrium or not, the same fraction is called the reaction quotient, Q. If Q is smaller than K_c, the reaction runs to the right. If Q is larger, it runs to the left. If they are equal, the mixture is at equilibrium.

Worked examples

Worked Example 1 — Writing the expression, with units

Write K_c, with its units, for: (a) N₂(g) + 3H₂(g) ⇌ 2NH₃(g); (b) CaCO₃(s) ⇌ CaO(s) + CO₂(g); (c) CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l); (d) Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq).

Think first. Which substances can change in concentration? Then find the units by cancelling mol/dm³ top and bottom.

(a) All gases: $K_{c} = \frac{[ NH _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}}$ . Units: $(mol/dm^{3})^{2}$ on top, $(mol/dm^{3})^{4}$ underneath, leaving $(mol/dm^{3})^{- 2}$ , which is dm⁶/mol².

(b) Both solids are left out: $K_{c} = [CO_{2}]$ , in mol/dm³.

(c) Four liquids mixed together, all changing: $K_{c} = \frac{[ CH _{3} COOC _{2} H _{5} ] [ H _{2} O ]}{[ CH _{3} COOH ] [ C _{2} H _{5} OH ]}$ . Two concentrations on top and two below, so K_c has no units.

(d) The water is only the solvent, so it is left out: $K_{c} = \frac{[ FeSCN ^{2 +} ]}{[ Fe ^{3 +} ] [ SCN ^{-} ]}$ , in dm³/mol.

Check. In each case the powers are the coefficients from the equation, and the units come from counting concentrations: (top count − bottom count) gives the power of mol/dm³. ✓ Some books write dm⁶/mol² as mol⁻² dm⁶; it is the same unit.

Worked Example 2 — Calculating K_c

An equilibrium mixture of H₂, I₂ and HI contains [H₂] = 0.11 mol/dm³, [I₂] = 0.11 mol/dm³ and [HI] = 0.78 mol/dm³. Find K_c. (Numbers chosen for practice.)

Think first. Write the expression first, then substitute. HI has a coefficient of 2.

$K_{c} = \frac{[ HI ] ^{2}}{[ H _{2} ] [ I _{2} ]} = \frac{0.7 8 ^{2}}{0.11 \times 0.11} = \frac{0.6084}{0.0121} \approx 50$ . Units: two concentrations on top, two below, so none.

Check. 50 is the value at about 443 °C, so this mixture is at equilibrium near that temperature. ✓ Forgetting the square would give $0.78 \div 0.0121 \approx 64$ , a classic wrong answer.

Worked Example 3 — An equilibrium found exactly

For N₂O₄(g) ⇌ 2NO₂(g), take K_c = 4.6 × 10⁻³ mol/dm³ at 25 °C. A flask starts with 0.100 mol/dm³ of N₂O₄ and no NO₂. Find the equilibrium concentrations.

Think first. Use a table with three rows: start, change, equilibrium. Let x mol/dm³ of N₂O₄ react. The equation turns one N₂O₄ into two NO₂, so NO₂ rises by 2x.

	N₂O₄	NO₂
Start	0.100	0
Change	−x	+2x
Equilibrium	0.100 − x	2x

So $\frac{( 2 x ) ^{2}}{0.100 - x} = 4.6 \times 1 0^{- 3}$ . Multiply out: $4 x^{2} + 0.0046 x - 0.00046 = 0$ . The quadratic formula gives the positive root $x \approx 0.0102$ .

Equilibrium: [N₂O₄] ≈ 0.100 − 0.0102 ≈ 0.090 mol/dm³ and [NO₂] ≈ 0.020 mol/dm³.

Check. $0.020 3^{2} \div 0.0898 \approx 4.6 \times 1 0^{- 3}$ . ✓ Some books skip the quadratic by treating 0.100 − x as 0.100. That gives x ≈ 0.0107, and putting it back in gives about 5.2 × 10⁻³, which is 12% too big. The shortcut is safe only when x is tiny compared with the starting amount, here less than about 5% of 0.100. Here x is 10% of it, so solve exactly.

Worked Example 4 — The ester promise

In the first lecture of this topic, Berthelot's flask of 1.00 mol of ethanoic acid and 1.00 mol of ethanol reached equilibrium with 0.67 mol of ester. The last lecture promised that with 2.00 mol of ethanol the ester rises to about 0.85 mol. Find K_c, then check the promise.

Think first. K_c has no units here, so the volume cancels. Amounts in moles can be used directly.

From the first flask, exactly two-thirds reacts: ester and water $\frac{2}{3}$ mol each, acid and ethanol $\frac{1}{3}$ mol each. So $K_{c} = \frac{( 2/3 ) ^{2}}{( 1/3 ) ^{2}} = 4$ .

Second flask: let x mol of ester form. Then $\frac{x ^{2}}{( 1 - x ) ( 2 - x )} = 4$ , which gives $3 x^{2} - 12 x + 8 = 0$ . The root below 1 is $x = (12 - 48) \div 6 \approx 0.845$ .

So the ester rises from 0.67 mol to about 0.85 mol. The promise is kept, with numbers.

Check. $0.84 5^{2} \div (0.155 \times 1.155) \approx 4.0$ . ✓ And x is below 1.00, as it must be, because only 1.00 mol of acid was there. Extra ethanol pulls more of the acid into ester, exactly as Le Chatelier predicted, and K_c says by how much.

Tables like the one in Worked Example 3 are where most marks are lost. The change row must follow the coefficients. Find the slip in the working below.

Spot the error — a start, change, equilibrium table

A flask starts with 1.00 mol/dm³ each of H₂ and I₂, and no HI. K_c = 50. A student finds the equilibrium concentration of HI. Tap the line that contains the mistake.

The student's working has one wrong line. Here it is, with the correction.

Start: [H₂] = 1.00, [I₂] = 1.00, [HI] = 0. Correct.
Change: H₂ −x, I₂ −x, HI +x. This is the mistake. The equation makes 2HI from one H₂ and one I₂, so HI rises by 2x.
Equilibrium: 1.00 − x, 1.00 − x, x. This follows from line 2, so it carries the error.
x² ÷ (1.00 − x)² = 50, so x ÷ (1.00 − x) ≈ 7.07, and x ≈ 0.876. The algebra is right for the wrong table.
Check of hydrogen atoms: 2 × 0.124 + 0.876 ≈ 1.12, but there were 2.00 at the start. Hydrogen has been lost, so a line above is wrong.

With +2x: 2x ÷ (1.00 − x) ≈ 7.07, so x ≈ 0.780 and [HI] ≈ 1.56 mol/dm³. Hydrogen atoms: 2 × 0.220 + 1.56 = 2.00. ✓

The atom check is quick and catches the most common slip. Count one kind of atom at the start and at equilibrium. If the numbers differ, the change row is wrong.

Three traps

Trap 1 — Putting in solids, or solvent water

Solids never go into K_c, and neither does water when it is the solvent. But water does go in when it is one of several liquids mixed together, as in an ester equilibrium. Ask: can this concentration change?

Trap 2 — Forgetting the powers

Every concentration is raised to its coefficient. Without the square on [HI], the value for the hydrogen iodide mixture comes out as 64, not 50, and it changes from mixture to mixture. The powers are what make K_c constant.

Trap 3 — "Adding more reactant raises K_c"

Concentration, pressure and catalysts change the mixture, never K_c. Only temperature changes K_c. If a question changes the temperature, the old value of K_c can no longer be used.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Write K_c, with units, for (a) 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) and (b) C(s) + H₂O(g) ⇌ CO(g) + H₂(g).

Hint 1
Products on top, coefficients as powers. Can the concentration of solid carbon change?

Hint 2
For the units, count the concentrations: top count minus bottom count gives the power of mol/dm³.

Full answer
(a) $K_{c} = \frac{[ SO _{3} ] ^{2}}{[ SO _{2} ] ^{2} [ O _{2} ]}$ . Two on top, three below, so the units are $(mol/dm^{3})^{- 1}$ = dm³/mol. (b) Carbon is a solid and is left out: $K_{c} = \frac{[ CO ] [ H _{2} ]}{[ H _{2} O ]}$ . Two on top, one below, so mol/dm³. Check: steam is a gas here, not a solvent, so it stays in. ✓
An equilibrium mixture for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) contains [SO₂] = 0.20, [O₂] = 0.10 and [SO₃] = 0.60 mol/dm³. Find K_c. Is the mixture mostly products or mostly reactants? (Numbers chosen for practice.)

Hint 1
Use your expression from Question 1(a).

Hint 2
Square 0.60 and 0.20 before you divide.

Full answer
$K_{c} = \frac{0.6 0 ^{2}}{0.2 0 ^{2} \times 0.10} = \frac{0.36}{0.0040} = 90$ dm³/mol. There is three times as much SO₃ as SO₂, so products are favoured. Check: $90 \times 0.2 0^{2} \times 0.10 = 0.36 = 0.6 0^{2}$ . ✓
At 443 °C, K_c = 50 for H₂ + I₂ ⇌ 2HI. Find K_c at the same temperature for (a) 2HI ⇌ H₂ + I₂ and (b) ½H₂ + ½I₂ ⇌ HI.

Hint 1
Write each expression and compare it with the original.

Hint 2
Reversing the equation turns the fraction upside down. Halving the coefficients halves every power.

Full answer
(a) $K_{c} = \frac{[ H _{2} ] [ I _{2} ]}{[ HI ] ^{2}} = \frac{1}{50} = 0.020$ . (b) $K_{c} = \frac{[ HI ]}{[ H _{2} ] ^{1/2} [ I _{2} ] ^{1/2}} = 50 \approx 7.07$ . Check: for the mixture in Worked Example 2, $0.11 \times 0.11 \div 0.7 8^{2} \approx 0.020$ and $0.78 \div 0.11 \times 0.11 \approx 7.1$ . ✓ So K_c always belongs to an equation written in a particular way.
A 1.00 dm³ flask at 443 °C starts with 1.00 mol of H₂ and 2.00 mol of I₂ (K_c = 50). Find the equilibrium concentrations. What percentage of the hydrogen is converted, and how does this compare with starting from 1.00 mol of each?

Hint 1
Change row: H₂ −x, I₂ −x, HI +2x. Then $(2 x)^{2} = 50 (1.00 - x) (2.00 - x)$ .

Hint 2
Multiply out to get $46 x^{2} - 150 x + 100 = 0$ . Use the root that is less than 1.00.

Full answer
$4 x^{2} = 50 (2 - 3 x + x^{2})$ gives $46 x^{2} - 150 x + 100 = 0$ , so $x = (150 - 4100) \div 92 \approx 0.934$ . (The other root, about 2.33, would use more hydrogen than there is.) Equilibrium: [H₂] ≈ 0.066, [I₂] ≈ 1.066, [HI] ≈ 1.869 mol/dm³. About 93% of the hydrogen is converted, compared with about 78% from equal amounts. Check: $1.86 9^{2} \div (0.066 \times 1.066) \approx 50$ ✓, and hydrogen atoms: $2 \times 0.066 + 1.869 \approx 2.00$ . ✓ An excess of the cheaper reactant converts more of the expensive one. Industry uses this idea all the time.

The bigger idea

The equilibrium law has a simple reason underneath, which the first lecture of this topic set up. At equilibrium the forward and reverse rates are equal. For a reaction that happens in one step, each rate is proportional to the concentrations of the colliding particles. Set the two rates equal and rearrange, and the ratio of the two rate constants appears. That ratio is K_c. Temperature changes rate constants, which is why it changes K_c. Concentrations do not, which is why they cannot.

K_c also connects to energy. At Pre-University level you will meet the link between K and the Gibbs free energy change of a reaction. A reaction with a very negative ΔG has a huge K and goes almost to completion. The same constant appears for acids, as K_a, and for gases, as K_p, which the next lecture builds from partial pressures.

Key concepts in this lesson

For aA + bB ⇌ cC + dD, the equilibrium constant is K_c = [C]^c[D]^d ÷ [A]^a[B]^b, using equilibrium concentrations in mol/dm³.
Solids, and water when it is the solvent, are left out, because their concentrations cannot change, while liquids mixed together all go in.
At a fixed temperature, K_c has the same value for every equilibrium mixture, so changing concentrations, pressure or adding a catalyst changes the mixture but not K_c.
Only temperature changes K_c, and for an exothermic reaction K_c falls as the temperature rises, as Bodenstein's hydrogen iodide data show.
Equilibrium amounts are found with a start, change, equilibrium table whose change row follows the coefficients, checked by counting atoms.