Chemical equilibria: reversible reactions, dynamic equilibrium

1. Overview

Many reactions never go to completion: the products react to re-form the reactants, and the mixture settles into a dynamic equilibrium. This section covers what that means, how Le Chatelier's principle predicts the effect of changing the conditions, and how the equilibrium constants $K_{\text{c}}$ and $K_{\text{p}}$ describe an equilibrium mixture in numbers. The Haber and Contact processes then show these ideas deciding the conditions used in industry.

Key Definitions

• Reversible reaction: a reaction that can go in both directions, because the products can react together to re-form the reactants; it is shown with the ⇌ sign.
• Dynamic equilibrium: the state reached by a reversible reaction in a closed system when the rates of the forward and reverse reactions are equal, so the concentrations of reactants and products stay constant (they are not usually equal).
• Closed system: a system that no substance can enter or leave, although energy can still be transferred between it and its surroundings.
• Le Chatelier's principle: if a change is made to a system at dynamic equilibrium, the position of equilibrium moves to minimise this change.
• Position of equilibrium: the relative amounts of reactants and products in an equilibrium mixture; when it moves to the right, more products form.
• Equilibrium constant, Kc: the equilibrium concentrations of the products, each raised to the power of its coefficient in the balanced equation and multiplied together, divided by the equilibrium concentrations of the reactants treated in the same way; for a given equation it changes only with temperature.
• Mole fraction: the amount in moles of one gas in a mixture divided by the total amount in moles of all the gases in the mixture; it has no units.
• Partial pressure: the pressure that one gas in a mixture would exert if it alone filled the whole container at the same temperature; it equals the mole fraction of that gas multiplied by the total pressure.
• Equilibrium constant, Kp: the same kind of expression as Kc, but using the equilibrium partial pressures of the gases only; for a given equation it changes only with temperature.

Content

Reversible reactions

A reversible reaction can go both ways: the products can react to re-form the reactants. The sign ⇌ shows this. Nitrogen and hydrogen react to form ammonia, and under the same conditions some ammonia breaks down into nitrogen and hydrogen again:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

In principle every reaction is reversible. When the reverse reaction is negligible, the reaction is treated as complete and written with →.

Dynamic equilibrium

Suppose a reaction A ⇌ B starts with A only, in a sealed container at constant temperature. Square brackets, as in [A], mean concentration in mol dm⁻³.

1. At first the forward rate is at its fastest, because [A] is at its highest. The reverse rate is zero, because there is no B.
2. As A is used up, the forward rate falls. As B builds up, the reverse rate rises.
3. Once the two rates are equal, each reaction remakes what the other uses up, so [A] and [B] stop changing.

This state is a dynamic equilibrium.

• Dynamic: both reactions carry on. If some of the A in the mixture is swapped for A containing a heavier isotope, the label soon appears in B too, although [A] and [B] stay the same.
• Constant, not equal: in the first diagram, [B] ends higher than [A]. Properties you can observe, such as colour and pressure, stay constant too.

Why a closed system is needed

A closed system is one that no substance can enter or leave. Energy can still flow in or out, so the temperature can be kept steady.

If a product escapes, the reverse reaction cannot keep pace, and the forward reaction runs on until a reactant is used up. For calcium carbonate heated strongly, CaCO₃(s) ⇌ CaO(s) + CO₂(g):

• Sealed container: CO₂ builds up until the reverse reaction is as fast as the forward one, and its pressure then stays constant.
• Open container: CO₂ escapes, so all the CaCO₃ eventually decomposes.

A solution in an open beaker can be treated as a closed system, as long as nothing in the equilibrium, such as a gas, escapes from it.

Le Chatelier's principle

Le Chatelier's principle: if a change is made to a system at dynamic equilibrium, the position of equilibrium moves to minimise this change.

• The position of equilibrium is the balance between reactants and products. "Moves to the right" means more products form.
• Minimise does not mean cancel. Add more of a reactant and some of it is used up, but its concentration still ends higher than before the addition.
• The principle predicts the direction of a shift, not its size.

Changing the concentration

• Adding a reactant or removing a product moves the position to the right.
• Adding a product or removing a reactant moves it to the left.

These rules apply to gases, dissolved substances and liquids mixed together. Adding or removing some of a pure solid, or of a pure liquid on its own, does not move the position (see Worked example 4).

Adding more A to A ⇌ B makes the forward reaction faster than the reverse until a new equilibrium is reached. The value of $K_{\text{c}}$ does not change.

Changing the pressure

Count the moles of gas on each side of the equation, ignoring solids, liquids and dissolved substances.

• Increasing the pressure (compressing the mixture) moves the position to the side with fewer moles of gas, because fewer gas molecules in the same volume exert a lower pressure.
• Decreasing the pressure moves it to the side with more moles of gas.
• With equal moles of gas on both sides, the position does not move.

Equilibrium	Moles of gas, left : right	Increasing the pressure
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)	4 : 2	moves right
CaCO₃(s) ⇌ CaO(s) + CO₂(g)	0 : 1	moves left
H₂(g) + I₂(g) ⇌ 2HI(g)	2 : 2	no change in position

Pumping unreactive argon into a container of fixed volume raises the total pressure but not the partial pressures of the reacting gases, so the position does not move.

Changing the temperature

A $\Delta H$ value given with an equation refers to the forward reaction; the reverse reaction has the same value with the opposite sign.

• Raising the temperature moves the position in the endothermic direction, which takes in heat and so minimises the rise.
• Lowering the temperature moves it in the exothermic direction.

Both reactions speed up when the temperature rises, but the endothermic one speeds up by a larger factor, because it has the larger activation energy, $E_{\text{A}}$. Temperature is the only change that also alters the value of K.

Adding a catalyst

A catalyst lowers $E_{\text{A}}$ for the forward and reverse reactions by the same amount, so both speed up by the same factor. Equilibrium is reached sooner, but the position of equilibrium and the value of K do not change. A catalyst never increases the equilibrium yield.

The equilibrium constant, Kc

For the general equation aA + bB ⇌ cC + dD:

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

Each concentration is the value at equilibrium. To deduce an expression:

1. Products go on top, reactants underneath.
2. Raise each concentration to the power of its coefficient: 2SO₂ gives $[\text{SO}_2]^2$, not $2[\text{SO}_2]$.
3. Include gases and dissolved substances. Leave out a pure solid, or a pure liquid on its own, because its concentration is fixed. Leave out water when it is the solvent of a dilute solution, but keep it when it is one of several liquids mixed together, as in esterification.
4. $K_{\text{c}}$ belongs to the equation as written. Reversing the equation gives $1/K_{\text{c}}$; doubling every coefficient squares $K_{\text{c}}$.

Units. Substitute mol dm⁻³ for every concentration and cancel. For ammonia:

$$\frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^3} = (\text{mol dm}^{-3})^{-2} = \text{mol}^{-2}\,\text{dm}^{6}$$

Equation	$K_{\text{c}}$	Units
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)	$\dfrac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$	mol⁻² dm⁶
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)	$\dfrac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}$	mol⁻¹ dm³
CH₃CH₂COOH(l) + CH₃OH(l) ⇌ CH₃CH₂COOCH₃(l) + H₂O(l)	$\dfrac{[\text{ester}][\text{H}_2\text{O}]}{[\text{acid}][\text{alcohol}]}$	no units
CaCO₃(s) ⇌ CaO(s) + CO₂(g)	$[\text{CO}_2]$	mol dm⁻³

A very large K means the equilibrium mixture is mostly products; a very small K means it is mostly reactants.

Mole fraction and partial pressure

In a mixture of gases, each gas supplies part of the total pressure.

• The mole fraction of gas A, written here as $x_{\text{A}}$, is $n_{\text{A}}/n_{\text{total}}$. It has no units, and the mole fractions add up to 1.
• The partial pressure of A is $p_{\text{A}} = x_{\text{A}} \times p_{\text{total}}$. The partial pressures add up to the total pressure.

This works because, at fixed volume and temperature, a gas's pressure is proportional to its amount ($pV = nRT$, topic 4.1). For 0.30 mol of CO₂ mixed with 0.90 mol of N₂ at 160 kPa, $x_{\text{CO}_2} = 0.30/1.20 = 0.25$, so $p_{\text{CO}_2} = 0.25 \times 160 = 40$ kPa.

The equilibrium constant, Kp

For a reaction involving gases, $K_{\text{p}}$ uses partial pressures in place of concentrations:

$$K_{\text{p}} = \frac{p_{\text{C}}^{\,c}\;p_{\text{D}}^{\,d}}{p_{\text{A}}^{\,a}\;p_{\text{B}}^{\,b}}$$

• Include gases only. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), $K_{\text{p}} = p_{\text{CO}_2}$, in Pa.
• Never use square brackets in $K_{\text{p}}$, because they mean concentration. Write $p_{\text{NH}_3}$ or p(NH₃).
• The units are the pressure unit (Pa, kPa or atm) raised to the power (moles of gas on the right minus moles of gas on the left). When that power is zero, as for H₂(g) + I₂(g) ⇌ 2HI(g), $K_{\text{p}}$ has no units. For ammonia, $K_{\text{p}} = \dfrac{p_{\text{NH}_3}^{\,2}}{p_{\text{N}_2}\,p_{\text{H}_2}^{\,3}}$, in Pa⁻² when the pressures are in Pa.

You do not need to convert between $K_{\text{p}}$ and $K_{\text{c}}$.

Worked example 1: Kc from starting amounts

Question: 0.600 mol of carbon monoxide and 0.400 mol of chlorine are sealed in a 4.00 dm³ container and heated to a fixed temperature. At equilibrium the mixture contains 0.250 mol of COCl₂. Calculate $K_{\text{c}}$, with its units.

CO(g) + Cl₂(g) ⇌ COCl₂(g)

Step 1: an I, C, E table (initial, change and equilibrium amounts). The ratio is 1 : 1 : 1, so 0.250 mol each of CO and Cl₂ reacted. For $K_{\text{c}}$, divide each equilibrium amount by the volume.

	CO(g)	Cl₂(g)	COCl₂(g)
initial / mol	0.600	0.400	0
change / mol	−0.250	−0.250	+0.250
equilibrium / mol	0.350	0.150	0.250
concentration / mol dm⁻³ (÷ 4.00)	0.0875	0.0375	0.0625

Step 2: substitute. $$K_{\text{c}} = \frac{[\text{COCl}_2]}{[\text{CO}][\text{Cl}_2]} = \frac{0.0625}{0.0875 \times 0.0375} = 19.05$$

Step 3: units. $\dfrac{\text{mol dm}^{-3}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})} = \text{mol}^{-1}\,\text{dm}^{3}$

Answer: $K_{\text{c}} = 19.0$ mol⁻¹ dm³ (3 s.f.).

Worked example 2: Kp from mole fractions

Question: 0.800 mol of sulfuryl chloride, SO₂Cl₂, is heated in a sealed container to a fixed temperature. At equilibrium the mixture contains 0.240 mol of Cl₂, and the total pressure is 250 kPa. Calculate $K_{\text{p}}$, with its units.

SO₂Cl₂(g) ⇌ SO₂(g) + Cl₂(g)

Step 1: amounts at equilibrium. 0.240 mol of Cl₂ formed, so 0.240 mol of SO₂ also formed and 0.240 mol of SO₂Cl₂ reacted.

• SO₂Cl₂: 0.800 − 0.240 = 0.560 mol; SO₂: 0.240 mol; Cl₂: 0.240 mol
• total amount of gas = 0.560 + 0.240 + 0.240 = 1.040 mol

Step 2: partial pressures. Each mole fraction uses the total amount of gas at equilibrium; multiply it by the total pressure. $$p_{\text{SO}_2\text{Cl}_2} = \frac{0.560}{1.040} \times 250 = 134.62\ \text{kPa} \qquad p_{\text{SO}_2} = p_{\text{Cl}_2} = \frac{0.240}{1.040} \times 250 = 57.69\ \text{kPa}$$ Check: 134.62 + 57.69 + 57.69 = 250.00 kPa.

Step 3: substitute. $$K_{\text{p}} = \frac{p_{\text{SO}_2}\,p_{\text{Cl}_2}}{p_{\text{SO}_2\text{Cl}_2}} = \frac{57.69 \times 57.69}{134.62} = 24.7\ \text{kPa}$$

Answer: $K_{\text{p}} = 24.7$ kPa (3 s.f.). The units are kPa × kPa ÷ kPa = kPa.

Worked example 3: amounts at equilibrium from Kc

Question: 0.500 mol of carbon dioxide and 0.500 mol of hydrogen are sealed in a 5.00 dm³ container and heated to a temperature at which $K_{\text{c}} = 0.640$. Calculate the amount of each gas at equilibrium.

CO₂(g) + H₂(g) ⇌ CO(g) + H₂O(g)

Step 1: the table. Let $y$ mol of CO form. The ratio is 1 : 1 : 1 : 1.

	CO₂(g)	H₂(g)	CO(g)	H₂O(g)
initial / mol	0.500	0.500	0	0
equilibrium / mol	0.500 − y	0.500 − y	y	y

Step 2: the expression. Each concentration is the amount ÷ 5.00, and the 5.00s cancel (2 mol of gas on each side): $$K_{\text{c}} = \frac{[\text{CO}][\text{H}_2\text{O}]}{[\text{CO}_2][\text{H}_2]} = \frac{y^2}{(0.500 - y)^2} = 0.640$$

Step 3: take the square root of both sides. Both sides are perfect squares, so no quadratic is needed. $$\frac{y}{0.500 - y} = \sqrt{0.640} = 0.800 \qquad y = 0.400 - 0.800y \qquad 1.800y = 0.400 \qquad y = 0.2222$$

Answer: 0.222 mol each of CO and H₂O, and 0.500 − 0.2222 = 0.278 mol each of CO₂ and H₂ (3 s.f.). Check: $0.2222^2 / 0.2778^2 = 0.640$.

What changes the value of K?

Change made	Position of equilibrium	Value of $K_{\text{c}}$ or $K_{\text{p}}$
Concentration of a reactant or product	moves to oppose the change	no change
Pressure, by changing the volume	moves towards fewer moles of gas when raised	no change
Temperature	moves in the endothermic direction when raised	changes
Catalyst	does not move (equilibrium is reached sooner)	no change

Only temperature changes K, and the direction depends on the forward reaction:

• Forward reaction exothermic: raising the temperature makes K smaller.
• Forward reaction endothermic: raising the temperature makes K larger.

How can the position move while K stays the same? Halving the volume of a Haber mixture doubles every partial pressure, which multiplies the expression for $K_{\text{p}}$ by $2^2/(2 \times 2^3) = \tfrac{1}{4}$. The value of the fraction is now below $K_{\text{p}}$, so ammonia forms until the fraction equals $K_{\text{p}}$ again.

Worked example 4: predicting changes for an unfamiliar equilibrium

Question: Steam reacts with very hot carbon. The value of $K_{\text{p}}$ increases as the temperature rises.

C(s) + H₂O(g) ⇌ CO(g) + H₂(g)

(a) Write the expression for $K_{\text{p}}$, with its units. (b) Deduce the sign of $\Delta H$ for the forward reaction. (c) State the effect of each change below on the position of equilibrium and on $K_{\text{p}}$.

(a) Carbon is a solid, so it is left out: $K_{\text{p}} = \dfrac{p_{\text{CO}}\,p_{\text{H}_2}}{p_{\text{H}_2\text{O}}}$, with units Pa × Pa ÷ Pa = Pa.

(b) A larger $K_{\text{p}}$ at a higher temperature means more products, so the forward reaction is the endothermic direction: $\Delta H$ is positive.

(c)

Change	Position	$K_{\text{p}}$	Reason
Pressure increased at constant temperature	moves left	no change	1 mol of gas on the left, 2 mol on the right; carbon is not a gas
Hydrogen removed as it forms	moves right	no change	more H₂ forms, replacing some of what was removed
More carbon added	no change	no change	a pure solid is not in the expression
Catalyst added	no change	no change	both rates rise by the same factor

Equilibrium in industry: the Haber and Contact processes

Both processes make a product from an exothermic reversible reaction between gases. Le Chatelier's principle shows which conditions give a higher yield, but the conditions chosen also depend on rate and cost.

The figures below are typical values; individual plants vary.

	Haber process	Contact process
Reaction	N₂(g) + 3H₂(g) ⇌ 2NH₃(g)	2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Forward reaction	exothermic	exothermic
Moles of gas, left → right	4 → 2	3 → 2
Temperature	about 400 to 450 °C	about 400 to 450 °C
Pressure	about 200 atm (about 2 × 10⁴ kPa)	about 1 to 2 atm (about 100 to 200 kPa)
Catalyst	iron	vanadium(V) oxide, V₂O₅

Temperature (both). A lower temperature would move the position to the right and give a higher yield, but the rate would be far too slow, even with a catalyst. About 400 to 450 °C is a compromise: a reasonable yield, made quickly.

Pressure (Haber). With fewer moles of gas on the right, a high pressure increases both the yield and the rate. But compressing gases uses a lot of energy, and high-pressure plant is expensive to build and to keep safe. About 200 atm balances yield and rate against cost.

Pressure (Contact). The equilibrium already lies far to the right at 1 to 2 atm, so almost all the SO₂ is converted, and a higher pressure would add cost for very little gain. Excess air supplies extra O₂, which moves the position to the right. The SO₃ is then used to make sulfuric acid.

Catalysts. Iron and V₂O₅ increase the rate, so equilibrium is approached in the short time the gases spend in the reactor. They do not change the equilibrium yield.

Recycling (Haber). Only a small fraction of the gases (typically around 15%) reacts on each pass. Cooling condenses the ammonia, which boils at a much higher temperature than nitrogen or hydrogen because of hydrogen bonding between its molecules. The unreacted N₂ and H₂ are recycled.

Key Equations

Equilibrium constant in terms of concentrations, for aA + bB ⇌ cC + dD ([X] = equilibrium concentration in mol dm⁻³; units found by cancelling): $$K_{\text{c}} = \frac{[\text{C}]^{c}\,[\text{D}]^{d}}{[\text{A}]^{a}\,[\text{B}]^{b}}$$

Equilibrium constant in terms of partial pressures, gases only ($p_{\text{X}}$ = equilibrium partial pressure in Pa, kPa or atm): $$K_{\text{p}} = \frac{p_{\text{C}}^{\,c}\;p_{\text{D}}^{\,d}}{p_{\text{A}}^{\,a}\;p_{\text{B}}^{\,b}}$$

Mole fraction ($n_{\text{A}}$ = amount of gas A in mol; $n_{\text{total}}$ = total amount of all the gases in mol; no units): $$x_{\text{A}} = \frac{n_{\text{A}}}{n_{\text{total}}}$$

Partial pressure ($p_{\text{total}}$ = total pressure, in the same unit as $p_{\text{A}}$): $$p_{\text{A}} = x_{\text{A}} \times p_{\text{total}} \qquad p_{\text{total}} = p_{\text{A}} + p_{\text{B}} + p_{\text{C}} + \dots$$

Reversed equation (same temperature): $$K(\text{reverse}) = \frac{1}{K(\text{forward})}$$

Common Mistakes to Avoid

• Mistake: saying the concentrations at equilibrium are equal, or leaving out the closed system. Fix: they are constant, not equal: in Worked example 1, [CO] = 0.0875 mol dm⁻³ but [COCl₂] = 0.0625 mol dm⁻³. Give all three points: closed system, equal rates, constant concentrations.
• Mistake: giving the definition of dynamic equilibrium when asked for Le Chatelier's principle, or writing "the system shifts". Fix: use the set wording, in which the position of equilibrium moves "to minimise this change".
• Mistake: saying K changes because the position has moved, or that temperature has no effect on K. Fix: only temperature changes K. Compressing a Haber mixture moves the position right with $K_{\text{p}}$ unchanged; heating it moves the position left and makes $K_{\text{p}}$ smaller.
• Mistake: putting amounts in mol into $K_{\text{c}}$ when the volume does not cancel. Fix: divide by the volume first. In Worked example 1, moles give $0.250/(0.350 \times 0.150) = 4.76$, out by a factor of 4.00 (the volume in dm³); the correct value is 19.0 mol⁻¹ dm³.
• Mistake: dividing by the wrong total for mole fractions, such as the starting amount. Fix: use the total amount of gas at equilibrium: 1.040 mol in Worked example 2. Using 0.800 mol gives $K_{\text{p}} = 32.1$ kPa instead of 24.7 kPa.
• Mistake: multiplying by a coefficient instead of using it as a power, or ignoring the mole ratio. Fix: for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), $K_{\text{c}}$ contains $[\text{SO}_2]^2$, not $2[\text{SO}_2]$, and if 0.10 mol of O₂ reacts, 0.20 mol of SO₂ reacts.
• Mistake: putting the wrong substances into K, or writing $K_{\text{p}}$ with square brackets. Fix: leave out pure solids (for calcium carbonate, $K_{\text{p}} = p_{\text{CO}_2}$), but keep water in an esterification $K_{\text{c}}$, where it is mixed with the other liquids. Write $p_{\text{CO}_2}$ or p(CO₂), never [CO₂], in $K_{\text{p}}$.
• Mistake: writing "fewer moles" without "of gas", or counting a solid. Fix: in Worked example 4 there is 1 mol of gas on the left and 2 mol of gas on the right, so a higher pressure moves the position left; the carbon does not count.

Exam Tips

• Define dynamic equilibrium with three points: closed system, equal forward and reverse rates, constant concentrations. Give Le Chatelier's principle in its set wording.
• Predict or explain a change in three parts: which way the position moves, why (name the endothermic direction, or count the moles of gas), and the effect on yield. For example: "moves right, because there are 4 mol of gas on the left and 2 mol on the right, so more ammonia forms".
• Deduce an expression with products over reactants and the coefficients as powers, then find the units by cancelling. When asked whether K changes, answer yes or no and give the reason.
• Calculate with an I, C, E table (initial, change, equilibrium), say which volume or total pressure you use, and carry unrounded values to the end. Give as many significant figures as the data justify (usually 3), with units or "no units".
• Explain the conditions in the Haber or Contact process by giving the effect of each condition on yield, rate and cost. "It is a compromise" says nothing unless you say what is being balanced.