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R2.3

How far? The extent of chemical change

Reactivity 2 · How much, how fast and how far? · SL and HL

🎯What you need to be able to do

  • Describe the characteristics of a physical or chemical system at dynamic equilibrium.
  • Deduce the equilibrium constant expression from the equation for a homogeneous reaction.
  • Interpret the magnitude of \( K \) as a measure of the extent of reaction, and relate the \( K \) values of forward and reverse reactions at the same temperature.
  • Apply Le Châtelier’s principle to predict and explain the effect of changes in concentration, pressure and temperature, including the effect on the value of \( K \).
  • AHL Calculate the reaction quotient \( Q \) and determine the direction in which a reaction will proceed to reach equilibrium.
  • AHL Solve problems involving \( K \) and initial and equilibrium concentrations, using the approximation \( [\text{reactant}]_{\text{initial}} \approx [\text{reactant}]_{\text{eqm}} \) when \( K \) is very small.
  • AHL Use \( \Delta G^{\ominus} = -RT\ln K \) to relate the equilibrium constant to the Gibbs energy change.

📚The chemistry

Dynamic equilibrium

A reversible reaction in a closed system reaches dynamic equilibrium when the rates of the forward and reverse reactions are equal. The four characteristics you must be able to state:

  • the system is closed — no matter enters or leaves;
  • the rates of the forward and reverse reactions are equal;
  • the concentrations of all species remain constant — constant, not necessarily equal;
  • it is dynamic: both reactions continue at the molecular level, and there is no observable change in macroscopic properties.
“At equilibrium the reaction has stopped” and “the concentrations are equal”. Both are wrong and both are common. Nothing stops — the two opposing reactions continue at identical rates, which is what dynamic means, and isotopic labelling experiments prove it. And the concentrations are constant, not equal; an equilibrium with \( K = 10^{6} \) is almost entirely products, and its concentrations are wildly unequal.

The equilibrium law

For a homogeneous reaction \( a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D} \):

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

Products on top, reactants underneath, each raised to the power of its coefficient. Two rules that follow from the definition:

  • Pure solids and pure liquids are omitted, because their concentration is fixed by their density and does not change. So for \( \mathrm{CaCO_3}(s) \rightleftharpoons \mathrm{CaO}(s) + \mathrm{CO_2}(g) \), \( K = [\mathrm{CO_2}] \) — which is why state symbols matter.
  • \( K \) depends only on temperature. It is unchanged by concentration, pressure or a catalyst.

What the magnitude of K tells you

\( K \ggg 1 \)
reaction goes essentially to completion; almost only products at equilibrium
\( K > 1 \)
products favoured
\( K = 1 \)
comparable amounts of reactants and products
\( K < 1 \)
reactants favoured
\( K \lll 1 \)
barely proceeds; almost only reactants at equilibrium

For the reverse reaction at the same temperature, \( K \) is the reciprocal: \( K_{\text{reverse}} = 1/K_{\text{forward}} \). If the equation is doubled, \( K \) is squared; if halved, take the square root. Always check which equation a quoted \( K \) belongs to.

Note carefully that \( K \) says nothing about rate. A reaction can have an enormous \( K \) and be immeasurably slow, because \( K \) is thermodynamics and rate is kinetics (R2.2).

Le Châtelier’s principle

If a system at equilibrium is subjected to a change, the position of equilibrium shifts so as to partially oppose that change. The word “partially” matters: the system never fully undoes what you did to it.

Increase a concentration
equilibrium shifts away from the species added, to consume it.
\( K \) unchanged.
Increase the pressure (decrease volume)
shifts towards the side with fewer moles of gas. No shift if the moles of gas are equal.
\( K \) unchanged.
Increase the temperature
shifts in the endothermic direction, absorbing the added energy.
\( K \) DOES change.
Add a catalyst
no shift at all — forward and reverse rates are increased equally, so equilibrium is reached sooner at the same position.
\( K \) unchanged.

Temperature is the only one of these that changes \( K \), and that is the single most examined point in this sub-topic. For an exothermic forward reaction, raising the temperature shifts the equilibrium left and decreases \( K \); for an endothermic forward reaction, raising the temperature shifts it right and increases \( K \).

The principle applies to heterogeneous equilibria too, including physical ones such as \( \mathrm{X}(g) \rightleftharpoons \mathrm{X}(aq) \) — the dissolution of carbon dioxide in a fizzy drink, which is why opening the bottle (lowering the pressure) drives \( \mathrm{CO_2} \) out of solution.

The industrial compromise

The Haber process, \( \mathrm{N_2}(g) + 3\mathrm{H_2}(g) \rightleftharpoons 2\mathrm{NH_3}(g) \), \( \Delta H = -92\ \mathrm{kJ\,mol^{-1}} \), is the standard illustration and is worth being able to argue through:

  • Pressure: 4 moles of gas become 2, so high pressure increases the yield. Limited in practice by the cost and safety of high-pressure plant — about 200 atm.
  • Temperature: the forward reaction is exothermic, so a low temperature increases the yield — but at low temperature the rate is uselessly slow. The compromise temperature of about 450 °C accepts a lower equilibrium yield in order to reach it in a usable time.
  • Catalyst: iron. It does not improve the yield — it improves the rate, which is what makes the compromise temperature tolerable.
  • Removal of product: ammonia is condensed out and unreacted gases are recycled, which continually shifts the equilibrium to the right.

AHL The reaction quotient

\( Q \) has exactly the same expression as \( K \), but is calculated with the concentrations present at any moment, not necessarily at equilibrium. Comparing the two tells you which way the reaction must go:

\( Q < K \)
too few products → net reaction proceeds forwards (to the right)
\( Q = K \)
the system is at equilibrium; no net change
\( Q > K \)
too many products → net reaction proceeds backwards (to the left)

AHL Equilibrium calculations

The standard method is an ICE table: Initial concentrations, Change (in the ratio of the coefficients, negative for reactants), and Equilibrium (the sum). Substitute the equilibrium row into the expression for \( K \).

The guide states that quadratic equations are not expected and only homogeneous equilibria will be assessed. What makes that possible is the approximation for a very small \( K \): when so little reactant is converted that the change is negligible, take

\[ [\text{reactant}]_{\text{eqm}} \approx [\text{reactant}]_{\text{initial}} \]

which removes the quadratic. State the assumption when you use it — and it is only justified when \( K \) is genuinely small, typically \( 10^{-3} \) or below.

AHL Equilibrium and Gibbs energy

\[ \Delta G^{\ominus} = -RT\ln K \]

The equilibrium constant and the standard Gibbs energy change are two measures of the same thing: how far a reaction goes. A negative \( \Delta G^{\ominus} \) corresponds to \( K > 1 \), and a positive \( \Delta G^{\ominus} \) to \( K < 1 \). Because the relationship is logarithmic, quite small changes in \( \Delta G^{\ominus} \) move \( K \) by orders of magnitude. This is treated fully in R1.4, and it also explains, at a deeper level than Le Châtelier, why temperature is the only variable that changes \( K \): temperature appears in the Gibbs relationship, and concentration and pressure do not.

✏️Worked example

Consider \( \mathrm{N_2O_4}(g) \rightleftharpoons 2\mathrm{NO_2}(g) \), for which \( \Delta H = +57\ \mathrm{kJ\,mol^{-1}} \).
(a) Write the expression for \( K_{\mathrm{c}} \).
(b) State and explain the effect on the position of equilibrium and on the value of \( K_{\mathrm{c}} \) of (i) increasing the pressure, (ii) increasing the temperature, (iii) adding a catalyst.
(c) AHL 2.00 mol of \( \mathrm{N_2O_4} \) is placed in a 1.00 dm3 vessel. At equilibrium 0.40 mol of \( \mathrm{N_2O_4} \) has dissociated. Calculate \( K_{\mathrm{c}} \).
(d) AHL At the same temperature a mixture contains \( [\mathrm{N_2O_4}] = 1.00 \) and \( [\mathrm{NO_2}] = 0.20\ \mathrm{mol\,dm^{-3}} \). Determine the direction in which the reaction proceeds.

(a) Both species are gases, so both appear, and the coefficient of \( \mathrm{NO_2} \) is 2:

\[ K_{\mathrm{c}} = \frac{[\mathrm{NO_2}]^{2}}{[\mathrm{N_2O_4}]} \]

(b) (i) Increasing the pressure. There is 1 mole of gas on the left and 2 on the right. The equilibrium shifts towards the side with fewer moles of gas, so it shifts to the left, partially opposing the increase in pressure. \( K_{\mathrm{c}} \) is unchanged, because \( K \) depends only on temperature.

(ii) Increasing the temperature. The forward reaction is endothermic (\( \Delta H \) is positive), so the equilibrium shifts in the endothermic direction to absorb the added energy — that is, to the right, producing more \( \mathrm{NO_2} \). Here \( K_{\mathrm{c}} \) does change: it increases, because there are now relatively more products at equilibrium. This is observable, since \( \mathrm{N_2O_4} \) is colourless and \( \mathrm{NO_2} \) is brown — the mixture darkens on heating.

(iii) Adding a catalyst. No shift in the position of equilibrium and no change in \( K_{\mathrm{c}} \). A catalyst lowers the activation energy of the forward and reverse reactions equally, so both rates increase by the same factor and equilibrium is simply reached faster.

(c) Build an ICE table. The volume is 1.00 dm3, so amounts in moles are numerically equal to concentrations.

Initial
\( [\mathrm{N_2O_4}] = 2.00 \)
\( [\mathrm{NO_2}] = 0 \)
Change
\( -0.40 \)
\( +2 \times 0.40 = +0.80 \)
Equilibrium
\( 2.00 - 0.40 = 1.60 \)
\( 0.80 \)
\[ K_{\mathrm{c}} = \frac{(0.80)^{2}}{1.60} = \frac{0.64}{1.60} = 0.40 \]

Units are not required for \( K_{\mathrm{c}} \) in this course. The value is less than 1, so at this temperature the equilibrium slightly favours \( \mathrm{N_2O_4} \).

(d) Calculate \( Q \) with the same expression:

\[ Q = \frac{(0.20)^{2}}{1.00} = \frac{0.040}{1.00} = 0.040 \]

\( Q = 0.040 \) is less than \( K = 0.40 \), so there are too few products relative to equilibrium. The reaction proceeds forwards, to the right, forming more \( \mathrm{NO_2} \) until \( Q \) has risen to 0.40.

Check it. In (c), substitute back: \( \dfrac{0.80^2}{1.60} = 0.40 \) ✓, and confirm the ICE arithmetic conserves nitrogen — initially \( 2 \times 2.00 = 4.00 \) mol of N atoms; at equilibrium \( 2 \times 1.60 + 0.80 = 4.00 \) ✓. In (d), sanity-check the direction without algebra: there is a lot of \( \mathrm{N_2O_4} \) and very little \( \mathrm{NO_2} \) compared with the equilibrium mixture in (c), so of course more \( \mathrm{NO_2} \) must form. If your \( Q \) comparison says otherwise, you have inverted the expression.
Saying that increasing the concentration of a reactant increases \( K \). It does not. Adding a reactant shifts the position of equilibrium to the right, so more product is formed — but the ratio defined by \( K \) is restored to exactly its previous value, which is why the shift happens. The only thing that changes \( K \) is temperature. A useful discipline: for every Le Châtelier question, answer two separate questions in two separate sentences — which way does the position shift, and does \( K \) change? The second answer is “no” unless the temperature was altered.

📝Practise

Work through these on paper, then reveal the answer.

1. State four characteristics of a system at dynamic equilibrium, and describe one piece of evidence that the reaction has not stopped.
(i) The system is closed — no substance enters or leaves. (ii) The rate of the forward reaction equals the rate of the reverse reaction. (iii) The concentrations of all species remain constant — constant, not equal. (iv) All macroscopic properties (colour, pressure, pH, density) are constant with time, so there is no observable change. Evidence that the reaction continues: isotopic labelling. If a reactant containing a heavier isotope is introduced into a system already at equilibrium, the label subsequently appears in the products even though no concentration changes. Since nothing macroscopic alters, the only explanation is that both reactions are still proceeding, at equal rates. This is what the word dynamic is doing in the name, and it is the distinction between equilibrium and a reaction that has simply finished.
2. Write the equilibrium constant expression for: (a) \( 2\mathrm{SO_2}(g) + \mathrm{O_2}(g) \rightleftharpoons 2\mathrm{SO_3}(g) \), (b) \( \mathrm{CaCO_3}(s) \rightleftharpoons \mathrm{CaO}(s) + \mathrm{CO_2}(g) \), (c) \( \mathrm{CH_3COOH}(aq) \rightleftharpoons \mathrm{CH_3COO^{-}}(aq) + \mathrm{H^{+}}(aq) \).
(a) \( K_{\mathrm{c}} = \dfrac{[\mathrm{SO_3}]^2}{[\mathrm{SO_2}]^2[\mathrm{O_2}]} \) — every species is a gas, and each is raised to the power of its coefficient. (b) \( K_{\mathrm{c}} = [\mathrm{CO_2}] \). Both solids are omitted, because the concentration of a pure solid is fixed by its density and does not change as the reaction proceeds; only the gas appears. This is precisely why state symbols are not optional. (c) \( K_{\mathrm{c}} = \dfrac{[\mathrm{CH_3COO^{-}}][\mathrm{H^{+}}]}{[\mathrm{CH_3COOH}]} \). Note that water does not appear even in the form written with \( \mathrm{H_2O} \) as a reactant, because as the solvent it is in vast excess and its concentration is effectively constant. This particular constant has a name — it is \( K_{\mathrm{a}} \), the acid dissociation constant (R3.1).
3. For \( 2\mathrm{SO_2}(g) + \mathrm{O_2}(g) \rightleftharpoons 2\mathrm{SO_3}(g) \), \( \Delta H = -196\ \mathrm{kJ\,mol^{-1}} \). Predict and explain the effect on the yield of SO3 and on \( K_{\mathrm{c}} \) of: (a) increasing the pressure, (b) increasing the temperature, (c) removing SO3 as it forms.
(a) Increasing pressure: there are 3 moles of gas on the left and 2 on the right, so the equilibrium shifts towards the side with fewer moles of gas — to the right. The yield of SO3 increases. \( K_{\mathrm{c}} \) is unchanged. (b) Increasing temperature: the forward reaction is exothermic, so the equilibrium shifts in the endothermic direction — to the left — to absorb the added energy. The yield decreases, and here \( K_{\mathrm{c}} \) does change: it decreases. (c) Removing SO3: the equilibrium shifts to the right to replace what was removed, so more SO3 is formed overall; \( K_{\mathrm{c}} \) is unchanged. This is the Contact process, and it shows the same industrial dilemma as the Haber process: thermodynamics wants a low temperature and kinetics wants a high one, so a compromise temperature of about 450 °C is used with a vanadium(V) oxide catalyst, which raises the rate without affecting the yield.
4. At 700 K, \( K_{\mathrm{c}} = 54 \) for \( \mathrm{H_2}(g) + \mathrm{I_2}(g) \rightleftharpoons 2\mathrm{HI}(g) \). (a) State what this tells you about the extent of reaction. (b) Calculate \( K_{\mathrm{c}} \) for \( 2\mathrm{HI}(g) \rightleftharpoons \mathrm{H_2}(g) + \mathrm{I_2}(g) \) and for \( \tfrac{1}{2}\mathrm{H_2}(g) + \tfrac{1}{2}\mathrm{I_2}(g) \rightleftharpoons \mathrm{HI}(g) \).
(a) \( K_{\mathrm{c}} = 54 \) is greater than 1, so at equilibrium the products are favoured: the mixture contains considerably more hydrogen iodide than hydrogen and iodine. It is not, however, anywhere near completion — a \( K \) of 54 is modest, and appreciable quantities of both reactants remain. (b) For the reverse reaction, \( K \) is the reciprocal: \( K = \dfrac{1}{54} = \mathbf{0.019} \). For the equation halved, \( K \) is the square root: \( K = \sqrt{54} = \mathbf{7.3} \). The general rules: reversing an equation inverts \( K \); multiplying an equation by \( n \) raises \( K \) to the power \( n \). This is why a quoted value of \( K \) is meaningless unless the equation it refers to is stated alongside it.
5. AHL 1.00 mol of \( \mathrm{H_2} \) and 1.00 mol of \( \mathrm{I_2} \) are placed in a 2.00 dm3 vessel. At equilibrium 0.75 mol of HI is present. Calculate \( K_{\mathrm{c}} \) for \( \mathrm{H_2} + \mathrm{I_2} \rightleftharpoons 2\mathrm{HI} \).
Work in concentrations, so divide every amount by 2.00 dm3. Initial: \( [\mathrm{H_2}] = [\mathrm{I_2}] = \dfrac{1.00}{2.00} = 0.500 \); \( [\mathrm{HI}] = 0 \). Equilibrium HI: \( \dfrac{0.75}{2.00} = 0.375\ \mathrm{mol\,dm^{-3}} \). Change: HI increased by 0.375, and the ratio is 1 : 1 : 2, so each reactant decreased by \( \dfrac{0.375}{2} = 0.1875 \). Equilibrium reactants: \( 0.500 - 0.1875 = 0.3125\ \mathrm{mol\,dm^{-3}} \) each. Then \( K_{\mathrm{c}} = \dfrac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]} = \dfrac{(0.375)^2}{(0.3125)(0.3125)} = \dfrac{0.1406}{0.09766} = \mathbf{1.44} \). Check by conserving atoms: initially 2.00 mol of H atoms; at equilibrium \( 2(0.3125 \times 2.00) + 0.75 = 1.25 + 0.75 = 2.00 \) ✓. The commonest error here is forgetting to divide the amounts by the volume — harmless when the volume is 1 dm3, fatal when it is not.
6. AHL For a reaction at 298 K, \( K = 4.0 \times 10^{-3} \). (a) Calculate \( \Delta G^{\ominus} \). (b) A mixture is prepared for which \( Q = 8.0 \times 10^{-5} \). State and explain which way it will proceed. (\( R = 8.31\ \mathrm{J\,K^{-1}\,mol^{-1}} \).)
(a) \( \Delta G^{\ominus} = -RT\ln K = -(8.31)(298)\ln(4.0 \times 10^{-3}) \). Now \( \ln(4.0 \times 10^{-3}) = -5.521 \), so \( \Delta G^{\ominus} = -(2476)(-5.521) = +13\,670\ \mathrm{J\,mol^{-1}} = \mathbf{+13.7\ kJ\,mol^{-1}} \). The positive sign is consistent with \( K < 1 \): under standard conditions the reactants are favoured. (b) \( Q = 8.0 \times 10^{-5} \) is smaller than \( K = 4.0 \times 10^{-3} \), so the mixture contains too few products relative to the equilibrium position. The reaction therefore proceeds forwards, to the right, forming more product until \( Q \) rises to equal \( K \). Note the apparent paradox worth understanding: \( \Delta G^{\ominus} \) is positive, yet the reaction still goes forward — because \( \Delta G^{\ominus} \) refers to standard conditions, and under the actual non-standard conditions \( \Delta G = \Delta G^{\ominus} + RT\ln Q \) is negative. A positive \( \Delta G^{\ominus} \) means \( K \) is small, not that nothing happens.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • PhET — Reversible Reactions, which shows both directions running simultaneously and makes the “dynamic” part of dynamic equilibrium visible rather than asserted.
  • RSC Learn Chemistry — the cobalt chloride and the \( \mathrm{NO_2}/\mathrm{N_2O_4} \) equilibrium demonstrations, both of which change colour and let you see Le Châtelier’s principle act on temperature and concentration.
  • Chemistry LibreTexts — a careful account of why the equilibrium constant depends only on temperature, for when the assertion starts to feel like something you have been told rather than something you understand.