R2.3 How far? The extent of chemical change

Reversible reactions and closed systems

Many equations are written with a single arrow, as if every particle of reactant turns into product. Many reactions can also run backwards: the products react together to re-form the reactants. A reversible reaction is shown with the double half-arrow ${{\xrightleftharpoons{}}}$.

Heating blue hydrated copper(II) sulfate drives off its water of crystallization and leaves white anhydrous copper(II) sulfate. Adding water to the white powder turns it blue again and releases heat:

The forward reaction is endothermic and the backward reaction is exothermic, by exactly the same amount of energy. This is true of every reversible reaction: if the forward reaction has an enthalpy change $\Delta H$, the backward reaction has $-\Delta H$.

Open and closed systems

Whether a reversible reaction can reach equilibrium depends on the system it takes place in. Calcium carbonate heated strongly in an open crucible decomposes completely into calcium oxide, because the carbon dioxide formed escapes into the air: ${{\mathrm{CaCO}{\vphantom{X}}_{\smash[t]{3}}\mskip2mu (\mathrm{s})\xrightleftharpoons{}\mathrm{CaO}\mskip2mu (\mathrm{s}) {}+{} \mathrm{CO}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})}}$. In a sealed container at the same temperature the carbon dioxide cannot escape. It builds up until it recombines with calcium oxide as fast as it is formed, and the decomposition stops short of completion.

A closed system need not be sealed in glass. A reaction in solution in a beaker behaves as a closed system as long as no species escapes as a gas or leaves as a precipitate. In an open system, a reversible reaction that releases a gas is pulled towards products as the gas escapes, and equilibrium is never reached.

Dynamic equilibrium

When hydrogen and iodine vapour are sealed in a flask at 700 K, they react to form hydrogen iodide:

At the start only hydrogen and iodine are present, so only the forward reaction occurs and its rate is at its highest. As the reactants are used up, their concentrations fall and the forward rate decreases. As soon as some hydrogen iodide has formed, the backward reaction starts; its rate grows as the concentration of hydrogen iodide rises. Eventually the two rates become equal. From then on, hydrogen iodide is formed and decomposed at the same rate, so no further change in any concentration can be detected. The graph of rate against time shows the two rates converging.

The word dynamic matters. At dynamic equilibrium both reactions are still taking place, with molecules continually reacting in both directions; the system only appears to have stopped because the two processes cancel. The term position of equilibrium describes the composition of the equilibrium mixture. An equilibrium that "lies to the right" contains mainly products; one that "lies to the left" contains mainly reactants.

The characteristics of a system at equilibrium

The last characteristic can be tested directly. A flask that starts with hydrogen iodide only decomposes part of it into hydrogen and iodine, and ends up with exactly the same composition as a flask that started with the equivalent amounts of hydrogen and iodine. The concentration–time graphs of the two experiments level off at the same values.

Physical and chemical equilibria

Dynamic equilibrium is not limited to chemical reactions. Any reversible physical change in a closed system reaches the same kind of balance between two opposing processes.

Liquid and vapour

Water in a stoppered flask evaporates: the most energetic molecules at the surface escape into the space above the liquid. The vapour molecules move randomly, and some strike the liquid surface and are recaptured (condensation). As the amount of vapour builds up, the rate of condensation rises until it equals the rate of evaporation:

The amount of vapour then stays constant, even though molecules continue to move in both directions. The diagram of a stoppered flask beside an open beaker shows why a closed system is needed.

A bottle of bromine shows the same equilibrium in colour: the space above the dark red liquid fills with orange-brown vapour, and the intensity of the colour stops changing once ${{\mathrm{Br}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{l})\xrightleftharpoons{}\mathrm{Br}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})}}$ is at equilibrium.

Solid and saturated solution

When excess solid is stirred with a solvent, it dissolves until the solution is saturated. A saturated solution in contact with undissolved solid is at dynamic equilibrium: ions or molecules leave the solid and are deposited on it at the same rate. For sodium chloride in water:

Gas and dissolved gas

A gas in contact with a liquid dissolves until the rate at which molecules enter the solution equals the rate at which they leave it. Carbonated drinks depend on this heterogeneous equilibrium, which responds to pressure changes as Le Châtelier's principle predicts:

Chemical equilibria

The same balance of rates applies to the chemical equilibria that recur throughout chemistry:

The equilibrium law and the equilibrium constant

The composition of an equilibrium mixture depends on how much of each substance was put in, so the equilibrium concentrations themselves are not fixed. A particular combination of them is. The table gives three equilibrium mixtures for the hydrogen iodide reaction at 700 K, made from different starting mixtures.

Equilibrium concentrations calculated from $K = 54$ at 700 K. Starting concentrations in $\text{mol dm}^{-3}$.

The simple ratio of product to reactants changes from one experiment to the next. When the concentration of hydrogen iodide is squared, matching its coefficient of 2 in the equation, the ratio is the same in every experiment, within rounding. This constant is the equilibrium constant, $K$, and the rule that produces it is the equilibrium law.

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

Square brackets mean the equilibrium concentration in $\text{mol dm}^{-3}$. When concentrations are used, the constant is often written $K_\text{c}$. Because the powers on the top and bottom of the fraction often differ, the units of $K$ change from reaction to reaction; for simplicity $K$ is treated as a pure number, without units.

Writing equilibrium constant expressions

1. Start from the balanced equation, with the correct coefficients.

2. Write the products on top of the fraction and the reactants underneath.

3. Raise each concentration to the power of its coefficient.

4. Multiply the terms together; never add them.

The magnitude of K and the extent of reaction

Because products are on top of the expression, a large value of $K$ means that the equilibrium mixture contains much more product than reactant, and a small value means the reverse. The size of $K$ therefore measures the extent of reaction: how far the reaction has gone towards products when equilibrium is reached. It says nothing about how quickly equilibrium is reached.

Values at 298 K calculated from standard Gibbs energies of formation (IB Chemistry data booklet, 2025, and NIST-JANAF tables, 1998); the value at about 1100 K from NIST-JANAF data (1998); esterification and ester hydrolysis from laboratory measurements.

For the simplest kind of equilibrium, ${{\mathrm{A}\xrightleftharpoons{}\mathrm{B}}}$, the fraction of product at equilibrium is $K/(1 + K)$. The bar chart of equilibrium compositions turns this into a picture: a hundredfold change in $K$ near 1 changes the mixture from mostly reactant to mostly product.

In practice, a reaction with $K$ greater than about $10^{10}$ is treated as going to completion, and one with $K$ smaller than about $10^{-10}$ as not happening at all. The very small value of $K$ for the formation of nitrogen monoxide from air at room temperature explains why the nitrogen and oxygen of the atmosphere coexist without reacting. At the temperatures inside a car engine, above 1500 K, $K$ is much larger, and enough nitrogen monoxide forms to become a serious pollutant.

K, acids and other equilibria

Weak acids give a useful example of $K$ as a measure of extent. Ethanoic acid dissociates in water as ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{COOH}\mskip2mu (\mathrm{aq}) {}+{} \mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}\mskip2mu (\mathrm{l})\xrightleftharpoons{}\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{COO}{\vphantom{X}}^{-}\mskip2mu (\mathrm{aq}) {}+{} \mathrm{H}{\vphantom{X}}_{\smash[t]{3}}\mathrm{O}{\vphantom{X}}^{+}\mskip2mu (\mathrm{aq})}}$, with $K = 1.7\times10^{-5}$ at 298 K. The value shows that only a tiny fraction of the acid molecules have given up a proton at equilibrium: the acid is weak. The equilibrium constant for the dissociation of an acid is called the acid dissociation constant, and the larger its value, the stronger the acid.

K depends only on temperature

At a given temperature, $K$ has a single value for a given equation. Changing the concentration of a reactant or product, the pressure, or adding a catalyst changes the composition of the equilibrium mixture or how fast it is reached, but the ratio in the expression returns to the same value of $K$. Only a change in temperature changes $K$. For the synthesis of ammonia, an exothermic reaction, $K$ falls steeply as the temperature rises:

Calculated from NIST-JANAF thermochemical tables (4th edition, 1998) for ammonia, standard pressure 100 kPa.

The direction of the change depends on the sign of $\Delta H$. For an exothermic forward reaction, $K$ decreases as temperature increases; for an endothermic forward reaction, $K$ increases. Le Châtelier's principle accounts for the direction.

Relating K values for related equations

The value of $K$ belongs to the equation as written. If the equation is reversed, the expression is turned upside down. For the hydrogen iodide equilibrium at 700 K:

The same reasoning shows what happens when the coefficients are multiplied. Doubling every coefficient squares every term in the expression, so the new constant is $K^2$. Halving every coefficient takes the square root, giving $\sqrt{K}$. In general, multiplying the coefficients by $n$ raises $K$ to the power $n$.

Le Châtelier's principle

An equilibrium mixture stays the same only while the conditions stay the same. If a reactant is added, the pressure is changed or the temperature is altered, the rates of the forward and backward reactions are no longer equal, and the composition changes until a new equilibrium is reached. Le Châtelier's principle predicts the direction of the change.

"Partly" is important. The shift reduces the effect of the change but never cancels it completely: adding a reactant shifts the equilibrium to use some of it up, but its concentration at the new equilibrium is still higher than before. The principle is qualitative. It predicts the direction of the shift, but not its size; that needs $K$.

Three questions should be answered separately for any change: which way the position of equilibrium shifts, whether the value of $K$ changes, and whether the rate at which equilibrium is reached changes.

Changing a concentration

Adding a reactant increases the rate of the forward reaction, because reactant particles collide more often. For a time the forward reaction is faster than the backward reaction, so more product forms, until the rates are equal again. The equilibrium has shifted to the right, using up some of the added reactant. Removing a product has the same effect, because the backward reaction slows down. Conversely, adding a product or removing a reactant shifts the equilibrium to the left.

The concentration–time graph for the Haber equilibrium shows what happens when extra nitrogen is added to a mixture already at equilibrium.

$K$ is unchanged by a change in concentration. The new equilibrium concentrations are different, but when they are put into the expression they give the same value of $K$ as before.

Colour changes that show the shift

Equilibria between coloured ions make the shifts visible in the laboratory. Chromate(VI) ions are yellow and dichromate(VI) ions are orange:

Adding a few drops of dilute sulfuric acid increases $[{{\mathrm{H}{\vphantom{X}}^{+}}}]$, so the equilibrium shifts to the right and the solution turns orange. Adding sodium hydroxide solution removes ${{\mathrm{H}{\vphantom{X}}^{+}}}$ ions (they react with ${{\mathrm{OH}{\vphantom{X}}^{-}}}$ to form water), so the equilibrium shifts to the left and the solution turns yellow again. The change can be reversed as often as the acid and alkali are added.

Iron(III) ions and thiocyanate ions form a deep blood-red complex ion:

Adding either iron(III) ions or thiocyanate ions makes the colour deeper as the equilibrium shifts to the right. Adding a substance that removes ${{\mathrm{Fe}{\vphantom{X}}^{3+}}}$ ions shifts it to the left and the red colour fades.

Changing the pressure

Pressure changes matter only for equilibria involving gases. At constant temperature, the pressure of a gas mixture is proportional to the total number of gas molecules in a given volume. Compressing an equilibrium mixture into a smaller volume increases the pressure, and the system opposes the increase by shifting towards the side with fewer gas molecules. Lowering the pressure shifts the equilibrium towards the side with more gas molecules.

Brown nitrogen dioxide and colourless dinitrogen tetroxide show the effect directly. When a gas syringe containing the equilibrium mixture is pushed in quickly, the colour first darkens, because the same number of brown ${{\mathrm{NO}{\vphantom{X}}_{\smash[t]{2}}}}$ molecules is packed into a smaller volume. Within seconds the colour becomes paler again, though not as pale as at the start, as some ${{\mathrm{NO}{\vphantom{X}}_{\smash[t]{2}}}}$ combines to form ${{\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{4}}}}$. The left-hand concentration–time sketch shows the same sequence.

A change in pressure does not change $K$, as long as the temperature is constant. It changes the composition of the equilibrium mixture because the concentrations all change at once, by the same factor, and this upsets the balance of the expression whenever the numbers of gas molecules on the two sides differ.

Changing the temperature

A change in temperature affects an equilibrium differently from the other changes, because it changes the value of $K$ itself. Every reversible reaction is exothermic in one direction and endothermic in the other. Raising the temperature favours the endothermic direction, which absorbs heat and so partly opposes the rise. Lowering the temperature favours the exothermic direction, which releases heat.

Nitrogen dioxide is again a clear example. The forward reaction is endothermic, because a bond must be broken to split ${{\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{4}}}}$ into two ${{\mathrm{NO}{\vphantom{X}}_{\smash[t]{2}}}}$ molecules:

Heating the sealed mixture shifts the equilibrium to the right and the brown colour deepens; cooling it in ice shifts it to the left and the colour fades. The photograph of one sealed sample at five temperatures shows the effect.

The concentration–time sketch for heating the ${{\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{4}}}}$–${{\mathrm{NO}{\vphantom{X}}_{\smash[t]{2}}}}$ mixture shows no sudden jump at the moment the temperature is raised: the concentrations change gradually as the system moves to the new equilibrium. Because $[{{\mathrm{NO}{\vphantom{X}}_{\smash[t]{2}}}}]$ rises and $[{{\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{4}}}}]$ falls, the value of $[{{\mathrm{NO}{\vphantom{X}}_{\smash[t]{2}}}}]^2/[{{\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{4}}}}]$ at the new equilibrium is larger: $K$ has increased.

The cobalt(II) chloride equilibrium is a colourful classroom demonstration of both temperature and concentration effects. The pink hexaaquacobalt(II) ion and the blue tetrachlorocobaltate(II) ion are in equilibrium, and the forward reaction is endothermic:

• Heating a purple mixture in a hot-water bath turns it blue (shift to the right, the endothermic direction); cooling it in ice turns it pink.

• Adding concentrated hydrochloric acid, a source of chloride ions, turns it blue.

• Adding water turns it pink. Water is a product, but the main effect is dilution: there are five dissolved particles on the left for every one on the right, so the equilibrium shifts towards the side with more dissolved particles, which partly restores their total concentration.

Adding a catalyst

A catalyst provides an alternative reaction pathway with a lower activation energy. For a reversible reaction, the forward and backward reactions pass through the same transition state, so lowering the energy barrier lowers the activation energy of both directions by the same amount. The energy profile with and without a catalyst shows this.

Both rates increase by the same factor, so they stay equal and the position of equilibrium does not change. The value of $K$ is unchanged too. What the catalyst changes is the time needed to reach equilibrium, which is shorter. In industry this matters a great deal: a catalyst allows an acceptable amount of product to be made in a short time, often at a lower temperature than would otherwise be needed.

Summary of the effects

Applying Le Châtelier's principle to other systems

The principle applies to any equilibrium, including heterogeneous ones and systems never met before. The method is always the same: identify the species whose concentration (or the condition) changes, decide which direction partly reverses that change, and state the consequence that can be observed or measured.

Gases dissolving in liquids

Carbonated drinks are bottled under a pressure of carbon dioxide several times atmospheric pressure, typically 300–400 kPa in a sealed bottle of cola at room temperature. Carbon dioxide in the gas space above the drink is in equilibrium with dissolved carbon dioxide:

Opening the bottle lets the gas escape and the pressure of carbon dioxide above the drink falls to that of the air, where carbon dioxide is only about 0.04 % of the gas. The equilibrium shifts to the left, towards the side with gas molecules, and dissolved carbon dioxide comes out of solution as bubbles. A drink left open goes "flat" as the system moves towards a new equilibrium with almost no dissolved gas. Dissolving carbon dioxide is exothermic, so a warm drink also holds less gas than a cold one.

Some of the dissolved carbon dioxide reacts with water to form carbonic acid, which makes fizzy drinks acidic:

When carbonated water is left open, the loss of ${{\mathrm{CO}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{aq})}}$ shifts these linked equilibria to the left as well, so $[{{\mathrm{H}{\vphantom{X}}^{+}}}]$ falls and the pH rises.

Oxygen transport at high altitude

Haemoglobin (Hb) in red blood cells carries oxygen from the lungs to the tissues. The binding can be summarised as an equilibrium:

At the summit of Mount Everest the air pressure is about a third of that at sea level, so less oxygen dissolves in the blood. The equilibrium shifts to the left, less oxyhaemoglobin forms, and the tissues receive less oxygen: climbers suffer breathlessness, headaches and fatigue. Over several weeks at altitude the body makes more haemoglobin. The higher concentration of Hb shifts the equilibrium back to the right, restoring the amount of oxygen carried. Athletes use the same effect when they train at high altitude.

Dilution, acids and other added substances

A reagent is rarely added on its own. Dilute hydrochloric acid, for example, is mostly water, and its effect on an equilibrium depends on which of its components matters more.

• Dilution. Adding water to an aqueous equilibrium lowers the concentration of every dissolved species at once. The equilibrium shifts towards the side with more dissolved particles. For the cobalt(II) chloride equilibrium (five dissolved particles on the left, one on the right), adding dilute hydrochloric acid turns the solution pink: the water it adds outweighs its chloride ions.

• Choice of acid catalyst in esterification. Concentrated sulfuric acid is used as the catalyst for making esters. A dilute acid would add water, a product, and shift the equilibrium to the left, lowering the yield; concentrated sulfuric acid also absorbs some of the water formed. An excess of the cheaper reactant (often the alcohol) shifts the equilibrium further to the right.

• Mixing bleach with acid. Household bleach contains chlorate(I) and chloride ions in an alkaline solution: ${{\mathrm{Cl}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{aq}) {}+{} 2\,\mathrm{OH}{\vphantom{X}}^{-}\mskip2mu (\mathrm{aq})\xrightleftharpoons{}\mathrm{OCl}{\vphantom{X}}^{-}\mskip2mu (\mathrm{aq}) {}+{} \mathrm{Cl}{\vphantom{X}}^{-}\mskip2mu (\mathrm{aq}) {}+{} \mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}\mskip2mu (\mathrm{l})}}$. An acid such as vinegar removes ${{\mathrm{OH}{\vphantom{X}}^{-}}}$ ions, so the equilibrium shifts to the left and toxic chlorine gas is released. This is why bleach must never be mixed with acidic cleaners.

Equilibrium in industry: the Haber and Contact processes

Industrial chemists need as much product as possible, as quickly and cheaply as possible. Le Châtelier's principle tells them which conditions give a high equilibrium yield; collision theory and the Maxwell–Boltzmann distribution tell them which conditions give a high rate. Where the two disagree, compromise conditions are chosen.

The Haber process for ammonia

The Haber process combines nitrogen from the air with hydrogen made from methane and steam:

The ammonia is used mainly to make fertilizers such as ammonium nitrate and urea, and also nitric acid, explosives and nylon. The graph of the equilibrium percentage of ammonia shows how its yield depends on the conditions.

The curves follow directly from Le Châtelier's principle:

• Pressure. Four moles of gas become two, so a high pressure shifts the equilibrium to the right and increases the yield. High pressure also increases the rate, because the gases are more concentrated. However, very high pressures need expensive thick-walled vessels and powerful compressors, use a great deal of energy, and increase the risk of leaks. Plants typically operate at about 100–250 atm, often around 200 atm.

• Temperature. The forward reaction is exothermic, so a low temperature gives a higher equilibrium yield (and a larger K). At low temperatures, however, the reaction is far too slow, even with a catalyst, and the iron catalyst works effectively only above about 400 °C. A compromise temperature of about 400–450 °C gives a reasonable yield at a reasonable rate.

• Catalyst. Iron, with small amounts of potassium and aluminium oxides as promoters, speeds up both reactions so that the mixture approaches equilibrium in the short time the gases spend in the converter. It has no effect on the equilibrium yield.

• Removing the product. The gas leaving the converter is cooled, and ammonia (boiling point −33 °C) condenses and is removed. The unreacted nitrogen and hydrogen are recycled.

Only about 15 % of the gas is converted on each pass, well below the equilibrium value of about 26 % at 200 atm and 450 °C, because the gases do not stay in the converter long enough to reach equilibrium. Recycling the unreacted gases means that about 98 % of the nitrogen and hydrogen fed in is eventually converted into ammonia.

The Contact process for sulfuric acid

Sulfuric acid is made by the Contact process. Sulfur is burned to sulfur dioxide, which is then oxidized to sulfur trioxide in a reversible, exothermic reaction:

As for ammonia, a low temperature favours the yield but slows the reaction, so the gases pass over a vanadium(V) oxide, ${{\mathrm{V}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{5}}}}$, catalyst at about 400–450 °C, in several beds with cooling between them. A high pressure would favour sulfur trioxide (three moles of gas become two), but the equilibrium already lies far to the right at these temperatures, so the process runs at only 1–2 atm: the small gain in yield would not pay for the compressors. Using excess air (oxygen) also shifts the equilibrium to the right. In modern plants, sulfur trioxide is absorbed between catalyst beds, which removes a product and drives the equilibrium further right, so that more than 99.5 % of the sulfur dioxide is converted.

The sulfur trioxide is absorbed in concentrated sulfuric acid rather than in water, because its reaction with water is so exothermic that it forms a corrosive mist of acid droplets.

The reaction quotient, Q (HL)

The equilibrium constant expression can be evaluated for any mixture, not only one at equilibrium. The result is the reaction quotient.

As a reaction proceeds, $Q$ changes; at equilibrium it no longer changes, and $Q = K$. Comparing $Q$ with $K$ therefore shows how far a mixture is from equilibrium and in which direction it must react to get there.

• $Q < K$: there is too little product compared with the equilibrium mixture. The forward reaction is faster than the backward reaction, so there is a net forward reaction until $Q$ rises to $K$.

• $Q = K$: the mixture is at equilibrium; the two rates are equal and there is no net change.

• $Q > K$: there is too much product. The backward reaction is faster, so there is a net backward reaction until $Q$ falls to $K$.

Using Q to explain Le Châtelier's principle

Le Châtelier's principle predicts the direction of a shift, and comparing $Q$ with $K$ gives the reason for it. Any change to a system at equilibrium either changes $Q$ (concentration and pressure changes) or changes $K$ (temperature changes). In both cases $Q \neq K$ for a time, and the system reacts until they are equal again.

Concentration

Adding nitrogen to the Haber equilibrium increases $[{{\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}}}]$, which is in the denominator of $Q = \dfrac{[{{\mathrm{NH}{\vphantom{X}}_{\smash[t]{3}}}}]^2}{[{{\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}}}][{{\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}}}]^3}$. $Q$ falls below $K$, so the net reaction is forward until $Q = K$ again: the equilibrium shifts to the right.

Pressure

Halving the volume of the Haber mixture doubles every concentration. The numerator of $Q$ contains two concentration terms and the denominator four, so the denominator increases more:

Now $Q < K$, so the system reacts in the forward direction, forming more ammonia, until $Q$ rises to $K$ again. For ${{\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g}) {}+{} \mathrm{I}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})\xrightleftharpoons{}2\,\mathrm{HI}\mskip2mu (\mathrm{g})}}$, with two gas molecules on each side, the factors of 2 cancel, $Q$ stays equal to $K$, and the equilibrium does not shift.

Dilution

Doubling the volume of the iron(III) thiocyanate equilibrium by adding water halves every concentration:

Now $Q > K$, so the net reaction is backward: the equilibrium shifts towards the side with more dissolved particles, and the red colour becomes paler than dilution alone would make it.

Temperature

A change in temperature does not change the concentrations at the moment it happens, so $Q$ is unchanged, but $K$ changes. For the exothermic Haber reaction, raising the temperature makes $K$ smaller. The existing mixture now has $Q > K_{\text{new}}$, so the net reaction is backward and some ammonia decomposes until $Q$ falls to the new, smaller $K$.

Calculating the composition of equilibrium mixtures (HL)

Equilibrium problems are solved by tracking the amount of every species from the start of the reaction to equilibrium. An ICE table (Initial, Change, Equilibrium) organizes the working:

1. Write the balanced equation and, under it, the initial amount or concentration of every species.

2. Write the change for each species using the coefficients: if $x$ mol of ${{\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}}}$ react, $3x$ mol of ${{\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}}}$ react and $2x$ mol of ${{\mathrm{NH}{\vphantom{X}}_{\smash[t]{3}}}}$ form. Reactants decrease ($-$) and products increase ($+$).

3. Add the change to the initial value to get the equilibrium amount.

4. Convert amounts to concentrations by dividing by the volume (unless the volumes cancel), then substitute into the expression for $K$.

Finding K from equilibrium data

Determining K for esterification by titration

The equilibrium constant for the esterification of ethanoic acid can be measured in a school laboratory. Known amounts of ethanoic acid and ethanol are mixed in a sealed tube with a little hydrochloric acid as a catalyst, and left for about a week at room temperature to reach equilibrium. The mixture is then titrated with standard sodium hydroxide solution to find the total amount of acid left. A separate titration of the same volume of hydrochloric acid alone (a blank) gives the amount of acid due to the catalyst, which is subtracted. The rest is ethanoic acid. Equimolar amounts of acid and alcohol give a mixture with about two-thirds of the acid converted, which corresponds to $K \approx 4$.

Finding equilibrium concentrations from K

When $K$ is known, an unknown change $x$ can be found from the expression. Quadratic equations are avoided either because the expression is a perfect square or because $K$ is so small that $x$ is negligible compared with the initial concentrations.

The approximation for very small values of K

When $K$ is very small, the reaction hardly proceeds, so the amount of reactant used up is tiny compared with the amount present at the start. The equilibrium concentration of each reactant can then be taken as equal to its initial concentration, the small K approximation:

This removes $x$ from the denominator and leaves a simple equation. The approximation is the same one used to calculate the pH of a weak acid solution, where only a tiny fraction of the acid dissociates. After solving, the assumption is checked: if $x$ is less than about 5 % of the initial concentration, it was justified.

Equilibrium and Gibbs energy (HL)

The change in Gibbs energy, $\Delta G$, shows whether a change is spontaneous at constant temperature and pressure: a reaction can proceed by itself in the direction in which $\Delta G$ is negative. The standard Gibbs energy change, $\Delta G^{\ominus}$, refers to the complete conversion of reactants into products with every substance in its standard state. Its value fixes the position of equilibrium, through an equation given in the data booklet:

Here $R = 8.31\ \text{J K}^{-1}\ \text{mol}^{-1}$ (data booklet), $T$ is the temperature in kelvin and $\Delta G^{\ominus}$ must be in $\text{J mol}^{-1}$. Rearranging gives $K = e^{-\Delta G^{\ominus}/RT}$.

Because the relationship is exponential, a modest value of $\Delta G^{\ominus}$ makes a large difference. At 298 K, $\Delta G^{\ominus} = -30\ \text{kJ mol}^{-1}$ gives $K \approx 2\times10^{5}$, and $+30\ \text{kJ mol}^{-1}$ gives $K \approx 5\times10^{-6}$. Reactions with $\Delta G^{\ominus}$ more negative than about $-60\ \text{kJ mol}^{-1}$ effectively go to completion. The graph of $K$ against $\Delta G^{\ominus}$ shows the shape of the relationship near $K = 1$.

When no value of $\Delta G^{\ominus}$ is given, it can be calculated from standard Gibbs energies of formation, in the same way as an enthalpy change from enthalpies of formation: products minus reactants, each multiplied by its coefficient.

The calculation also works in reverse: a measured value of $K$ gives $\Delta G^{\ominus}$ for the reaction at that temperature.

Gibbs energy before equilibrium is reached

For a mixture that is not at equilibrium, the Gibbs energy change depends on its composition through $Q$ (data booklet):

The sign of $\Delta G$ shows which direction is spontaneous for that mixture. When $Q < K$, $\Delta G$ is negative and the forward reaction is favoured. When $Q > K$, $\Delta G$ is positive, so the backward reaction is favoured. As the mixture reacts, $Q$ approaches $K$ and $\Delta G$ approaches zero. At equilibrium $\Delta G = 0$ and $Q = K$, so $0 = \Delta G^{\ominus} + RT\ln K$, which is where $\Delta G^{\ominus} = -RT\ln K$ comes from. The graph of the Gibbs energy of the mixture against its composition shows the same idea as a valley, with equilibrium at the bottom.

The full treatment of Gibbs energy, including $\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}$ and the effect of temperature on spontaneity, belongs with entropy and spontaneity. Combining the two equations also explains why $K$ changes with temperature.

Applying the ideas in exam questions

Paper 1A: an equilibrium constant expression

Paper 1A: K for a related equation

Paper 1A: changes that affect K

Paper 1B: the effect of temperature on a colour equilibrium

(Paper 1B style) A student investigated the equilibrium ${{\mathrm{Fe}{\vphantom{X}}^{3+}\mskip2mu (\mathrm{aq}) {}+{} \mathrm{SCN}{\vphantom{X}}^{-}\mskip2mu (\mathrm{aq})\xrightleftharpoons{}[\mathrm{FeSCN}]{\vphantom{X}}^{2+}\mskip2mu (\mathrm{aq})}}$, in which only ${{[\mathrm{FeSCN}]{\vphantom{X}}^{2+}}}$ is coloured (red). Equal volumes of the same iron(III) nitrate and potassium thiocyanate solutions were mixed in five tubes, and each tube was left in a water bath at a different temperature for 10 minutes. The absorbance of each mixture was then measured quickly with a colorimeter. Each absorbance is the mean of three readings.

Paper 2: the Contact process

Paper 2: methanol synthesis (HL)

Self-check

• To do:

  I can describe a reversible reaction and explain why equilibrium can be reached only in a closed system.

• To do:

  I can explain how dynamic equilibrium is reached in terms of the rates of the forward and backward reactions, and sketch rate–time and concentration–time graphs.

• To do:

  I can state the characteristics of a system at equilibrium, for physical and chemical systems.

• To do:

  I can deduce the equilibrium constant expression for a homogeneous equilibrium from its equation.

• To do:

  I can interpret the size of K (K ≪ 1, K < 1, K = 1, K > 1, K ≫ 1) in terms of the position of equilibrium and the extent of reaction.

• To do:

  I can state that K depends only on temperature, and relate K for a reverse or scaled equation to K for the original.

• To do:

  I can apply Le Châtelier's principle to changes in concentration, pressure and temperature, stating the shift, the effect on K and the reason.

• To do:

  I can explain why a catalyst affects neither the position of equilibrium nor K.

• To do:

  I can apply Le Châtelier's principle to heterogeneous and unfamiliar systems, such as carbonated drinks, haemoglobin, dilution and colour equilibria.

• To do:

  I can explain the conditions used in the Haber and Contact processes as compromises between yield, rate and cost.

• To do:

  I can calculate Q and use it to predict the direction of reaction and which rate is greater (HL).

• To do:

  I can use Q to explain the effects of concentration, pressure and temperature changes on an equilibrium (HL).

• To do:

  I can use ICE tables to calculate K from initial and equilibrium data, and equilibrium concentrations from K (HL).

• To do:

  I can use the approximation [reactant]eqm ≈ [reactant]initial when K is very small, and check it (HL).

• To do:

  I can calculate K from ΔG⦵ and ΔG⦵ from K using ΔG⦵ = −RT ln K, and relate the sign of ΔG⦵ to the position of equilibrium (HL).

• To do:

  I can use ΔG = ΔG⦵ + RT ln Q to decide which direction is spontaneous before equilibrium is reached (HL).

Ignore state symbols. Accept → in place of ⇌.

(b) Each species takes its coefficient as a power, including water, which is a gas here:

(c) All four substances are in one liquid mixture, so the equilibrium is homogeneous and water appears in the expression like any other product: