Equilibria

Reversible reactions and dynamic equilibrium

A reversible reaction is one in which the products can react together to re-form the reactants, so the reaction can go in either direction under the same conditions. We write it with the symbol `<->`: the forward reaction goes from left to right and the reverse reaction goes from right to left. Heating ammonium chloride is a simple example: NH4Cl(s) <-> NH3(g) + HCl(g). The white solid breaks down into two colourless gases, and if the gases are cooled they recombine to give the white solid again.

When a reversible reaction takes place in a closed system, at a constant temperature, it settles into dynamic equilibrium. At equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction, so the concentrations of the reactants and the products remain constant. The word dynamic matters: both reactions are still happening, all the time, at equal rates. Equilibrium does not mean the amounts of reactants and products are equal - a mixture at equilibrium can be almost all product or almost all reactant.

A closed system is one in which no matter can enter or leave, although energy can still be transferred in or out (a sealed flask can be heated). The closed system is essential, because both reactions must be able to take place at the same time and at the same rate. If ammonia is allowed to escape from the Haber reaction, or if hydrogen chloride escapes when ammonium chloride is heated in an open tube, the reverse reaction can never keep pace with the forward reaction, and equilibrium is never established.

The same equilibrium mixture is reached whether you start with the reactants or with the products, provided the temperature is the same. Starting with hydrogen and iodine, or starting with hydrogen iodide alone, gives the same amounts of all three substances at equilibrium. Temperature must also be kept constant: the position of equilibrium is fixed by the temperature, and the mixture will adjust to a new position if the temperature changes.

• At equilibrium the rate of the forward reaction equals the rate of the reverse reaction.
• The concentrations of reactants and products stay constant, but they are not usually equal to each other.
• Both reactions continue all the time; equilibrium is dynamic, not static.
• The same equilibrium mixture is reached from either direction, at a given temperature.
• Only a closed system can reach equilibrium, because nothing must be able to escape.

Le Chatelier's principle: how an equilibrium responds to change

Le Chatelier's principle states that if a change is made to a system at dynamic equilibrium, the position of equilibrium moves to minimise this change. The position of equilibrium describes how much of the mixture is reactant and how much is product. When the temperature, the concentration or the pressure is changed, the system is disturbed, and it responds by moving in whichever direction opposes the disturbance.

Temperature. If you raise the temperature, the system absorbs the extra heat by favouring the endothermic direction, so the position of equilibrium moves that way. If the forward reaction is exothermic, raising the temperature moves the equilibrium to the left and lowers the yield of product. Lowering the temperature does the opposite and moves the equilibrium in the exothermic direction. Temperature is also the only change that alters the value of the equilibrium constant.

Pressure. If you increase the pressure by compressing the mixture, the system responds by moving to the side with fewer molecules of gas, because that reduces the pressure again. In N2(g) + 3H2(g) <-> 2NH3(g) there are four moles of gas on the left and two on the right, so a high pressure favours ammonia. Decreasing the pressure moves the equilibrium to the side with more gas molecules. Adding an unreactive gas such as argon at constant volume has no effect at all, because the partial pressures of the reacting gases are unchanged.

Concentration. Adding more of a reactant makes the forward reaction go faster than the reverse reaction, so the position of equilibrium moves to the right until the two rates are equal again. Removing a product as it forms has the same effect: in the Haber process, ammonia is liquefied out of the gas mixture, so the equilibrium is being pulled to the right all the time. Changing the volume of the container is simply another way of changing the pressure.

Catalyst. A catalyst speeds up the forward and the reverse reactions equally, so it does not change the position of equilibrium and does not change the yield. It only makes the system reach equilibrium sooner. This is exactly what makes a catalyst so valuable in industry: it allows a plant to work at a lower, compromise temperature, where the equilibrium yield is better, and still produce ammonia or sulphur trioxide at an acceptable rate.

Equilibrium constants: Kc, mole fractions and Kp

For a general equilibrium aA + bB <-> cC + dD, the equilibrium constant in terms of concentration, Kc, is Kc = [C]^c [D]^d / ([A]^a [B]^b). The square brackets mean concentration in mol dm-3, and the powers are the numbers written in front of each species in the equation. Only equilibrium concentrations go into the expression, never the starting concentrations.

Kc is a constant at a given temperature, for an equation written in a stated direction. Its units depend on the powers: for H2(g) + I2(g) <-> 2HI(g) the expression is [HI]^2 / ([H2] x [I2]), and the units cancel to give a quantity with no units at all, because there are two moles of gas on each side of the equation.

For equilibria involving gases it is more convenient to work with partial pressures, because the volume then no longer matters. The mole fraction of a gas is the amount of that gas divided by the total amount of all the gases present: mole fraction = amount of the gas / total amount of all the gases, in mol. The partial pressure of a gas is the pressure it would exert if it alone occupied the container: partial pressure = mole fraction x total pressure. Partial pressures are measured in atm or in Pa.

Kp is the equilibrium constant written in terms of partial pressures. For N2(g) + 3H2(g) <-> 2NH3(g), Kp = p(NH3)^2 / (p(N2) x p(H2)^3). Here the units are atm^2 / atm^4 = atm^-2, so this Kp must always be quoted with its units. You will not be asked to use the relationship between Kp and Kc.

The size of the constant tells you where the equilibrium lies. A large K (much greater than 1) means the equilibrium mixture is mainly products, so the forward reaction goes almost to completion; a very small K means the mixture is mainly reactants. What changes K is the temperature, and only the temperature. Adding a catalyst, changing the pressure or removing a product shifts the position of equilibrium, but the value of Kc or Kp stays exactly the same.

Working out what is in an equilibrium mixture

Most equilibrium problems are solved with a small table of amounts. Write down what you start with, use the equation to work out how much has reacted, then add or subtract to find the amounts at equilibrium. Finally divide by the volume to turn amounts into concentrations, because Kc is written in terms of concentrations.

Worked example. 0.200 mol of hydrogen and 0.200 mol of iodine are sealed in a 2.00 dm3 flask and left at 700 K until equilibrium is reached. The mixture is found to contain 0.320 mol of hydrogen iodide. Two moles of HI are formed from one mole of H2, so 0.320 / 2 = 0.160 mol of H2 has reacted, leaving 0.200 - 0.160 = 0.040 mol of H2 and 0.040 mol of I2. The concentrations at equilibrium are [HI] = 0.320 / 2.00 = 0.160 mol dm-3 and [H2] = [I2] = 0.040 / 2.00 = 0.020 mol dm-3. Substituting into the expression gives Kc = (0.160)^2 / (0.020 x 0.020) = 0.0256 / 0.00040 = 64, with no units.

If you are given Kc and asked for the composition, work backwards. 1.00 mol of H2 and 1.00 mol of I2 are sealed in a 1.00 dm3 flask at a temperature where Kc = 4.00 for H2(g) + I2(g) <-> 2HI(g). If x mol of each has reacted, the equilibrium concentrations are (1.00 - x) for each reactant and 2x for HI, so Kc = (2x)^2 / (1.00 - x)^2 = 4.00. Taking the square root of both sides gives 2x / (1.00 - x) = 2.00, so x = 0.500 mol and the flask contains 0.500 mol of H2, 0.500 mol of I2 and 1.00 mol of HI. Cambridge will not require you to solve a quadratic equation, so an expression that is a perfect square like this one is always solved by taking the square root.

For a Kp calculation you need the amounts as well, because the mole fractions come from them: total amount of gas = the sum of the amounts of all the gases; mole fraction = amount of gas / total amount; partial pressure = mole fraction x total pressure. The partial pressures of all the gases in the mixture always add up to the total pressure, which is a quick way of checking your answer.

Industry: the Haber process and the Contact process

The Haber process makes ammonia from nitrogen and hydrogen: N2(g) + 3H2(g) <-> 2NH3(g). The forward reaction is exothermic, with an enthalpy change of about -92 kJ mol-1. Nitrogen is obtained from the air by the fractional distillation of liquid air, and hydrogen is usually made from natural gas. Ammonia is the starting material for nitric acid and for nitrogen-containing fertilisers, so the process runs on an enormous scale.

A high yield of ammonia needs a low temperature, because the forward reaction is exothermic, and a high pressure, because four moles of gas become two. A low temperature, however, makes the reaction far too slow, and a very high pressure needs enormously strong vessels and costs a great deal of energy to compress the gases. The plant therefore runs at about 400-450 °C and about 200 atm, with an iron catalyst. These are compromise conditions: the best balance between the yield, the rate of production and the cost. The catalyst does not change the yield, but it makes the rate acceptable at the compromise temperature, so the process can use a temperature low enough to give a reasonable equilibrium yield.

Ammonia is also removed from the mixture as it forms. The gases leaving the reactor are cooled, and ammonia liquefies first because it has the highest boiling point, while the unreacted nitrogen and hydrogen stay as gases and are recycled back into the reactor. Removing a product keeps pulling the position of equilibrium to the right, which is why the process can keep converting the gases even though only a moderate fraction reacts on each pass.

The Contact process makes sulphur trioxide, which is used to manufacture sulfuric acid: 2SO2(g) + O2(g) <-> 2SO3(g). This forward reaction is also exothermic, with an enthalpy change of about -196 kJ mol-1, so the same argument applies: a higher temperature would lower the yield, but the rate has to be fast enough to be economic, so the converter runs at about 400-450 °C over a vanadium(V) oxide, V2O5, catalyst. Three moles of gas become two, so a higher pressure would also improve the yield, but the equilibrium already lies so far to the right at ordinary pressure that the extra expense is not worth it, and the plant works at only about 1-2 atm. The sulphur trioxide is absorbed in concentrated sulfuric acid to make oleum, which is then diluted carefully to give sulfuric acid.

• Haber process: N2 + 3H2 <-> 2NH3, about 400-450 °C, about 200 atm, iron catalyst.
• Haber process: ammonia is liquefied out and the unreacted gases are recycled, which keeps shifting the position of equilibrium to the right.
• Contact process: 2SO2 + O2 <-> 2SO3, about 400-450 °C, about 1-2 atm, vanadium(V) oxide catalyst.
• A higher pressure improves the yield, because there are fewer molecules of gas on the right, but it raises the cost of the plant and of compression.
• A lower temperature improves the yield, because both forward reactions are exothermic, but it makes the rate too slow.

Bronsted-Lowry acids and bases

The common acids you must be able to name and write formulas for are hydrochloric acid, HCl; sulfuric acid, H2SO4; nitric acid, HNO3; and ethanoic acid, CH3COOH. The common alkalis are sodium hydroxide, NaOH; potassium hydroxide, KOH; and ammonia, NH3. An alkali is a base that is soluble in water, so every alkali is a base, but not every base is an alkali: copper(II) oxide is a base, yet it is insoluble and so it is not an alkali.

In the Bronsted-Lowry theory, an acid is a proton (H+ ion) donor and a base is a proton acceptor. A proton is a hydrogen ion, and every acid-base reaction is a transfer of a proton from the acid to the base. When hydrogen chloride dissolves in water, the HCl molecule donates a proton to a water molecule: HCl + H2O -> H3O+ + Cl-, so HCl is the acid and water is the base. In aqueous solution we normally write the H3O+ simply as H+(aq), and it is these H+(aq) ions that make a solution acidic.

A strong acid is one that is fully dissociated in aqueous solution, so every molecule of the acid gives up its proton and none of the original acid remains: HCl(aq) -> H+(aq) + Cl-(aq). Hydrochloric acid, sulfuric acid and nitric acid are strong acids. A weak acid is only partially dissociated, so most of the molecules stay whole and an equilibrium is set up: CH3COOH(aq) <-> H+(aq) + CH3COO-(aq). Ethanoic acid is a weak acid, and in a 0.100 mol dm-3 solution only a small fraction of the molecules has donated a proton.

Bases follow the same pattern. Sodium hydroxide and potassium hydroxide are strong bases: they are ionic solids that are fully dissociated into their ions in water, giving a high concentration of OH-(aq) ions. Ammonia is a weak base: it accepts a proton from water and sets up an equilibrium, NH3(aq) + H2O(l) <-> NH4+(aq) + OH-(aq), so only a small fraction of the ammonia molecules is protonated and the concentration of OH-(aq) ions is low.

The pH scale runs from 0 to 14, although solutions can have values outside that range. Pure water has pH 7, acidic solutions have pH below 7 and alkaline solutions have pH above 7. The lower the pH, the greater the concentration of H+(aq) ions, so a solution of pH 1 is strongly acidic and a solution of pH 12 is strongly alkaline.

Strong and weak acids: the differences you can measure

Because a strong acid is fully dissociated, 0.100 mol dm-3 hydrochloric acid contains 0.100 mol dm-3 of H+(aq) ions and has a pH of about 1. A weak acid of the same concentration contains only a small fraction of H+(aq) ions, so 0.100 mol dm-3 ethanoic acid has a pH of about 2.9. Strength and concentration are different ideas: a concentrated solution of a weak acid can easily have a lower pH than a dilute solution of a strong acid.

The difference shows up in any test that depends on the concentration of ions. A pH meter gives a direct reading, so it shows the strong acid at about pH 1 and the weak acid at about pH 2.9. Universal indicator gives colours: red in the strong acid and orange or yellow in the weak acid. A conductivity test, or a bulb with two carbon electrodes, shows the strong acid conducting the electricity much better, because a fully dissociated acid provides many more ions to carry the current, while the weak acid of the same concentration is a poor conductor.

Both acids react with a reactive metal such as magnesium: Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g). The strong acid reacts much faster, because the concentration of H+(aq) ions is much higher, so there are far more productive collisions each second. The weak acid fizzes more slowly, but if the same amount in moles of each acid is taken, the same total volume of hydrogen is eventually produced: the equilibrium CH3COOH <-> H+ + CH3COO- moves to the right as the H+ ions are used up, so the weak acid acts as a reservoir of H+ ions.

Neutralisation happens when an acid reacts with a base, and in aqueous solution the reaction that matters is the same every time: H+(aq) + OH-(aq) -> H2O(l). A salt is formed as well as water, the salt being made of the cation from the base and the anion from the acid. The equations for the common cases are HCl + NaOH -> NaCl + H2O; H2SO4 + 2KOH -> K2SO4 + 2H2O; and HNO3 + KOH -> KNO3 + H2O. Ethanoic acid with sodium hydroxide gives sodium ethanoate: CH3COOH + NaOH -> CH3COONa + H2O.

pH titration curves and choosing an indicator

A titration is used to find the volume of one solution that reacts exactly with a known volume of another. If you measure the pH after each addition of alkali from a burette and plot pH against the volume added, you get a pH titration curve. The point where the acid and the alkali have been mixed in exactly the ratio shown by the equation is the equivalence point, and on the curve it appears as the steep, almost vertical section.

The shape of the curve depends on the strengths of the acid and the alkali. A strong acid with a strong alkali starts at about pH 1 and rises slowly, then shoots up almost vertically through pH 7 and flattens off at about pH 12, so the vertical section covers a very wide range of pH. A weak acid with a strong alkali starts at about pH 3 and rises more gently, and its vertical section lies above pH 7, at about pH 8.7 for ethanoic acid with sodium hydroxide, because the salt formed in the reaction is basic. A strong acid with a weak alkali has its vertical section below pH 7, at about pH 5 to 6, because the salt formed is acidic. A weak acid with a weak alkali has no sharp vertical section at all, so no indicator can locate the end-point accurately and a pH meter must be used instead.

An indicator is itself a weak acid, and its acid form and its ion form have different colours, so it changes colour over a range of about two pH units. For a titration to be accurate, that colour-change range must lie on the steep, vertical part of the curve, so that the end-point and the equivalence point coincide. Methyl orange changes from red to yellow over about pH 3.1 to 4.4, so it suits a strong acid titrated with a weak alkali. Phenolphthalein changes from colourless to pink over about pH 8.3 to 10.0, so it suits a weak acid titrated with a strong alkali. Either indicator can be used for a strong acid with a strong alkali, because the vertical section there runs from about pH 3 to pH 11 and covers both ranges.