Chemical Equilibrium

Irreversible reactions are reactions in which the reactants are completely consumed and converted into products. Such a reaction stops when the limiting reactant is used up.

Reversible reactions can proceed in both directions — forward and reverse — without changing the concentration of reactants and products under the existing conditions. They never stop. The double arrow (⇌) tells that the reaction is reversible. Examples:

• N_2(g) + 3H_2(g) ⇌ 2NH_3(g)
• 2NO_2(g) ⇌ N₂O_4(g)
• PCl_5(g) ⇌ PCl_3(g) + Cl_2(g)

When a reversible reaction takes place in a closed container (nothing enters or escapes), the components change as some are consumed and others are formed, until a stage called chemical equilibrium is reached: the concentrations of reactants and products become constant and the rates of the forward and reverse reactions become equal.

8.1.1 Macroscopic events are phenomena that can be observed with the naked eye without considering the individual particles or molecules involved — for example change in colour, evolution or absorption of heat, formation of a precipitate, evolution of a gas, change in volume or pressure, and change in the composition of a substance.

8.1.2 Microscopic events are phenomena that cannot be observed with the naked eye — for example collisions between molecules, breaking and forming of bonds, rearrangement of atoms in molecules, and loss or gain of electrons.

At equilibrium a reversible reaction may appear static, but at the microscopic level it never stops: molecules are continuously changing. Reactant molecules convert into products, while at the same time product molecules react to go back to reactants. This is why equilibrium is called dynamic.

Consider steam and carbon monoxide: H₂O_(g) + CO_(g) ⇌ H_2(g) + CO_2(g). At the start, with the maximum number of collisions per second between H₂O and CO, the forward reaction has its maximum rate and gradually slows down as H₂O and CO are used up. As H₂ and CO₂ accumulate, the reverse reaction starts slowly, then increases, and finally becomes constant.

Eventually both reactions proceed at the same rate. The reaction is then in chemical equilibrium and the concentrations of reactants and products become constant. Unless the system is somehow disturbed, no further change in concentrations will occur.

State of chemical equilibrium: the state of a reversible reaction in which the composition of the reaction mixture does not change and the forward and reverse rates are equal.

The book's graphs (Fig. 8.1, 8.2) show concentration vs time. After equilibrium is established, concentrations become constant — but they need not be equal: (a) [Reactant] = [Product], (b) [Reactant] < [Product], or (c) [Reactant] > [Product] are all possible. What is constant is the value of each concentration, not their equality with each other.

📌 Note (book's Quick Check 8.1): in the reaction H₂O_(g) + CO_(g) ⇌ H_2(g) + CO_2(g), if the concentration of the products is higher at equilibrium, the reactant curves fall lower and the product curves level off higher on a concentration–time graph; on a rate–time graph the forward rate falls from its maximum and the reverse rate rises from zero until the two meet and run flat together.

8.3 Relation between macroscopic and microscopic events. Macroscopic events are the result of multiple simultaneous microscopic events. Understanding the microscopic events helps us explain and predict the observed macroscopic changes in an equilibrium system. When a change in equilibrium occurs, the microscopic events that explain it are the collisions and formation of new bonds between particles or molecules. These collisions change the rates of forward and reverse reactions, which are affected by activation energy and external factors. These microscopic events collectively appear as macroscopic changes observable by the naked eye.

8.4 Dynamic equilibrium between two physical states. Dynamic equilibrium is a state in a reversible process where the rate of change in one direction is equal to the rate of change in the opposite direction. Although continuous changes take place, no net change is observed in the system as a whole. When a reversible phase change takes place, there is also a dynamic equilibrium between the two physical states — for example, at 0 °C ice and water coexist in equilibrium: water changes into ice whereas ice converts back into water at the same rate.

Relationship between dynamic equilibrium and vapour pressure. When H₂O_(l) is placed in a closed container, some liquid molecules near the surface have enough energy to escape into the gaseous phase — this is evaporation. As more molecules escape, the pressure exerted by the gas molecules on the walls increases. At the same time, vapour molecules can collide with the liquid surface and be captured again — this is condensation: Liquid ⇌ Gas.

As evaporation and condensation continue, their rates eventually become equal and dynamic equilibrium is achieved: the number of liquid molecules going into the gas phase equals the number of gas molecules returning to the liquid phase. At dynamic equilibrium the vapour pressure remains constant at a given temperature, as long as the system is undisturbed. As temperature increases, the average kinetic energy of the liquid molecules increases, the rate of evaporation increases, and so the vapour pressure increases; the rate of condensation also increases at the higher temperature, and finally the system reaches a new dynamic equilibrium with a higher equilibrium vapour pressure. In short, dynamic equilibrium in terms of vapour pressure describes the state at which the rates of evaporation and condensation are equal.

📌 Note (book's Quick Check 8.2): at 100 °C water and its vapour are in dynamic equilibrium (the boiling point, where vapour pressure equals atmospheric pressure), and at 0 °C ice and water are likewise in dynamic equilibrium — in each case the two opposing phase changes proceed at the same rate.

8.5 Conditions for equilibrium. Equilibrium should be studied when the following conditions are fulfilled:

1. The state of equilibrium only applies to reversible reactions.
2. Equilibrium can be established only if the reaction vessel is closed. No part of the reactants or products is allowed to escape out. The book's Fig. 8.3 shows CaCO_3(s) ⇌ CaO_(s) + CO_2(g): equilibrium is established in the closed system, but cannot be established in the open system because CO₂ escapes.

8.6 Characteristics of chemical equilibrium.

1. At the stage of chemical equilibrium, the concentrations of reactants and products remain constant.
2. The state of equilibrium in a reversible reaction can be approached from either side — whether we start with reactants or products.
3. A catalyst does not change the equilibrium position and the equilibrium constant of the reaction. It helps to attain the equilibrium earlier.
4. The value of the equilibrium constant does not depend upon the initial concentrations of reactants; rather it is constant and depends on temperature only.

📌 Keep in Mind (book's side note): if pure solids or pure liquids are involved in an equilibrium system, their concentrations are not included in the equilibrium constant expression. This is because the change in concentration of any pure solid or liquid has no effect on the equilibrium constant.

8.7 Types of equilibrium. With respect to the physical states of reactants and products, there are two types:

8.7.1 Homogeneous equilibrium — an equilibrium system in which all of the reactants and products are in the same phase. Examples: N_2(g) + 3H_2(g) ⇌ 2NH_3(g); 2SO_2(g) + O_2(g) ⇌ 2SO_3(g); CH₃COOH_(l) + C₂H₅OH_(l) ⇌ CH₃COOC₂H_5(l) + H₂O_(l).

8.7.2 Heterogeneous equilibrium — an equilibrium in which the reactants and products are in more than one phase. Examples: CaCO_3(s) ⇌ CaO_(s) + CO_2(g); C_(s) + H₂O_(g) ⇌ CO_(g) + H_2(g); 3Fe_(s) + 4H₂O_(g) ⇌ Fe₃O_4(s) + 4H_2(g).

📌 Note (book's Quick Check 8.3): the forward rate slows down as equilibrium is approached because the reactants are being used up, so fewer collisions per second occur between them; in a capped soda-water bottle, gaseous CO₂ is in equilibrium with aqueous CO₂ (HCO₃⁻ and H⁺): CO_2(g) + H₂O_(l) ⇌ HCO₃⁻_(aq) + H⁺_(aq). While the cap is on, the system is closed and equilibrium holds; when the cap is removed, CO₂ escapes, so the equilibrium shifts to the left (reverse) to replace it — this is why the drink goes flat and fizzes.

Position of equilibrium refers to the relative amounts of products and reactants present in an equilibrium mixture.

In 1864, the Norwegian chemists C. Guldberg (1836–1902) and P. Waage (1833–1900) carefully measured the compositions of many reaction systems at equilibrium. They discovered that for any reversible reaction, the ratio of the product of the equilibrium concentrations of the products (raised to their coefficients in the balanced chemical equation) to the product of the equilibrium concentrations of the reactants (raised to their coefficients) is always a constant under a given set of conditions. This is the Law of Mass Action.

Consider the general reaction: aA + bB ⇌ cC + dD, where A and B are reactants, C and D are products, and a, b, c, d are the stoichiometric coefficients. Then:

K_c = [C]^c[D]^d / [A]^a[B]^b

The rate of the chemical reaction is directly proportional to the active masses of the reactants, raised to the coefficients present in the balanced equation. By active mass, we mean the molar concentration, expressed in moles dm⁻³ and shown by square brackets [ ].

Derivation. For A + B ⇌ C + D: Rate of forward reaction ∝ [A][B] = k_f[A][B], where k_f is the forward rate constant. Rate of reverse reaction ∝ [C][D] = k_r[C][D], where k_r is the reverse rate constant. At equilibrium the forward and reverse rates are equal: k_f[A][B] = k_r[C][D], so k_f/k_r = [C][D]/[A][B]. The left side is the ratio of two constants, so it gives another constant called the equilibrium constant (K_c): K_c = k_f/k_r = [C][D]/[A][B]. This expression is known as the equilibrium constant expression. For aA + bB ⇌ cC + dD it becomes K_c = [C]^c[D]^d/[A]^a[B]^b.

8.8.1 Equilibrium constant expressions of some important reactions.

• (i) N_2(g) + 3H_2(g) ⇌ 2NH_3(g) — K_c = [NH₃]² / ([N₂][H₂]³)
• (ii) 2N₂O_5(g) ⇌ 4NO_2(g) + O_2(g) — K_c = [NO₂]⁴[O₂] / [N₂O₅]²

Sample Problem 8.1 (book). Ethanol reacts with ethanoic acid to form ethyl ethanoate and water: CH₃COOH_(l) + C₂H₅OH_(l) ⇌ CH₃COOC₂H_5(l) + H₂O_(l). 500 cm³ of the reaction mixture at equilibrium contained 0.235 mol of ethanoic acid and 0.0350 mol of ethanol together with 0.182 mol of ethyl ethanoate and 0.182 mol of water. Step 1: concentrations = moles / 0.5 dm³: 0.470, 0.070, 0.364, 0.364 mol dm⁻³. Step 2: K_c = [CH₃COOC₂H₅][H₂O] / ([CH₃COOH][C₂H₅OH]). Step 3: K_c = (0.364 × 0.364) / (0.470 × 0.070) = 4.03 (to 3 significant figures). Step 4: units — mol dm⁻³ cancel out, so K_c has no units. Therefore, K_c = 4.03.

Sample Problem 8.2 (book). Propanone reacts with hydrogen cyanide: CH₃COCH_3(l) + HCN_(g) ⇌ CH₃C(OH)(CN)CH_3(l). A mixture of 0.0500 mol dm⁻³ propanone and 0.0500 mol dm⁻³ hydrogen cyanide is left to reach equilibrium at room temperature; at equilibrium the concentration of the product is 0.0233 mol dm⁻³. For every mole of product formed, 1 mole of each reactant is consumed, so [CH₃COCH₃] = [HCN] = 0.0500 − 0.0233 = 0.0267 mol dm⁻³. K_c = 0.0233 / (0.0267 × 0.0267) = 0.0233 / 0.000712 = 32.7 dm³ mol⁻¹.

8.8.2 Units of equilibrium constants. When the number of moles of reactants and products in a balanced chemical equation are equal, then units are cancelled, and the equilibrium constant does not have any unit — as in the ester formation reaction above. When the number of moles are unequal, the value of the equilibrium constant depends upon the units used. For ammonia synthesis, N_2(g) + 3H_2(g) ⇌ 2NH_3(g): K_c = [NH₃]² / ([N₂][H₂]³) = [mol dm⁻³]² / ([mol dm⁻³][mol dm⁻³]³) = 1 / [mol dm⁻³]² = dm⁶ mol⁻².

📌 Note (book's Quick Check 8.4, worked): (a) K_c expressions — (i) Sn²⁺_(aq) + 2Fe³⁺_(aq) ⇌ Sn⁴⁺_(aq) + 2Fe²⁺_(aq): K_c = [Sn⁴⁺][Fe²⁺]²/([Sn²⁺][Fe³⁺]²); (ii) 4NH_3(g) + 5O_2(g) ⇌ 4NO_(g) + 6H₂O_(g): K_c = [NO]⁴[H₂O]⁶/([NH₃]⁴[O₂]⁵); (iii) PCl_5(g) ⇌ PCl_3(g) + Cl_2(g): K_c = [PCl₃][Cl₂]/[PCl₅]. (b) For H_2(g) + CO_2(g) ⇌ H₂O_(g) + CO_(g) with initial [H₂] = [CO₂] = 10.00 mol dm⁻³ and equilibrium [CO] = 9.47 mol dm⁻³: by stoichiometry [H₂O] = 9.47, [H₂] = [CO₂] = 10.00 − 9.47 = 0.53 mol dm⁻³, so K_c = (9.47 × 9.47)/(0.53 × 0.53) ≈ 319 (no units, equal moles both sides).

There are four different types of quantities which may be used to calculate the equilibrium constants of a reversible reaction. For the general reaction aA + bB ⇌ cC + dD:

• (1) Concentrations in mol dm⁻³: K_c = [C]^c[D]^d / [A]^a[B]^b
• (2) Partial pressures (p) for gaseous reactants and products: K_p = p_C^c·p_D^d / (p_A^a·p_B^b)
• (3) Number of moles (n): K_n = n_C^c·n_D^d / (n_A^a·n_B^b)
• (4) Mole fractions (X): K_x = X_C^c·X_D^d / (X_A^a·X_B^b)

The relationships between these equilibrium constants are:

K_p = K_c(RT)^Δn    K_p = K_x(P)^Δn    K_p = K_n(N)^Δn

where R = general gas constant, T = absolute temperature of the system, P = pressure of the system, N = total number of moles of reactants and products, and Δn = difference of number of moles of products and reactants in the balanced chemical equation (products − reactants). It depends upon the value of Δn whether one equilibrium constant is bigger or smaller than the other. If the number of moles of reactants and products in a balanced chemical equation are equal, all the constants have equal values: Δn = 0, then K_p = K_c = K_x = K_n. Hence, whichever concentration units are used, the equilibrium concentrations are the same — the constants are numerically equal when Δn = 0.

Sample Problem 8.3 (book). N_2(g) and H_2(g) combine to form NH_3(g). The value of K_c at 500 °C is 6.0 × 10⁻². Calculate K_p. Solution: N₂ + 3H₂ ⇌ 2NH₃; K_c = 6.0 × 10⁻²; T = 500 + 273 = 773 K; R = 0.0821 dm³ atm K⁻¹ mol⁻¹; K_p = K_c(RT)^Δn; Δn = 2 − 4 = −2. K_p = 6.0 × 10⁻² × (0.0821 × 773)⁻² = 6.0 × 10⁻² / (63.5)² = 0.06 / 4032.25 = 1.4 × 10⁻⁵. So the value of K_p is less than K_c.

Sample Problem 8.4 (book). In the reaction 2SO_2(g) + O_2(g) ⇌ 2SO_3(g), the equilibrium partial pressures at constant temperature are SO₂ = 1.0 × 10⁶ Pa, O₂ = 7.0 × 10⁶ Pa, SO₃ = 8.0 × 10⁶ Pa. Step 1: K_p = p²_SO3 / (p²_SO2 × p_O2). Step 2: K_p = (8.0 × 10⁶)² / ((1.0 × 10⁶)² × (7.0 × 10⁶)) = 9.1 × 10⁻⁶ Pa⁻¹.

📌 Note (book's Quick Check 8.5, worked): (a) Comparing K_c and K_p magnitudes — for ammonia synthesis Δn = −2, so K_p = K_c(RT)⁻², i.e. K_p is much smaller than K_c at ordinary temperatures; for PCl₅ dissociation, PCl₅ ⇌ PCl₃ + Cl₂, Δn = +1, so K_p = K_c(RT), i.e. K_p is larger than K_c. (b) For N₂ + 3H₂ ⇌ 2NH₃ at constant temperature with total pressure 2.0 × 10⁷ Pa, partial pressure of nitrogen 1.5 × 10⁷ Pa and of hydrogen 0.4 × 10⁷ Pa: partial pressure of ammonia = 2.0 × 10⁷ − 1.5 × 10⁷ − 0.4 × 10⁷ = 0.1 × 10⁷ Pa. K_p = (0.1 × 10⁷)² / ((1.5 × 10⁷)(0.4 × 10⁷)³) = 1.0 × 10¹² / 9.6 × 10²⁶ ≈ 1.0 × 10⁻¹⁵ Pa⁻².

8.10 Position of equilibrium and reaction conditions. The position of equilibrium refers to the relative amounts of products and reactants present in an equilibrium mixture. If a system in equilibrium is disturbed:

• If the concentration of products is increased relative to the reactants, we say that the position of equilibrium has shifted to the left.
• If the concentration of products is decreased relative to the reactants, we say that the position of equilibrium has shifted to the right.

8.11 Le Chatelier's principle can be stated as follows: "If a system in equilibrium is disturbed, it behaves in such a way as to nullify or counteract the effect of that disturbance." Le Chatelier's principle describes what happens to a system when something momentarily takes it away from equilibrium. Its applications are: (i) effect of change in concentration, (ii) effect of change in pressure, (iii) effect of change in temperature, (iv) effect of catalyst on equilibrium. (Henry-Louis Le Chatelier, 1850–1936, a French chemist, proposed the principle — a significant achievement in chemistry.)

8.12 The effect of change in concentrations. If the system is at the equilibrium position and a certain amount of one of the reactants or products is added or removed, the equilibrium position is disturbed. There are four possibilities: (i) adding reactant(s), (ii) removing reactant(s), (iii) adding product(s), (iv) removing product(s).

8.12.1 Hydrolysis of BiCl₃. Consider the reversible reaction in which BiCl₃ reacts with water to give a white insoluble compound, BiOCl: BiCl₃ + H₂O ⇌ BiOCl + 2HCl. The expression for this reaction can be written as K_c = [BiOCl][HCl]² / ([BiCl₃][H₂O]). When water is added to BiCl₃ solution, it becomes cloudy due to the formation of white precipitate of BiOCl. At equilibrium, a certain amount of BiOCl and HCl has been produced and a certain number of moles of BiCl₃ are left behind. If a few moles of BiCl₃ are added at equilibrium, the reaction is pushed to the forward direction.

The addition of BiCl₃ or H₂O disturbs the equilibrium position. To re-establish the equilibrium, more BiOCl and HCl are produced — in other words, the reaction is pushed in the forward direction. K_c remains constant, but the equilibrium position has changed.

The addition of BiOCl or HCl, or both, disturbs the equilibrium position. To re-establish the equilibrium, more BiCl₃ is produced and the reaction moves in the reverse direction. If a certain amount of BiCl₃ is removed at equilibrium stage, the reaction will move in the reverse direction to compensate that change. A new equilibrium position is established, K_c remaining constant. If a certain amount of BiOCl or HCl or both is removed at equilibrium stage, the reaction will move in the forward direction to compensate that change. A new equilibrium position is established; however, K_c remains constant.

The general rule behind all four cases: adding a substance shifts equilibrium away from that side; removing a substance shifts equilibrium towards that side — always to counteract the disturbance, while K_c itself does not change.

8.13 The effect of change in pressure or volume. The effect of change of volume or pressure is not applicable when (a) the system does not involve gaseous components, or (b) the number of moles of reactants and products are equal in a gaseous phase homogeneous equilibrium (n_reactants = n_products). This effect can only be discussed if the gaseous moles of reactants and products are unequal (n_reactants ≠ n_products). Consider the synthesis of ammonia by Haber's process: N_2(g) + 3H_2(g) ⇌ 2NH_3(g).

• (a) Increase in pressure or decrease in volume: for the forward reaction, the number of moles and hence volume decreases. At equilibrium stage the volume occupied by the mixture is less than the volume at the initial stage under the given conditions. So, if pressure is increased at equilibrium, more ammonia is produced — the reaction is shifted to the forward direction.
• (b) Decrease in pressure or increase in volume: if the pressure is decreased or volume is increased, the reaction will move to that side where the reaction occupies greater volume. Ammonia synthesis is shifted to the reverse direction.

📌 Note (book's Quick Check 8.6, worked): changing volume or pressure only changes the equilibrium position, not the equilibrium constant. (i) 2H_2(g) + O_2(g) ⇌ 2H₂O_(g), increasing P: moles 3 → 2, shift forward (fewer moles). (ii) 2CO_(g) + O_2(g) ⇌ 2CO_2(g), increasing V (pressure falls): moles 3 → 2, shift reverse (more moles). (iii) N₂O_4(g) ⇌ 2NO_2(g), increasing P: moles 1 → 2, shift reverse (fewer moles). (iv) 2SO_2(g) + O_2(g) ⇌ 2SO_3(g), increasing V: moles 3 → 2, shift reverse (more moles).

8.14 The effect of change in temperature. Le Chatelier's principle can be used to predict the direction of a reaction with a change in temperature. Temperature is the only factor that also affects the value of the equilibrium constant (K_c).

For an exothermic reaction, an increase in temperature (adding heat) favours the reverse reaction. A decrease in temperature (removing heat) favours the forward reaction. Example: 2SO_2(g) + O_2(g) ⇌ 2SO_3(g), ΔH° = −198 kJ. An increase in temperature shifts the reaction from right to left: the concentration of SO₃ decreases and the concentrations of SO₂ and O₂ increase, so the value of the equilibrium constant decreases. A decrease in temperature favours the forward reaction and the formation of the product, so the value of the equilibrium constant increases. For this reaction, K_c is 2.8 × 10² at 1000 K, whereas at 298 K the value of K_c is 1 × 10²⁶ — a dramatic fall with temperature, exactly as expected for an exothermic reaction.

For an endothermic reaction, an increase in temperature (adding heat) favours the forward reaction. A decrease in temperature (removing heat) favours the reverse reaction. Example: N₂O_4(g) ⇌ 2NO_2(g), ΔH° = +57.2 kJ. An increase in temperature shifts the reaction from left to right: the concentration of N₂O₄ decreases and NO₂ increases, so K_c increases. A decrease in temperature shifts it from right to left and favours the reactant. K_c for this reaction is 7.7 × 10⁻⁵ at 273 K, whereas it is 0.4 at 373 K.

A catalyst is that substance which increases the rate of a chemical reaction without itself being consumed in the reaction. The catalyst does not change the equilibrium position and the equilibrium constant of a chemical reaction. It speeds up the rate at which equilibrium is attained. It means that the yield of the chemical reaction remains the same.

A catalyst provides a new path of lower activation energy for a reaction. If we plot a graph between reaction coordinate on x-axis and potential energy of the chemical reaction on y-axis, then the maxima of the curve become lower in the presence of a catalyst, as shown in the book's Fig. 8.4. Because it lowers the activation energy of the forward and reverse reactions by the same amount, both rates increase equally, so equilibrium is reached sooner but at exactly the same position.

It is one of the basic needs of industry to have maximum yield of the product. We can apply the concept of equilibrium in order to have the maximum yield from the industrial synthesis of ammonia.

8.16.1 Synthesis of ammonia by Haber's process. Ammonia synthesis from nitrogen and hydrogen is a reversible process: N_2(g) + 3H_2(g) ⇌ 2NH_3(g), ΔH = −46.11 kJ mol⁻¹. The maximum yield of ammonia can be achieved by the following ways:

1. Decrease the concentration of ammonia by removing it from the reaction vessel from time to time. Equilibrium will shift to the forward direction in accordance with Le Chatelier's principle.
2. Since four moles of reactants combine to give two moles of products, the reaction happens with decreasing volume. High pressure will shift the equilibrium position to the right to give more and more ammonia.
3. This is an exothermic reaction. By decreasing temperature, it will shift to the forward direction according to Le Chatelier's principle.

The book's Table 8.1 and Fig. 8.5 show how prominent the effect of temperature and pressure on the yield of ammonia is: the most complete conversion is 98.3% at 473 K (200 °C) and 1000 atmospheric pressure. At 200 °C the yield is being favoured but the rate of reaction becomes very slow and the process becomes uneconomical. So the temperature is raised to a moderate level, i.e. 400 °C, and a catalyst is used to increase the rate. If we want to achieve the same rate without a catalyst, then we require much higher temperature, which lowers the yield.

Optimum conditions to get best yield of ammonia: (a) Pressure between 200–300 atmospheres; (b) Temperature around 673 K (400 °C); (c) Pieces of iron crystals present in a fused mixture of MgO, Al₂O₃ and SiO₂ as catalyst.

8.16.2 Preparation of sulphur trioxide. To manufacture H₂SO₄, sulphur trioxide gas is produced from SO₂ and O₂ in a reversible process: 2SO_2(g) + O_2(g) ⇌ 2SO_3(g), ΔH = −97.9 kJ mol⁻¹. According to Le Chatelier's principle, a high pressure and low temperature are the essential conditions to have better yield of SO₃. At low temperature, the equilibrium constant for the formation of SO₃ is large but equilibrium is achieved very slowly. As the temperature is raised, the rate increases but the yield of SO₃ drops off. The book's Table 8.2 shows, at 1 atmosphere: at 200 °C, K_c = 5500 and mole % of SO₃ = 98; at 300 °C, 690 and 91; at 400 °C, 160 and 75; at 500 °C, 55 and 61; at 600 °C, 25 and 46; at 700 °C, 13 and 31.

Optimum conditions. In order to have a best yield of SO₃ within a reasonable time, a mixture of SO_2(g) and O_2(g) (air) at one atmospheric pressure is passed over a solid catalyst, such as V₂O₅. But due to the exothermic reaction, the temperature of the gas increases to 600 °C. The equilibrium mixture is recycled at low temperature of 400–500 °C to increase the yield of SO_3(g). Recycling the unreacted gases in both processes is also how industry maximises yield and minimises waste: reactants that did not convert are sent back through the reactor instead of being discarded.

📌 Note (book's Quick Check 8.7): from the book's K_p–temperature tables — for the exothermic ammonia synthesis and SO₃ formation, the proportion of products falls as temperature increases (K_p falls), while for the endothermic N₂O₄ ⇌ 2NO₂ system the proportion of products rises as temperature increases (K_p rises). At 600 K the book's table gives K_p = 3.1 × 10⁻³ atm⁻¹ for ammonia synthesis, 1.5 × 10¹ atm for N₂O₄ dissociation, and 3.2 × 10¹ atm⁻¹ for SO₃ formation; converting with K_c = K_p/(RT)^Δn (Δn = −2, +1, −1 respectively, T = 600 K, R = 0.0821 dm³ atm K⁻¹ mol⁻¹, RT = 49.26) gives K_c ≈ 7.5 dm⁶ mol⁻² (ammonia), ≈ 0.30 mol dm⁻³ (N₂O₄), and ≈ 1577 dm³ mol⁻¹ (SO₃).