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6th Year HL · Chemistry
Chemical transformations · Chemical equilibrium

Le Châtelier's principle and industrial processes

A fertiliser plant cannot simply wait for ammonia to form at room temperature: it would take forever. Nor can it raise the temperature without limit to speed things up, because that pushes the equilibrium the wrong way. Designing an industrial reactor is largely about predicting, with Le Châtelier's principle, which way an equilibrium will move before deciding on the operating conditions.

Le Châtelier's principle

A sealed flask containing an equilibrium mixture of $\mathrm{NO_2}$ (a reddish-brown gas) and $\mathrm{N_2O_4}$ (a colourless gas) goes almost clear in iced water and darkens in hot water. How does a closed system respond when we change its conditions?

Energy profile diagram of an endothermic reaction: the products are at a higher energy than the reactants, so heat must be absorbed. energy progress of reaction ΔH > 0 activation energy reactants products
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The following gas-phase chemical equilibrium is established in a closed vessel at $298\,\text{K}$: $$\mathrm{N_2O_4(g)} \rightleftharpoons 2\,\mathrm{NO_2(g)} \quad…

The following gas-phase chemical equilibrium is established in a closed vessel at $298\,\text{K}$:

$$\mathrm{N_2O_4(g)} \rightleftharpoons 2\,\mathrm{NO_2(g)} \quad \Delta H^\circ = +57.2\,\text{kJ}\times\text{mol}^{-1}$$

If the system is heated at constant volume until it reaches $350\,\text{K}$, answer the following questions with justification:

  1. a) In which direction does the chemical equilibrium shift to counteract the increase in temperature?
  2. b) What happens to the numerical value of the equilibrium constant $K_c$: does it increase, decrease or remain unchanged? Justify why only temperature is able to change its value.
Solve

Concentration, pressure and volume

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The process of calcination of limestone in a closed furnace at $1073\,\text{K}$ corresponds to the heterogeneous equilibrium: $$\mathrm{CaCO_3(s)} \rightleftharpoons…

The process of calcination of limestone in a closed furnace at $1073\,\text{K}$ corresponds to the heterogeneous equilibrium:

$$\mathrm{CaCO_3(s)} \rightleftharpoons \mathrm{CaO(s)} + \mathrm{CO_2(g)} \quad \Delta H^\circ = +178\,\text{kJ}\times\text{mol}^{-1}$$

Explain with reasoning how each of the following independent actions at $1073\,\text{K}$ affects the equilibrium:

  1. a) The introduction of an additional mass of $50\,\text{g}$ of pure $\mathrm{CaCO_3(s)}$ into the vessel, keeping the volume and temperature unchanged.
  2. b) An isothermal expansion of the vessel that doubles the total volume available to the gases. Assume that the two pure solid phases remain present.
Solve
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In a cylinder fitted with a movable piston at a constant temperature of $298\,\text{K}$, a gaseous mixture described by the reaction…

In a cylinder fitted with a movable piston at a constant temperature of $298\,\text{K}$, a gaseous mixture described by the reaction $\mathrm{N_2O_4(g)} \rightleftharpoons 2\,\mathrm{NO_2(g)}$ is at equilibrium. The initial volume of the system is $V_1 = 2.0\,\text{L}$ and it contains $0.20\,\text{mol}$ of $\mathrm{N_2O_4}$ and $0.40\,\text{mol}$ of $\mathrm{NO_2}$. By compressing the piston, the volume is suddenly reduced to $V_2 = 1.0\,\text{L}$ while the temperature is kept unchanged.

The following illustration shows schematically the initial state of the system at equilibrium and the situation immediately after the piston is compressed, before the equilibrium readjusts:

Equilibrium V₁ = 2.0 L · T = 298 K Compression rapid Instantaneous V₂ = 1.0 L · T = 298 K N₂O₄ (g) NO₂ (g)
  1. a) Calculate the numerical value of the constant $K_c$ in the initial state.
  2. b) Determine the value of the reaction quotient $Q_c$ immediately after compression to $V_2 = 1.0\,\text{L}$, compare it with $K_c$ and deduce in which direction the system will shift to reach the new equilibrium. For the calculations, use the ideal model and the teaching values provided.
Solve

Catalysts and inert gases: what does not shift the equilibrium

Two rigid vessels: after argon is added at constant volume, the number of reacting particles per unit volume is unchanged. V and T constant + Ar same reacting concentrations
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In the catalysis stage of the industrial contact process for obtaining sulfuric acid, the following reversible reaction takes place: $$2\,\mathrm{SO_2(g)} +…

In the catalysis stage of the industrial contact process for obtaining sulfuric acid, the following reversible reaction takes place:

$$2\,\mathrm{SO_2(g)} + \mathrm{O_2(g)} \rightleftharpoons 2\,\mathrm{SO_3(g)} \quad \Delta H^\circ = -198\,\text{kJ}\times\text{mol}^{-1}$$

Explain the effect of the following independent changes on the position of the equilibrium at constant temperature:

  1. a) The addition of divanadium pentoxide, $\mathrm{V_2O_5(s)}$, used as a solid catalyst in the reactor.
  2. b) The continuous and selective removal of $\mathrm{SO_3(g)}$ from the reactor as it forms.
Solve
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Phosphorus pentachloride is introduced into a closed reactor and heated to $523\,\text{K}$, establishing the following gaseous equilibrium: $$\mathrm{PCl_5(g)}…

Phosphorus pentachloride is introduced into a closed reactor and heated to $523\,\text{K}$, establishing the following gaseous equilibrium:

$$\mathrm{PCl_5(g)} \rightleftharpoons \mathrm{PCl_3(g)} + \mathrm{Cl_2(g)}$$

Analyse, with justification, the effect on the amount of $\mathrm{PCl_5(g)}$ of introducing an amount of inert gas (argon, $\mathrm{Ar}$) under the following different thermodynamic conditions:

  1. a) The noble gas is injected while the volume of the vessel and the temperature are kept strictly constant.
  2. b) The noble gas is injected while the vessel is allowed to expand so that the total pressure of the mixture and the temperature remain constant.
Solve

Industrial compromise: yield against rate

Graph showing equilibrium yield falling and reaction rate rising as temperature increases, with a compromise temperature marked in between. temperature equilibrium yield rate of reaction compromise T

Back to the start: the Haber process uses about $400\text{–}450\,^\circ\text{C}$ and $150\text{–}200$ atm with an iron catalyst. The pressure is high enough to push the equilibrium towards $\mathrm{NH_3}$, and the temperature is high enough for a fast reaction without losing too much yield.

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The synthesis of ammonia by the Haber-Bosch process is based on the exothermic reaction: $$\mathrm{N_2(g)} + 3\,\mathrm{H_2(g)} \rightleftharpoons 2\,\mathrm{NH_3(g)}…

The synthesis of ammonia by the Haber-Bosch process is based on the exothermic reaction:

$$\mathrm{N_2(g)} + 3\,\mathrm{H_2(g)} \rightleftharpoons 2\,\mathrm{NH_3(g)} \quad \Delta H^\circ = -92.2\,\text{kJ}\times\text{mol}^{-1}$$

Industrially, the process is operated neither at room temperature nor at unlimitedly high pressures, but under conditions of thermal and mechanical compromise ($T \approx 400\text{--}450\,^\circ\text{C}$ and $P \approx 150\text{--}200\,\text{atm}$) in the presence of promoted iron catalysts. Justify these industrial conditions by considering:

  1. a) The choice of high working pressures in light of the cost and safety considerations of the plant.
  2. b) The need to use an intermediate temperature ($450\,^\circ\text{C}$) in light of the analysis of equilibrium yield versus reaction rate.
Solve
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In an industrial Haber-Bosch plant, ammonia is synthesised at $700\,\text{K}$ and $200\,\text{atm}$ total according to the equilibrium: $$\mathrm{N_2(g)} +…

In an industrial Haber-Bosch plant, ammonia is synthesised at $700\,\text{K}$ and $200\,\text{atm}$ total according to the equilibrium:

$$\mathrm{N_2(g)} + 3\,\mathrm{H_2(g)} \rightleftharpoons 2\,\mathrm{NH_3(g)} \quad \Delta H^\circ = -92.2\,\text{kJ}\times\text{mol}^{-1}$$

The feed gas contains impurities of methane and argon from natural gas and air, which do not react. To avoid losses of reactants, the unreacted exit gases are continuously recycled to the reactor after condensing the $\mathrm{NH_3}$.

  1. a) Deduce how the progressive accumulation of these inert gases in the closed recycling loop affects the partial pressure of the reactants and the molar yield of formation of $\mathrm{NH_3}$, assuming that the total mechanical pressure ($200\,\text{atm}$) and the temperature are kept fixed. Explain why it is essential to install a continuous gas purge.
  2. b) The synthesis reaction releases a large amount of heat energy. If the reactor did not have an efficient cooling system and were allowed to operate adiabatically, explain how the temperature of the catalyst bed, the constant $K_p$ and the final conversion of reactants would vary in a coupled way.
Solve

Combined effects: when pressure and temperature compete

Diagram of two opposing effects on the yield of an exothermic reaction: higher temperature lowers it and higher pressure raises it. ↑ T yield ↓ ↑ P yield ↑ net effect: which one wins?
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In the gas-phase synthesis of methanol according to the equilibrium: $$\mathrm{CO(g)} + 2\,\mathrm{H_2(g)} \rightleftharpoons \mathrm{CH_3OH(g)} \quad \Delta H^\circ =…

In the gas-phase synthesis of methanol according to the equilibrium:

$$\mathrm{CO(g)} + 2\,\mathrm{H_2(g)} \rightleftharpoons \mathrm{CH_3OH(g)} \quad \Delta H^\circ = -90.7\,\text{kJ}\times\text{mol}^{-1}$$

a chemical engineering team monitors the methanol yield in a pilot reactor. In a first experiment at $500\,\text{K}$ and $50\,\text{atm}$, the mole fraction of $\mathrm{CH_3OH}$ at equilibrium is $x_1 = 0.22$.

Subsequently, the state of the system is modified simultaneously, operating at a new temperature $T_2 = 560\,\text{K}$ and a new total pressure $P_2 = 120\,\text{atm}$.

  1. a) Analyse the opposing effects of the change in temperature and in pressure on the position of the equilibrium, stating in which direction each of these changes acts separately.
  2. b) From the experimental value obtained under the second set of conditions, $x_2 = 0.18$, deduce which of the two effects has prevailed in the overall response of the system and justify formally whether the equilibrium constant $K_p$ at $560\,\text{K}$ is greater than, less than or equal to that at $500\,\text{K}$.
Solve
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The water-gas shift reaction is key to regulating the molar ratio in synthesis gas: $$\mathrm{CO(g)} + \mathrm{H_2O(g)} \rightleftharpoons \mathrm{CO_2(g)} +…

The water-gas shift reaction is key to regulating the molar ratio in synthesis gas:

$$\mathrm{CO(g)} + \mathrm{H_2O(g)} \rightleftharpoons \mathrm{CO_2(g)} + \mathrm{H_2(g)} \quad \Delta H^\circ = -41.2\,\text{kJ}\times\text{mol}^{-1}$$

At $1000\,\text{K}$, the equilibrium constant is $K_c = 1.00$.

  1. a) Starting from an equimolar mixture of $1.00\,\text{mol}$ of $\mathrm{CO}$ and $1.00\,\text{mol}$ of $\mathrm{H_2O}$ in a reactor of $10.0\,\text{L}$ at $1000\,\text{K}$, calculate the degree of conversion of $\mathrm{CO}$ at equilibrium, $\alpha_1$.
  2. b) To increase the conversion of $\mathrm{CO}$ without changing the temperature, it is decided to feed the reactor with an excess of steam, using $3.00\,\text{mol}$ of $\mathrm{H_2O}$ for every $1.00\,\text{mol}$ of $\mathrm{CO}$. Determine quantitatively the new degree of conversion $\alpha_2$, confirm the direction of the change using Le Chatelier's principle and explain why a change in volume at constant temperature would not alter either of the two results. For the calculations, use the ideal model and the teaching values provided.
Solve

Summary

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