Shifting Equilibria: Le Châtelier’s Principle - Summary

Key Concepts and Summary

Systems at equilibrium can be disturbed by changes to temperature, concentration, and, in some cases, volume and pressure. The system’s response to these disturbances is described by Le Châtelier’s principle: An equilibrium system subjected to a disturbance will shift in a way that counters the disturbance and re-establishes equilibrium. A catalyst will increase the rate of both the forward and reverse reactions of a reversible process, increasing the rate at which equilibrium is reached but not altering the equilibrium mixture’s composition (K does not change).

Practice Problems: Le Châtelier’s Principle

The following equation represents a reversible decomposition:
$$CaCO_3\;(s)\rightleftharpoons CaO\;(s)+CO_2\;(g)$$

Under what conditions will decomposition in a closed container proceed to completion so that no CaCO₃ remains?

Solution

The amount of CaCO₃ must be so small that $P_{CO_2}$ is less than $K_P$ when the CaCO₃ has completely decomposed. In other words, the starting amount of CaCO₃ cannot completely generate the full $P_{CO_2}$ required for equilibrium.

Explain how to recognize the conditions under which changes in volume (i.e. overall volume of the reaction container) will affect gas-phase systems at equilibrium.

Solution

A change in volume of the reaction container implies a change in pressure – all other variables being constant, a decrease in container volume will cause an increase in pressure in that container. More specifically, the decrease in volume will cause an increase in the partial pressures of all componentns in the container, as well as an increase in the overall pressure. In order for a change in volume of the reaction container to affect the equilibrium position, there needs to be an overall change in the total number of moles of gas from products to reactants. You can also check whether a change will occur by looking at the Q for the reaction – if multiplying all pressures by a constant (see these examples for more).

What property of a reaction can we use to predict the effect of a change in temperature on the value of an equilibrium constant?

Solution

The change in enthalpy may be used. If the reaction is exothermic, the heat produced can be thought of as a product. If the reaction is endothermic the heat added can be thought of as a reactant. Additional heat would shift an exothermic reaction back to the reactants but would shift an endothermic reaction to the products. Cooling an exothermic reaction causes the reaction to shift toward the product side; cooling an endothermic reaction would cause it to shift to the reactants’ side.

The following reaction occurs when a burner on a gas stove is lit: $$CH_4\;(g)+2O_2\;(g)\rightleftharpoons CO_2\;(g)+2H_2O\;(g)$$

Is an equilibrium among CH₄, O₂, CO₂, and H₂O established under these conditions? Explain your answer.

Solution

No, it is not at equilibrium. Because the system is not confined, products continuously escape from the region of the flame; reactants are also added continuously from the burner and surrounding atmosphere.

A necessary step in the manufacture of sulfuric acid is the formation of sulfur trioxide, SO₃, from sulfur dioxide, SO₂, and oxygen, O₂, shown here. At high temperatures, the rate of formation of SO₃ is higher, but the equilibrium amount (concentration or partial pressure) of SO₃ is lower than it would be at lower temperatures. $$SO_2\;(g)+O_2\;(g)\rightleftharpoons 2SO_3\;(g)$$

(a) Does the equilibrium constant for the reaction increase, decrease, or remain about the same as the temperature increases?

Solution

The question states that at high temperatures the equilibrium amount of SO₃ (the product) is lower. Therefore the value of K is decreasing as temperature is increased.

(b) Is the reaction endothermic or exothermic?

Solution

Since K decreases at high temperatures, the reaction is exothermic.

Suggest four ways in which the concentration of hydrazine, N₂H₄, could be decreased in an equilibrium described by the following equation: $$N_2\;(g)+2\, H_2\;(g)\rightleftharpoons N_2H_4\;(g)\qquad ΔH=95\;kJ$$

Solution

N₂H₄ is a product in this reaction; we should try and shift the equilibrium to the left. Changes that would cause this shift:

Remove some reactants (N₂ or H₂)
Decrease the temperature (the reaction is endothermic)
Increase the volume of the reaction conainer (1 mol of gas in products vs 3 in reactants)

Adding N₂H₄ will not meet the requirements. While it would shift the reaction to the left, the final concentration of N₂H₄ would still be higher than the initial value after re-establishing equilibrium.

Suggest four ways in which the concentration of PH₃ could be increased in an equilibrium described by the following equation: $$P_4\;(g)+6H_2\;(g)\rightleftharpoons 4PH_3\;(g)\qquad ΔH=110.5\;kJ$$

Solution

PH₃ is a product in this reaction; we should try and shift the equilibrium to the right. Changes that would cause this shift:

Add more reactants (P₄ or H₂)
Increase the temperature (the reaction is endothermic)
Decrease the volume of the reaction conainer (4 mol of gas in products vs 7 in reactants)

Since PH₃ is the only product in this reaction, and we are trying to increase its concentration, removing PH₃ will not work. While it would shift the reaction to the right, the final concentration of PH₃ would not be higher, since some was removed.

How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?

(a) $2NH_3\;(g)\rightleftharpoons N_2\;(g)+3H_2\;(g)\qquad ΔH=92\;kJ$

Solution

This reaction is endothermic. Increasing T will cause a shift right (towards the products).
Δ n = +2. There are more mol of gas on the product side; decreasing volume will cause a shift left (towards the reactants).

(b) $N_2\;(g)+O_2\;(g)\rightleftharpoons 2NO\;(g)\qquad ΔH=181\;kJ$

Solution

This reaction is endothermic. Increasing T will cause a shift right (towards the products).
Δ n = 0. Decreasing volume will have no effect on the equilibrium.

Solution

This reaction is exothermic. Increasing T will cause a shift left (towards the reactants).
Δ n = +1. There are more mol of gas on the product side; decreasing volume will cause a shift left (towards the reactants)

(d) $CaCO\;(s)+CO_2\;(g)\rightleftharpoons CaCO_3\;(s)\qquad ΔH=-176\;kJ$

Solution

This reaction is exothermic. Increasing T will cause a shift left (towards the reactants).
Δ n = -2. There are more mol of gas on the reactant side; decreasing volume will cause a shift right (towards the products).

How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?

(a)$2H_2O(g)\rightleftharpoons 2H_2(g)+O_2(g)\qquad ΔH=484\;kJ$

Solution

This reaction is endothermic. Increasing T will cause a shift right (towards the products).
Δ n = +1. There are more mol of gas on the product side; decreasing volume will cause a shift left (towards the reactants).

(b)$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)\qquad ΔH=-92.2\;kJ$

Solution

This reaction is exothermic. Increasing T will cause a shift left (towards the reactants).
Δ n = -1. There are more mol of gas on the reactant side; decreasing volume will cause a shift right (towards the products).

(c)$2Br(g)\rightleftharpoons Br_2(g)\qquad ΔH=-224\;kJ$

Solution

(d)$H_2(g)+I_2(s)\rightleftharpoons 2HI(g)\qquad ΔH=53\;kJ$

Solution

This reaction is endothermic. Increasing T will cause a shift right (towards the products).
Δ n = 0.Decreasing volume will have no effect on the equilibrium position.

Methanol can be prepared from carbon monoxide and hydrogen at high temperature and pressure in the presence of a suitable catalyst.

(a) Write the expression for the equilibrium constant ($K_c$) for the reversible reaction $$2H_2\;(g)+CO\;(g)\rightleftharpoons CH_3OH\;(g)\qquad ΔH=-90.2\;kJ/mol$$

Solution

$$K_c=\frac{[CH_3OH]}{[H_2]^2[CO]}$$

(b) What will happen to the concentrations of H₂, CO, and CH₃OH at equilibrium if more H₂ is added?

Solution

H₂ is a reactant; adding more will cause the reaction to shift right (towards the products). The concentration of CH₃OH will increase, and the concentration of CO will decrease as the system reaches a new equilibrium state. The concentration of H₂ will also decrease from its ‘new’ value, but will remain higher than it was before more was added.

Solution

CO is a product; removing it will cause the reaction to shift left (towards the reactants). The concentration of CH₃OH will decrease, and the concentration of H₂ will increase as the system reaches a new equilibrium state. The concentration of CO will also increase from its ‘new’ value, but will remain lower than it was before more was removed.

(d) What will happen to the concentrations of H₂, CO, and CH₃OH at equilibrium if CH₃OH is added?

Solution

CH₃OH is a product; adding more will cause the reaction to shift left (towards the reactants). The concentrations of H₂ and CO will increase as the system reaches a new equilibrium state. The concentration of CH₃OH will also decrease from its ‘new’ value, but will remain higher than it was before more was added.

(e) What will happen to the concentrations of H₂, CO, and CH₃OH at equilibrium if the temperature of the system is increased?

Solution

The reaction is exothermic (from the ΔH provided above). If temperature is increased, the value of K at the higher temperature will be smaller. At the new equilibrium position, the concentration of CH₃OH will be lower, and the concentrations of H₂ and CO will be higher than in the initial state.

(f) What will happen to the concentrations of H₂, CO, and CH₃OH at equilibrium if more catalyst is added?

Solution

A catalyst has no effect on the state at equilibrium – the concentrations will remain unchanged. The reaction will be able to reach equilibrium more quickly with added catalyst.

Nitrogen and oxygen react at high temperatures.

(a) Write the expression for the equilibrium constant (K_c) for the reversible reaction $$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)\qquad ΔH=181\;kJ$$

Solution

$$K_C = \frac{[NO]^2}{[N_2]^2[O_2]^2}$$

(b) What will happen to the concentrations of N₂, O₂, and NO at equilibrium if more O₂ is added?

Solution

O₂ is a reactant; adding more will cause the reaction to shift right (towards the products). The concentration of NO will increase, and the concentration of N₂ will decrease as the system reaches a new equilibrium state. The concentration of O₂ will also decrease from its ‘new’ value, but will remain higher than it was before more was added.

Solution

N₂ is a reactant; removing it will cause the reaction to shift left (towards the reactants). The concentration of NO will decrease, and the concentration of O₂ will increase as the system reaches a new equilibrium state. The concentration of N₂ will also increase from its ‘new’ value, but will remain lower than it was before some was removed.

(d) What will happen to the concentrations of N₂, O₂, and NO at equilibrium if NO is added?

Solution

NO is a product; adding more will cause the reaction to shift left (towards the reactants). The concentrations of N₂ and O₂ will increase as the system reaches a new equilibrium state. The concentration of NO will decrease from its ‘new’ value, but will remain higher than it was before more was added.

(e) What will happen to the concentrations of N₂, O₂, and NO at equilibrium if the volume of the reaction vessel is decreased?

Solution

There are 2 mol of gas in the products, and 2 mol of gas in the reactants (Δn – 0). Since there is no change in the number of moles of gas between the products and reactants, changing the volume of the reaction container will not cause any shift in the equilibrium position. However while the number of moles of gas remained the same, the total volume will have decreased — all concentrations will increase because of this change.

(f) What will happen to the concentrations of N₂, O₂, and NO at equilibrium if the temperature of the system is increased?

Solution

This reaction is endothermic – as temperature is increased, the value of K will increase. This means the reaction will shift right (towards the products) as the system reaches a new equilibrium state. Concentrations of N₂ and O₂ will decrease, and the concentration of NO will increase.

(g) What will happen to the concentrations of N₂, O₂, and NO at equilibrium if a catalyst is added?

Solution

A catalyst has no effect on the state at equilibrium – the concentrations will remain unchanged. The reaction will be able to reach equilibrium more quickly with added catalyst.

Water gas, a mixture of H₂ and CO, is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon.

(a) Write the expression for the equilibrium constant for the reversible reaction: $$C(s)+H_2O(g)\rightleftharpoons CO(g)+H_2(g)\qquad ΔH=131.30\;kJ$$

Solution

$$K_C = \frac{[CO][H_2]}{[H_2 O]}$$

(b) What will happen to the concentration of each reactant and product at equilibrium if more C is added?

Solution

C is a solid, and does not appear in the expressions for K or Q – it does not affect the equilibrium position. As long as sufficient C (s) was present in the reaction vessel to reach equilibrium initially, adding more will not cause any change to the equilibrium position nor the concentrations of any species in the vessel.

Solution

H₂O (g) is a reactant: removing some will cause the reaction to shift left (towards the reactants) in order to establish a new equilibrium. The concentrations of CO and H₂ will decrease. C (s) does not have a defined concentration since it is solid; but the amount of C (s) will be higher at the new equilibrium position. The concentration of H₂O will increase vs the ‘new’ value as the new equilibrium is reached, but it will remain lower than the initial value (before any was removed).

(d) What will happen to the concentration of each reactant and product at equilibrium if CO is added?

Solution

CO (g) is a product: adding some will cause the reaction to shift left (towards the reactants) in order to establish a new equilibrium. The concentration H₂ will decrease and the concentration of H₂O will increase. C (s) does not have a defined concentration since it is solid; but the amount of C (s) will be higher at the new equilibrium position. The concentration of CO will decrease vs the ‘new’ value as the new equilibrium is reached, but it will remain higher than the initial value (before any was added).

(e) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?

Solution

This reaction is endothermic. Increasing the temperature will cause an increase in the value of K, meaning the reaction will shift right (towards the products) to re-establish equilibrium. The concentrations of CO and H₂ will increase, and the concentration of H₂O will decrease. While the concentration of C (s) is not defined, the amount of C(s) will decrease.

Pure iron metal can be produced by the reduction of iron(III) oxide with hydrogen gas.

(a) Write the expression for the equilibrium constant (K_c) for the reversible reaction $$Fe_2O_3(s)+3H_2(g)\rightleftharpoons 2Fe(s)+3H_2O(g)\qquad ΔH=98.7\;kJ$$

Solution

$$K_C = \frac{[H_2 O]^3}{[H_2]^3}$$

(b) What will happen to the concentration of each reactant and product at equilibrium if more Fe is added?

Solution

Fe is a solid, and does not appear in the expressions for K or Q – it does not affect the equilibrium position. As long as sufficient Fe (s) was present in the reaction vessel to reach equilibrium initially, adding more will not cause any change to the equilibrium position nor the concentrations of any species in the vessel.

Solution

H₂O is a product in the reaction; removing some will cause the reaction to shift to the right (towards the products) in order to re-establish equilibrium. The concentration of H₂ will decrease. The concentration of H₂O will increase, but will remain smaller than the initial value (before any was removed). The solids in the reaction do not have a defined concentration, but the amount of Fe₂O₃ will decrease and the amount of Fe (s) will increase.

(d) What will happen to the concentration of each reactant and product at equilibrium if H₂ is added?

Solution

H₂ is a reactant in the reaction; adding some will cause the reaction to shift to the right (towards the products) in order to re-establish equilibrium. The concentration of H₂O will increase. The concentration of H₂ will decrease, but will remain larger than the initial value (before any was added). The solids in the reaction do not have a defined concentration, but the amount of Fe₂O₃ will decrease and the amount of Fe (s) will increase.

(e) What will happen to the concentration of each reactant and product at equilibrium if the volume of the reaction vessel is decreased?

Solution

There are 3 mol of gas in the products, and 3 mol of gas in the reactants (Δn = 0). Since there is no change in the number of moles of gas between the products and reactants, changing the volume of the reaction container will not cause any shift in the equilibrium position. However while the number of moles of gas remained the same, the total volume will have decreased — concentrations of H₂ adn H₂O will increase because of this change. (The solids will be unaffected.)

(f) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?

Solution

This reaction is endothermic. Increasing the temperature will cause an increase in the value of K, meaning the reaction will shift right (towards the products) to re-establish equilibrium. The concentration of H₂ will decrease and the concentration of H₂O will increase. While the concentration of Fe₂O₃ (s) and Fe (s) are not defined, the amount of Fe₂O₃ will decrease, and the amount of Fe will increase.

Ammonia is a weak base that reacts with water according to this equation:

$$NH_3(aq)+3H_2O(l)\rightleftharpoons NH_4^+(aq)+OH^-(aq)$$

Will any of the following increase the percent of ammonia that is converted to the ammonium ion (NH₄⁺) in water?

(a) Addition of NaOH

(b) Addition of HCl

Solution

Only (b) will cause an increase in the amount of NH₄⁺ in solution.
Action (c) will increase the total amount of NH₄⁺ in solution, but not specifically via conversion from NH₃ as requested.

Acetic acid is a weak acid that reacts with water according to this equation:

$$CH_3CO_2H(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+CH_3CO_2^-(aq)$$

Will any of the following increase the percent of acetic acid that reacts and produces acetate ions $\left( CH_3CO_2^- \right)$?

(a) Addition of HCl

(b) Addition of NaOH

Solution

Only action (b) will produce a shift to the right (causing acetic acid to convert to acetate ions), by reacting with the $H_3 O^+$ and effectively removing it from the solution. This occurs via the classic neutralization reaction:$$H_3O^+ + OH^- \rightarrow 2\, H_2 O$$

Suggest two ways in which the equilibrium concentration of Ag⁺ can be reduced in a solution of Na⁺, Cl⁻, Ag⁺, and $NO_3^-$, in contact with solid AgCl. $$Na^+(aq)+Cl^-(aq)+Ag^+(aq)+NO_3^-(aq)\rightleftharpoons AgCl(s)+Na^+(aq)+NO_3^-(aq)\qquad ΔH=-65.9\;kJ$$

Solution

Since only species that appear in the expession for K (and Q) will affect the equilibrium position, it will help to consider the net ionic equation and write the expression for K: $$K_C = \frac{1}{[Ag^+][Cl^-]}$$ Therefore if we want to reduce the $[Ag^+]$ in this reaction, we can:

Add more $Cl^-$ to the container
Decrease the temperature (the reaction is exothermic)

How can the pressure of water vapor be increased in the following equilibrium? $$H_2O(l)\rightleftharpoons H_2O(g)\qquad ΔH=41\;kJ$$

Solution

Since the only other species present is water as a pure liquid, adding or removing reagents will not help us achieve the goal of this question. We are left with two options:

Change the temperature: as an endothermic reaction, increasing the temperature will cause the reaction to shift right (towards teh products), producing more water vapour.
Change the pressure: Increasing the volume of the reaction vessel will cause a decrease in partial pressures, and the reaction will shift right (towards teh products), producing more water vapour.

A solution is saturated with silver sulfate and contains excess solid silver sulfate:

$$Ag_2SO_4(s)\rightleftharpoons 2Ag^+(aq)+SO_4^{2-}(aq)$$

A small amount of solid silver sulfate containing a radioactive isotope of silver is added to this solution. Within a few minutes, a portion of the solution phase is sampled and tests positive for radioactive Ag⁺ ions. Explain this observation.

Solution

Equilibrium is a dynamic process – although the concentrations/amounts of the species will remain (nearly) constant, both the forward and reverse reactions are constantly occurring. The isotope-labelled solid will be dissociating to form ions in solution (producing the observed radioactive $Ag^+$ ions) just as the ions initially present in solution will form new $Ag_2 SO_4 \text{(s)}$.