Module 5.2 Factors that Affect Equilibrium

How does the addition of a catalyst affect a reversible reaction?

• It increases the activation energy of the forward reaction only.

• It decreases the activation energy of the forward reaction only.

• It increases the activation energy of both the forward and reverse reactions.

• It decreases the activation energy of both the forward and reverse reactions.

Answer explanation

The correct answer is that the addition of a catalyst decreases the activation energy of both the forward and reverse reactions. A catalyst works by providing an alternative reaction pathway with a lower activation energy, and it affects both directions of a reversible reaction equally. It speeds up both the forward and reverse reactions without altering the position of equilibrium. The other options are incorrect because a catalyst does not only affect one direction, nor does it increase the activation energy in any case.

Methanol can be produced from the reaction of carbon monoxide and hydrogen, according to the following equation:

CO(g) + 2H₂(g) ↔ CH₃OH(g) Δ_rH° = −90 kJ mol⁻¹

Which set of conditions will produce the maximum yield of methanol?

• Low pressure and low temperature

• Low pressure and high temperature

• High pressure and low temperature

• High pressure and high temperature

Answer explanation

The correct answer is high pressure and low temperature. Since the reaction produces fewer gas molecules (3 moles of reactants to 1 mole of product), increasing the pressure favors the formation of methanol. Additionally, the reaction is exothermic (ΔrH° = −90 kJ mol−1), so lowering the temperature shifts the equilibrium toward the products to counteract the loss of heat. The other options are incorrect because low pressure or high temperature would shift the equilibrium toward the reactants, reducing the yield of methanol.

Compounds X, Y and Z are in equilibrium. The diagram shows the effects of temperature and pressure on the equilibrium yield of compound Z.

Which equation would be consistent with this data?

• X(g) + 3Y(g) ↔ 2Z(g) ΔH > 0

• X(g) + 3Y(g) ↔ 2Z(g) ΔH < 0

• 2X(g) ↔ 2Y(g) + Z(g) ΔH > 0

• 2X(g) ↔ 2Y(g) + Z(g) ΔH < 0

Nitrogen dioxide reacts to form dinitrogen tetroxide in a sealed flask according to the following equation.

2NO₂(g) ↔ N₂O₄ (g) ΔH = −57.2 kJ mol⁻¹

Which graph best represents the rates of both the forward and reverse reactions when an equilibrium system containing these gases is cooled at time t?

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Consider this system in a fixed volume at constant temperature.

PCl₅(s) ↔ PCl₃(l) + Cl₂(g)

This system is initially at equilibrium. A small amount of solid PCl₅ is added.

Which statement is correct?

• The amount of Cl₂ will increase.

• The amount of PCl₃ will decrease.

• The amount of Cl₂ will not change.

• The amount of PCl₅ will increase then decrease.

Answer explanation

The correct answer is that the amount of Cl2 will not change. Since PCl5 is a solid, adding more solid does not affect the equilibrium position, as solids do not appear in the expression for equilibrium constants. Therefore, the concentrations of PCl3 and Cl2, which are in the liquid and gas phases respectively, remain unchanged. The other options are incorrect because changes in the amount of solid PCl5 do not directly affect the amounts of the liquid or gaseous components once equilibrium is established.

Nitrogen dioxide can react with itself to produce dinitrogen tetroxide.

2NO₂(g) ↔ N₂O₄(g) Keq = 0.010

In an experiment, 100.0 cm³ of NO₂ is placed in a syringe. The plunger is then pushed in until the volume is 50.0 cm³, while maintaining a constant temperature. The system is allowed to return to equilibrium.

Which statement is true for the system at equilibrium?

• The value of Keq has increased.

• The ratio [NO₂] / [N₂O₄] has decreased.

• The concentration of N₂O₄ has decreased.

• The concentrations of NO₂ and N₂O₄ have doubled.

Answer explanation

The correct answer is that the ratio [NO2] / [N2O4] has decreased. Since the volume of the system has been halved, the concentrations of both NO2 and N2O4 will increase. Because Keq favors the formation of N2O4 (a low value of 0.010 indicates that more N2O4 forms), compressing the volume increases the formation of N2O4 as the system shifts to reduce the number of gas molecules. The other options are incorrect because Keq is constant at a given temperature, the concentration of N2O4 will increase rather than decrease, and the concentrations of the gases do not simply double due to the shift in equilibrium.

Carbon dioxide reacts with hydrogen gas to form carbon monoxide and water vapour in a sealed flask, according to the following equation.

CO₂(g) + H₂(g) ↔ CO(g) + H₂O(g)

A temperature change was imposed on the equilibrium system at time t and the rates of both the forward and reverse reactions were monitored.

Which of the following correctly identifies the nature of both temperature change at time t and the ΔH of the forward reaction?

• Decrease in temperature and positive ΔH of the forward reaction.

• Decrease in temperature and negative ΔH of the forward reaction.

• Increase in temperature and positive ΔH of the forward reaction.

• Increase in temperature and negative ΔH of the forward reaction.

Nitrogen dioxide (a brown gas) and dinitrogen tetroxide (a colourless gas) are both forms of oxides of nitrogen. They are in equilibrium according to the equation

2NO₂(g) ↔ N₂O₄(g).

An equilibrium mixture of the two gases at room temperature is light brown but at higher temperatures the colour becomes a much deeper brown.

What conclusion can be drawn from this observation?

• The reverse reaction in the equation is endothermic.

• The forward reaction in the equation is endothermic.

• The brown colour is due to the strong nitrogen–oxygen bonds in NO₂.

• The equilibrium concentration of N₂O₄ is not dependent on temperature.

Answer explanation

The correct answer is that the forward reaction in the equation is exothermic. As the temperature increases, the system becomes a deeper brown, indicating an increase in NO2 concentration, which has a brown color. According to Le Chatelier's Principle, increasing the temperature shifts the equilibrium to favor the endothermic direction. Since the reaction shifts toward the production of more NO2 at higher temperatures, the reverse reaction (formation of NO2) must be endothermic, meaning the forward reaction (formation of N2O4) is exothermic. The other options are incorrect because the color change is related to the equilibrium shift, not the bond strength in NO2, and the equilibrium concentration of N2O4 is temperature-dependent.

Three gases X, Y and Z were mixed in a closed container and allowed to reach equilibrium. A change was imposed at time T and the equilibrium was re-established. The concentration of each gas is plotted against time.

Which reaction does the graph represent?

• X(g) + Y(g) ↔ 2Z(g)

• 2X(g) ↔ Y(g) + Z(g)

• 2X(g) ↔ Y(g) + 3Z(g)

• X(g) ↔ Y(g) + Z(g)

The following equilibrium was established in a container.

2SO₂(g) + O₂(g) ↔ 2SO₃(g) ΔH = -198 kJ mol^–1

Which of the following would increase the yield of SO₃(g)?

• Increasing the volume

• Increasing the temperature

• Removing the product as it is formed

• Keeping temperature and volume constant