Question

Predict the direction in which each of the following equilibria will shift if the pressure on the system is decreased by expansion. (a) \(\mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g)\) (b) \(\mathrm{CI} \mathrm{F}_{5}(g) \rightleftharpoons \mathrm{Cl} \mathrm{F}_{3}(g)+\mathrm{F}_{2}(g)\) (c) \(\mathrm{HBr}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\)

Short answer

Question: Predict the direction of the following chemical equilibria when pressure on the system is decreased by expansion: a) Ni(s) + 4 CO(g) ⇌ Ni(CO)4(g) b) ClF5(g) ⇌ ClF3(g) + F2(g) c) HBr(g) ⇌ 1/2 H2(g) + 1/2 Br2(g) Answer: When pressure is decreased by expansion, the equilibrium for the given reactions will shift as follows: (a) to the left (b) to the right (c) no shift

Step-by-step solution

a) Ni(s) + 4 CO(g) ⇌ Ni(CO)4(g)

First, count the number of moles of gas on each side of the equation. On the left side of the equilibrium, there are 4 moles of CO gas. On the right side, there is 1 mole of Ni(CO)4 gas. Since there are more moles of gas on the left side, the equilibrium will shift to the left when pressure is decreased by expansion.

b) ClF5(g) ⇌ ClF3(g) + F2(g)

Now, count the number of moles of gas in this reaction. On the left side of the equilibrium, there is 1 mole of ClF5 gas. On the right side, there are 1 mole of ClF3 gas and 1 mole of F2 gas, for a total of 2 moles of gas. Since there are more moles of gas on the right side, the equilibrium will shift to the right when pressure is decreased by expansion.

c) HBr(g) ⇌ 1/2 H2(g) + 1/2 Br2(g)

Finally, count the number of moles of gas in this last reaction. On the left side of the equilibrium, there is 1 mole of HBr gas. On the right side, there are 1/2 mole of H2 gas and 1/2 mole of Br2 gas, for a total of 1 mole of gas. Since there is an equal number of moles of gas on both sides, the equilibrium is not influenced by the decrease in pressure, and there will be no shift in the equilibrium in this case. In summary, when pressure is decreased by expansion, the equilibrium for the given reactions will shift as follows: (a) to the left (b) to the right (c) no shift

Key concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry that describes a state in which the concentrations of reactants and products remain constant over time, because the forward and reverse reactions proceed at the same rate. At equilibrium, the system is balanced, even though reactions are still occurring in both directions.

When we write chemical equations using the double arrow (\(\rightleftharpoons\)), it symbolizes that the reaction can proceed in both the forward and reverse directions. Equilibrium does not imply that the reactants and products are in equal concentrations, but that their rates of formation are equal. This means no net change in the concentrations of the substances involved.

Factors such as temperature, concentration, and pressure can affect the position of equilibrium. According to Le Chatelier's Principle, if an external change is applied to a system at equilibrium, the system will adjust itself to counteract that change and re-establish equilibrium. Understanding how equilibrium responds to changes like pressure is crucial for predicting the behavior of chemical reactions.

Pressure Effects on Equilibrium

Pressure effects on equilibrium are explained by Le Chatelier's Principle, which states that a change in pressure can shift the position of equilibrium to favor the side with fewer gas molecules. This principle is particularly essential for reactions involving gases, because gases are compressible and occupy the entire volume of their container.


When the pressure of a system at equilibrium is decreased by expansion:

• If there are more moles of gas on one side of the reaction, the equilibrium will shift towards the side with more moles of gas to counteract the decrease in pressure.
• For example, in the reaction \(\mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g)\), there are more gas moles on the left, so the equilibrium will shift left.
• If the moles of gas are equal on both sides, such as in \(\mathrm{HBr}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\), there will be no shift upon pressure decrease.

These shifts are due to the system seeking to increase pressure by adjusting to a state with larger volume occupied by gases, illustrating the dynamic nature of chemical equilibria.

Gas Moles in Reactions

In chemical reactions involving gases, it is important to understand the role of gas moles and how they influence equilibrium. The equilibrium concept revolves around balancing reactants and products, and gas moles play a crucial role here, especially when considering pressure changes.

The number of gas moles is decisive when predicting shifts in equilibrium due to pressure changes because:

• Each mole of gas occupies a similar volume at a given temperature and pressure.
• When counting moles, compare both sides of the equation. Reactions with unequal moles of gas between reactants and products will have different responses to pressure changes.

In a reaction like \(\mathrm{CI} \mathrm{F}_{5}(g) \rightleftharpoons \mathrm{Cl} \mathrm{F}_{3}(g)+ \mathrm{F}_{2}(g)\), increasing the pressure favors the side with fewer moles of gas, which is \(\mathrm{CI} \mathrm{F}_{5}(g)\) in this case, while decreasing pressure shifts it towards the side with more moles of gas.

Being aware of the moles of gas aids in predicting the behavior of reactions under various conditions, which is vital in industrial applications and laboratory settings.