Question

What effect does increasing the volume of the system have on the equilibrium condition in each of the following reactions? (a) \(\mathrm{C}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})\) (b) \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CaCO}_{3}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

Short answer

Increasing the volume of the system will cause the equilibrium to shift to the right for the first and third reactions, and remain the same for the second reaction.

Step-by-step solution

Analyze the first reaction

The first reaction is \(\mathrm{C}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})\). On the left side of the equation, there is one mole of gas, and on the right, there are two moles. Therefore, if the volume is increased, the equilibrium will shift to the right.

Analyze the second reaction

The second reaction is \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CaCO}_{3}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\). On the left side of the equation, there is one mole of gas, while on the right, there is also one mole. Hence, if the volume is increased, there will be no shift in the equilibrium as both sides contain the same amount of gas.

Analyze the third reaction

The third reaction is \(4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\). On the left side of the equation, we have nine moles of gas, while on the right, there are ten moles. Thus, if the volume is increased, the equilibrium will shift to the right.

Key concepts

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental guideline in chemistry, which predicts how a system at equilibrium reacts when it is subjected to changes. Imagine it like a see-saw; if you push it on one side, it tilts and tries to reach a new balance point.

In the context of chemical reactions, this principle states that if a system at equilibrium is disturbed, the system will adjust itself to counteract the effect of the disturbance. For instance, if you change the concentration, temperature, or volume, the equilibrium will shift in a direction that helps to minimize the change.

When you increase the volume of the system:

• The pressure decreases,
• The system counteracts this by shifting the equilibrium towards the side with more moles of gas.

Understanding Le Chatelier's Principle helps in predicting the behavior of various chemical reactions under different conditions.

Reaction Stoichiometry

Reaction stoichiometry is all about the relationships between the quantities of reactants and products in a chemical reaction. It's like a recipe that tells you how much of each ingredient (in this case, moles of substances) is needed and what you will end up with.

In a balanced chemical equation, the stoichiometry provides you with the proportion of reactants to products.
This is crucial for predicting the outcome of changes at equilibrium. For example:

• In the reaction \(\mathrm{C}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})\)
• The stoichiometry shows 1 mole of gaseous reactant producing 2 moles of gaseous product, indicating a volume shift if equilibrium changes.

The stoichiometry helps us understand how these shifts occur because it shows us which direction the reaction will favor when disturbed, such as in an increased volume scenario.

Equilibrium Shift due to Volume Change

Equilibrium shift due to changes in volume is an application of both Le Chatelier's Principle and reaction stoichiometry. When the volume of a gaseous system is increased, the pressure decreases.

The system compensates by favoring the direction with more moles of gas, thus increasing pressure again. Here's how it works:

• Increases in volume cause the equilibrium to shift toward the side with more gas molecules.
• If volume decreases, the equilibrium will shift toward the side with fewer gas molecules.

Let's consider the reaction \\(4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).

When volume increases, this reaction shifts to the right, as it produces more gas molecules (10 vs. 9 moles), in an effort to counteract the reduction in pressure.

Understanding these shifts is vital for predicting how reactions will behave under conditions of varying volume, which is a common consideration in industrial and laboratory settings.