Question

What effect does an increase in pressure have on each of the following systems at equilibrium? The temperature is kept constant, and, in each case, the reactants are in a cylinder fitted with a movable piston. (a) \(\mathrm{A}(s) \rightleftarrows 2 \mathrm{~B}(s)\) (b) \(2 \mathrm{~A}(l) \rightleftarrows \mathrm{B}(l)\) (c) \(\mathrm{A}(s) \rightleftarrows \mathrm{B}(g)\) (d) \(\mathrm{A}(g) \rightleftarrows \mathrm{B}(g)\) (e) \(\mathrm{A}(g) \rightleftarrows 2 \mathrm{~B}(g)\)

Short answer

Increasing pressure will have no effect on (a) and (b), shift left for (c) and (e), and have no effect on (d).

Step-by-step solution

Analyze System (a)

The system involves only solids: \( \mathrm{A}(s) \rightleftarrows 2 \mathrm{~B}(s) \). Changes in pressure do not affect solids as they have negligible volume change. Thus, increasing pressure will have no effect on this equilibrium.

Analyze System (b)

In this system, \( 2 \mathrm{~A}(l) \rightleftarrows \mathrm{B}(l) \), only liquids are involved. Similar to solids, liquids are largely incompressible, and changes in pressure have minimal effect on their volume. Hence, the equilibrium will not be affected by changes in pressure.

Analyze System (c)

This system involves a solid and a gas: \( \mathrm{A}(s) \rightleftarrows \mathrm{B}(g) \). Increasing pressure on the system with a gas shifts the equilibrium towards the side with fewer moles of gas. Since the reactant \(\mathrm{A} (s)\) is a solid with no effect from pressure change, increasing pressure will shift the equilibrium towards the left, favoring the production of more solid \(\mathrm{A}\).

Analyze System (d)

Both reactants and products are gases: \( \mathrm{A}(g) \rightleftarrows \mathrm{B}(g) \). The reaction has equal moles of gases on both sides (1 mole of \(\mathrm{A}(g)\) and 1 mole of \(\mathrm{B}(g)\)). Increasing the pressure does not shift the equilibrium in any direction as there is no change in the number of moles of gas.

Analyze System (e)

For the reaction \( \mathrm{A}(g) \rightleftarrows 2 \mathrm{~B}(g) \), there is an increase in the number of gas moles from reactant to product. Increasing pressure will shift the equilibrium toward the left, favoring the formation of \(\mathrm{A}(g)\) since it has fewer moles of gas.

Key concepts

Pressure Effect

When we talk about the pressure effect on chemical equilibrium, it's essential to understand that pressure primarily affects gases. Solids and liquids are mostly incompressible, so changes in pressure do not noticeably impact their equilibrium.

In gas reactions, increasing pressure usually causes the equilibrium to shift towards the side with fewer moles of gas, minimizing the pressure change according to Le Chatelier's Principle.

For example:

• In the reaction \( \mathrm{A}(s) \rightleftarrows \mathrm{B}(g) \), increasing the pressure shifts the equilibrium towards solid \( \mathrm{A} \) because solids are unaffected by pressure changes.
• In the gaseous reaction \( \mathrm{A}(g) \rightleftarrows 2 \mathrm{~B}(g) \), increasing the pressure favors the formation of \( \mathrm{A}(g) \), as it results in fewer moles of gas.

Le Chatelier's Principle

Le Chatelier's Principle is fundamental in predicting how a system at equilibrium responds to changes in concentration, temperature, and pressure. It states that if a dynamic equilibrium is disturbed, the system will adjust to minimize and counter the disturbance.

For pressure changes, this principle is particularly useful:

• If pressure increases, the system shifts to the side with fewer moles of gas to reduce pressure.
• If pressure decreases, the equilibrium shifts towards the side with more moles of gas.

This principle does not apply to solids and liquids as they are mostly unaffected by pressure changes. Le Chatelier's Principle helps us understand real-world chemical reactions and predict how changes affect equilibrium.

Gas Laws

Gas laws, such as Boyle's Law, play a significant role in understanding the impact of pressure on gases. Boyle's Law states that the volume of a gas is inversely proportional to its pressure at constant temperature.

This relationship is crucial when evaluating equilibrium changes in chemical reactions involving gases. Lowering the volume of a gas under constant conditions increases the pressure, which can shift equilibrium according to Le Chatelier's Principle.

Consider reactions where the number of gaseous moles on each side is different:

• In \( \mathrm{A}(g) \rightleftarrows 2 \mathrm{~B}(g) \), increasing pressure by decreasing volume leads to a shift towards \( \mathrm{A}(g) \), with fewer moles.
• If no change in mole number is present, as in \( \mathrm{A}(g) \rightleftarrows \mathrm{B}(g) \), the equilibrium position remains unchanged.

Equilibrium Constant

The equilibrium constant \( K \) is a measure of the ratio of concentrations of products to reactants at equilibrium. For gaseous reactions, the equilibrium constant can also be expressed in terms of partial pressures, known as \( K_p \).

It's important to note that the equilibrium constant itself does not change with pressure. However, changes in pressure can shift equilibrium positions, thereby altering concentrations, but \( K \) remains constant unless the temperature changes.

For a balanced equation \( \mathrm{A}(g) \rightleftarrows 2 \mathrm{~B}(g) \):

• Pressuring this system will shift the equilibrium to favor \( \mathrm{A}(g) \), altering concentrations but not \( K_p \).
• In \( \mathrm{A}(g) \rightleftarrows \mathrm{B}(g) \), no pressure effect changes the ratio since the number of moles on both sides is equal.

Understanding the equilibrium constant allows us to calculate and predict concentration changes under varying conditions.