Question

State what happens to the concentration of the indicated substance when the total pressure on each of the following equilibria is increased (by compression): (a) \(\mathrm{NO}_{2}\) (g) in \(2 \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{~s}) \rightleftharpoons 2 \mathrm{PbO}\) (s) \(+4 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}\) (g) (b) \(\mathrm{NO}(\mathrm{g})\) in \(3 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons 2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{NO}(\mathrm{g})\) (c) \(\mathrm{HI}(\mathrm{g})\) in \(2 \mathrm{HCl}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{~s}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (d) \(\mathrm{SO}_{2}\) (g) in \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) (e) \(\mathrm{NO}_{2}\) (g) in \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\)

Short answer

The concentration of the indicator substance changes as follows: (a) NO2 concentration decreases, (b) NO concentration increases, (c) HI concentration decreases, (d) SO2 concentration decreases, and (e) NO2 concentration does not change.

Step-by-step solution

Analyzing reactions and Le Châtelier's principle

For each equilibrium reaction, the principle of Le Châtelier states that if the conditions of a system at equilibrium are changed, the system responds in such a way as to partially counteract the effect of the change. Increasing the total pressure of the system (by compression) will push the equilibrium towards the side with fewer gas particles.

Assessing the Effect on Reaction (a)

The equilibrium reaction for (a) can be rewritten to highlight the gas participants: 2 Pb(NO3)2(s) ⇌ 4 NO2(g) + O2(g). Initially, there are zero molecules of gas, and on the right side there are 5 (4 NO2 and 1 O2). Increasing pressure will shift the equilibrium to the left due to Le Châtelier's principle, thus decreasing the concentration of NO2.

Assessing the Effect on Reaction (b)

For reaction (b), there are 3 molecules of gas on the left (3 NO2) and only 1 on the right (NO). Increase in pressure shifts the equilibrium towards the right, increasing the concentration of NO.

Assessing the Effect on Reaction (c)

In reaction (c), we have 2 gas molecules on the left (2 HCl) and 3 on the right (2 HI and 1 Cl2). Increasing pressure will shift the equilibrium towards the left side, reducing the concentration of HI.

Assessing the Effect on Reaction (d)

Looking at reaction (d), there are 3 gas molecules on the left (2 SO2 and 1 O2) and 2 gas molecules on the right (2 SO3). Increased pressure will favor the formation of the product SO3, thus reducing the concentration of SO2.

Assessing the Effect on Reaction (e)

For reaction (e), the number of gas molecules is equal on both sides (2 NO and 1 O2 on the left, 2 NO2 on the right). An increase in pressure has no effect on the equilibrium position, so the concentration of NO2 does not change.

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is critical for deciphering the behavior of reactions under various conditions. At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, resulting in constant concentrations of reactants and products over time. However, this does not mean the quantities of reactants and products are equal; it simply indicates their ratio remains unchanged in a closed system.

It's essential to recognize that equilibrium is a dynamic process. Even though concentration levels don't appear to change, reactions continue to occur. Think of it as a bustling city – while the population size may stay steady, people are continually moving in and out, akin to reactants converting to products and vice versa.

Effect of Pressure on Equilibrium

The effect of pressure on equilibrium arises from Le Châtelier's Principle, which declares that if a system at equilibrium experiences a change in condition, the system shifts to counteract the adjustment. In the context of gas-phase reactions, increasing pressure typically drives the equilibrium towards the side with fewer gas molecules. This is because a higher pressure compresses the gases, and the system can reduce pressure by favoring the side with fewer gas volumes.

Imagine a crowded room where everyone wants more space (akin to gases with high pressure). If people (gas molecules) can leave the room (reactions shifting towards fewer gas molecules), the overall crowding diminishes (pressure decreases), thus achieving a new equilibrium.

Equilibrium Shift

An equilibrium shift helps us predict how a chemical system at equilibrium will respond to a change in conditions. When pressure increases, the equilibrium will shift towards the side with fewer moles of gas to reduce the additional pressure. Conversely, when pressure decreases, the system favors the side with more gas moles to increase pressure.

Think of this as a see-saw balanced with weights on both ends. Adding weight (increasing pressure) to one side causes a shift (equilibrium shift) to regain a new kind of balance. The number of moles of gas corresponds to the weights on our see-saw – the side with fewer moles (weights) gains favor when pressured.

Gas Phase Reactions

Gas phase reactions provide a unique perspective on equilibrium and Le Châtelier's Principle due to the compressibility of gases. These reactions involve molecules in the gaseous state and can be influenced significantly by changes in pressure. Because gases occupy space, an alteration in pressure – for example, within a container – can drive the equilibrium position to change as it seeks to minimize that pressure.

In the classroom, you might imagine gas phase reactions as different groups of students engaged in a debate. When the moderator (pressure) asks for a more quiet discussion (reduces volume), some groups (gas molecules) may need to adjust – some students may speak less or even leave the discussion (reaction shifts), balancing the overall volume of the conversation.