Question

Ozone, \(\mathrm{O}_{\mathfrak{3}}\), decomposes to molecular oxygen in the stratosphere according to the reaction \(2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g)\). Would an increase in pressure favor the formation of ozone or of oxygen?

Short answer

An increase in pressure will favor the formation of ozone \((\mathrm{O}_3)\), as the reaction will shift to the left (towards the reactants) to minimize the effect of the pressure increase, according to Le Chatelier's Principle.

Step-by-step solution

Understand Le Chatelier's Principle

Le Chatelier's Principle states that if a system at equilibrium is subjected to change in concentration, pressure, or temperature, the equilibrium will shift to counteract the change and establish a new equilibrium.

Identify the change in the system

In this case, the change in the system is an increase in pressure. According to Le Chatelier's Principle, the reaction will shift to minimize the effect of the pressure increase.

Analyze the effect of pressure on the reaction

The given reaction is: \( 2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g) \) On the left-hand side of the reaction (reactants), there are 2 moles of gaseous ozone, and on the right-hand side of the reaction (products), there are 3 moles of gaseous oxygen.

Apply Le Chatelier's Principle to determine the favored side

Since an increase in pressure will be minimized by a decrease in the total number of gas molecules, the reaction will shift in the direction that has fewer moles of gas. In this case, the reaction will shift to the left (towards the reactants). Hence, an increase in pressure will favor the formation of ozone \( (\mathrm{O}_{3})\).

Key concepts

Chemical Equilibrium

Imagine a crowded dance floor where dancers move in and out with a certain rhythm. In a similar way, chemical equilibrium is like a dance between reactants and products. It occurs when the rate of the forward reaction, where reactants turn into products, is equal to the rate of the reverse reaction, where products revert back to reactants. It's important to note that when a chemical system is at equilibrium, it's dynamic - reactions don't stop happening; they occur at equal rates, maintaining a consistent concentration of substances involved.

Understanding equilibrium is crucial because it helps us predict how a reaction will respond to changes. This is where Le Chatelier's Principle comes into play, as it helps us understand that if a disturbance is introduced, like a change in pressure, the equilibrium will shift to counteract the change.

Stratospheric Ozone Decomposition

The decomposition of ozone in the stratosphere might seem like a distant concept, but it's surprisingly close to everyday life – it is important for understanding environmental science and protecting our planet. Ozone, or \(\mathrm{O}_{3}\), in the stratosphere absorbs harmful ultraviolet radiation from the sun, playing a critical role in shielding life on Earth.

The reaction given in the exercise, \(2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g)\), shows that two molecules of ozone decompose to form three molecules of molecular oxygen. While this reaction is natural and necessary for the formation of ozone, understanding it can also lend insight into environmental phenomena such as the depletion of the ozone layer, in which human activities can inadvertently accelerate this process.

Reaction Shift Due to Pressure Change

When you're in an elevator, and more people get in, you can feel the space getting tighter; similarly, increasing the pressure in a reaction's environment is like filling up that elevator. Le Chatelier's Principle suggests that the reaction will try to 'make more room.' How does it do this? By shifting toward the side with fewer moles of gas. In the stratospheric ozone decomposition reaction, this would mean favoring the side with the fewer gas molecules.

In our example, \(2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g)\), when pressure increases, the system will move towards the side with 2 moles of O3, the side that takes up less space. So, in response to a pressure increase, the system favors the production of ozone (\