Problem 9 - Chemistry in 30 days

Question

Which of the following gaseous reactions would be favoured by reducing the pressure (a) \(\mathrm{PCl}_{5} \Rightarrow \mathrm{PCl}_{3}+\mathrm{Cl}_{2}\) (b) \(2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\) (c) \(\mathrm{PCl}_{3}+\mathrm{Cl}_{2}=\mathrm{PCl}_{3}\) (d) \(\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI}\)

Step-by-Step Solution

Answer

Reaction (a) \\(\mathrm{PCl}_{5} \Rightarrow \mathrm{PCl}_{3}+\mathrm{Cl}_{2}\\) is favored by reducing pressure.

Step 1: Understanding Le Chatelier's Principle

Le Chatelier's principle states that if a stress is applied to a system at equilibrium, the system shifts in the direction that reduces the stress. When pressure is reduced, equilibrium will shift to favor the side with more moles of gas.

Step 2: Applying the Principle to Reaction (a)

In reaction (a) \(\mathrm{PCl}_{5} \Rightarrow \mathrm{PCl}_{3}+\mathrm{Cl}_{2}\), the reactant side has 1 mole of gas and the product side has 2 moles of gas. Reducing pressure will shift equilibrium to the right.

Step 3: Analyzing Reaction (b)

In reaction (b) \(2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\), the reactant side has 3 moles of gas and the product side has 2 moles of gas. Reducing pressure favors the side with more moles, hence no shift to right for favoring reaction (b).

Step 4: Evaluating Reaction (c)

In reaction (c) \(\mathrm{PCl}_{3}+\mathrm{Cl}_{2}=\mathrm{PCl}_{3}\), both sides have 1 mole of gas, so reducing pressure will not favor any particular direction due to equal moles of gas.

Step 5: Considering Reaction (d)

In reaction (d) \(\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI}\), both sides have 2 moles of gas, hence reducing pressure will not favor any shift in equilibrium.

Step 6: Conclusion

Out of all reactions mentioned, reaction (a) would be favored upon reducing pressure as it shifts towards the side with a greater number of gas moles (from 1 to 2).

Key Concepts

Gaseous Reactions

In chemistry, gaseous reactions involve chemical reactions where the reactants, products, or both are in the gaseous state. These reactions are often sensitive to changes in conditions such as temperature and pressure.
The molecules in gases are farther apart and move freely compared to solids and liquids. This makes gaseous reactions particularly dynamic.

• The position of equilibrium in gaseous reactions can be influenced easily by external factors.
• The type of gases involved and their respective molecular weights affect the reaction process.
• Gaseous-state reactions can also occur more rapidly compared to other states due to the increased energy and movement of molecules.

Understanding the behavior of gaseous reactions is essential when applying principles like Le Chatelier's Principle to predict changes in equilibrium.

Equilibrium Shifts

Equilibrium shifts occur when a reaction at dynamic equilibrium is subjected to a change in conditions, causing the reaction to adjust itself to maintain balance. This adjustment is guided by Le Chatelier's Principle.
When equilibrium is disturbed, the system will shift in the direction that counteracts the change. Some factors affecting equilibrium shifts include:

• Changes in concentration of reactants or products
• Alterations in temperature (e.g., heating generally shifts equilibrium in endothermic directions)
• Changes in pressure, particularly important in gaseous systems.

An equilibrium shift can affect the yield of products and is key in industrial chemical processes for optimizing output.

Pressure Changes

Pressure changes have a significant effect on the position of equilibrium in reactions involving gases. According to Le Chatelier's Principle, if the pressure on a system at equilibrium is decreased or increased, the system will move in a direction that opposes this change.
In gaseous reactions:

• If pressure is increased, the equilibrium shifts towards the side with fewer moles of gas, reducing pressure.
• If pressure is decreased, it shifts towards the side with more moles of gas, increasing pressure.

This concept is particularly useful in reactions where the number of gas moles is different on each side of the equation. For example, in the reaction \(\mathrm{PCl}_5 \Rightarrow \mathrm{PCl}_3+\mathrm{Cl}_2\), reducing pressure favors the production side with more moles of gas.

Moles of Gas

The number of moles of gas plays a crucial role in determining the direction of equilibrium shifts during pressure changes in gaseous reactions. The balance of gas moles on each side of a reaction equation provides insight into how the system will respond to modifications in pressure.
Key points to remember include:

• In any gaseous reaction, look at the mole count on each side—this will guide understanding of which direction is favored.
• More moles typically mean more volume and thus more influence under changed pressure conditions.
• Moles are not necessarily equal to molecules given the same volume since different gases may occupy different spaces (Ideal Gas Law).

Evaluating moles of gas helps predict how reactions will shift and can be a tool to optimize reaction conditions in practical applications.