Problem 191 - A Complete Resource Book in Chemistry for JEE Main

Question

What is the equilibrium expression for the reaction [2004] \(\mathrm{P}_{4}(\mathrm{~s})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{P}_{4} \mathrm{O}_{10}(\mathrm{~s}) ?\) (a) \(\mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{P}_{4} \mathrm{O}_{10}\right]}{5\left[\mathrm{P}_{4}\right]\left[\mathrm{O}_{2}\right]^{5}}\) (b) \(\mathrm{K}_{c}=\frac{1}{\left[\mathrm{O}_{2}\right]^{5}}\) (c) \(\mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{P}_{4} \mathrm{O}_{10}\right]}{\left[\mathrm{P}_{4}\right]\left[\mathrm{O}_{2}\right]^{5}}\) (d) \(\mathrm{K}_{\mathrm{c}}=\left[\mathrm{O}_{2}\right]^{5}\)

Step-by-Step Solution

Answer

The correct answer is (b) \( \mathrm{K}_{c} = \frac{1}{\left[\mathrm{O}_{2}\right]^{5}} \).

Step 1: Understand Reaction Components

The given reaction is \( \mathrm{P}_4(\mathrm{s}) + 5 \mathrm{O}_2(\mathrm{g}) \rightleftharpoons \mathrm{P}_4 \mathrm{O}_{10}(\mathrm{s}) \). It involves phosphorus as a solid, oxygen as a gas, and diphosphorus pentoxide as a solid.

Step 2: Identify Gaseous and Aqueous Species

In equilibrium expressions, we only include species in the gaseous or aqueous state because solids and pure liquids do not appear in the expression. For this reaction, the gaseous species is \( \mathrm{O}_2 \).

Step 3: Write Equilibrium Expression

The equilibrium constant expression for a reaction is based on the concentrations of the reactants and products in their gas or aqueous state. For the given reaction, since \( \mathrm{P}_4 \) and \( \mathrm{P}_4 \mathrm{O}_{10} \) are solids, they are not included in the expression. Thus, the equilibrium expression is based solely on \( \mathrm{O}_2 \): \[ \mathrm{K}_{c} = \frac{1}{[\mathrm{O}_2]^5} \]

Step 4: Match to Given Options

Compare the derived expression \( \mathrm{K}_{c} = \frac{1}{[\mathrm{O}_2]^5} \) to the given options. This expression matches option (b).

Key Concepts

Chemical Equilibrium

In chemical reactions, reaching a state of equilibrium is like finding balance on a see-saw. Imagine a condition where the reactants and products coexist peacefully. At this point, the rate of the forward reaction (reactants turning into products) is equal to the rate of the reverse reaction (products turning back into reactants).
Thus, there’s no net change in the concentration of either. This balance is what we call **Chemical Equilibrium**. It symbolizes stability in a chemical reaction, even though molecules continue to react and form products. The concentration of reactants and products remains constant because their reaction rates are equal. It's similar to maintaining a constant temperature in a room by balancing the speed of incoming and outgoing air flows.
For example, in the reaction of phosphorus and oxygen, once equilibrium is achieved, oxygen molecules continuously react, while the formed diphosphorus pentoxide can revert to its original reactants, all without changing the overall concentration of each substance.

Equilibrium Constant

The **Equilibrium Constant (K"]) is a critical number that helps us understand how far a reaction proceeds before reaching equilibrium. It’s essentially a snapshot of the concentration ratio of products to reactants at equilibrium. However, it only includes gases and solutes, not solids or pure liquids, because their concentrations don’t change significantly during the reaction.
This constant is denoted by \(K_c\) when dealing with concentrations in a solution. A large \(K_c\) value means the reaction heavily favors the formation of products, while a small \(K_c\) indicates reactants are favored. In our specific case of the phosphorus and oxygen reaction, the equilibrium constant takes the form \(K_c = \frac{1}{[O_2]^5}\) because only oxygen is in the gaseous state, thus involving in the equilibrium expression. Solid reactants and products like \(\mathrm{P}_4\) or \(\mathrm{P}_4\mathrm{O}_{10}\) are excluded from this expression.

Gaseous Species

In equilibrium expressions, gaseous species play a star role. They are the only substances considered in these expressions due to their easily measurable variations in concentration. In chemical reactions, molecules in the gas phase collide more often and react more, contributing significantly to the occurrence of equilibrium.
For instance, in the given exercise involving phosphorus and oxygen, the gaseous species is oxygen (\(\mathrm{O}_2\)). Its concentration is included in the equilibrium expression, represented as \(1 / [O_2]^5\). The gases, unlike solids and liquids, provide us with precise clues on shifting equilibrium positions and determining reaction extents.
Remember, only the behaviors and concentrations of gases change prominently, so tracking their changes effectively maps out the dynamics of the system at equilibrium.

Solid Reactants and Products

When considering equilibrium expressions, it is crucial to recognize that **solid reactants and products** are not included. This might seem odd at first, but it makes total sense.
Solids have a constant concentration because their density doesn't change with the reaction progression. This characteristic is why they are omitted from the equilibrium constant calculations. Imagine trying to measure water's saltiness without worrying about how full or empty the salt shaker is! That's how solids behave in equilibrium expressions, constant and unchanging.
In our reaction \(\mathrm{P}_4(\mathrm{s}) + 5 \mathrm{O}_2(\mathrm{g}) \rightleftharpoons \mathrm{P}_4 \mathrm{O}_{10}(\mathrm{s})\), both phosphorus and diphosphorus pentoxide are solids. Their roles in reactions are crucial, but their unchanging nature in concentration means they stay out of our \(K_c\) calculation, focusing us on beings in a state of flux, like gases.