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

Question

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

Step-by-Step Solution

Answer

The correct equilibrium expression is option (b): \( K_c=\frac{1}{[\mathrm{O}_2]^5} \).

Step 1: Understand the Reaction

The chemical equation for the equilibrium state is \( \mathrm{P}_{4}(\mathrm{~s})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{P}_{4} \mathrm{O}_{10}(\mathrm{~s}) \). This equation tells us that one mole of solid phosphorus \( \mathrm{P}_{4} \) reacts with five moles of gaseous oxygen \( \mathrm{O}_{2} \) to produce one mole of solid tetraphosphorus decaoxide \( \mathrm{P}_{4} \mathrm{O}_{10} \).

Step 2: Identify Phases and Their Impact

The substances involved have different phases: \( \mathrm{P}_{4} \) and \( \mathrm{P}_{4} \mathrm{O}_{10} \) are in solid state, while \( \mathrm{O}_{2} \) is a gas. In equilibrium expressions, the concentration of pure solids and pure liquids are not included, as they are constant and do not affect the position of equilibrium.

Step 3: Write the Equilibrium Expression

In general, the equilibrium constant expression, \( K_c \), for a reaction \( aA + bB \rightleftharpoons cC + dD \) is given by\[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]For our reaction, only \( \mathrm{O}_{2} \), which is in the gas phase, appears in the expression. Therefore, the expression for the equilibrium constant \( K_c \) will be:\[ K_c = \frac{1}{[\mathrm{O}_2]^5} \]This is because \( \mathrm{O}_{2} \) is raised to the power of 5 due to its coefficient in the balanced equation.

Step 4: Match with Given Options

Now compare the derived expression with the provided options:(a) \( \mathrm{K}_{c}=\frac{\left[\mathrm{P}_{4} \mathrm{O}_{10}\right]}{5\left[\mathrm{P}_{4}\right]\left[\mathrm{O}_{2}\right]^{5}} \)(b) \( \mathrm{K}_{c}=\frac{1}{\left[\mathrm{O}_{2}\right]^{5}} \)(c) \( \mathrm{K}_{c}=\frac{\left[\mathrm{P}_{4} \mathrm{O}_{10}\right]}{\left[\mathrm{P}_{4}\right]\left[\mathrm{O}_{2}\right]^{5}} \)(d) \( \mathrm{K}_{c}=\left[\mathrm{O}_{2}\right]^{5} \).We can see that the correct expression matches with option b.

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry that describes the state of a reaction in which the forward and reverse reactions occur at the same rate. When a chemical reaction reaches equilibrium, the concentrations of the reactants and products remain constant over time, though they are not necessarily equal. This means that the system has reached a balance, resulting in no net change in the concentrations of the components.

If we consider the reaction from the exercise: \[ \mathrm{P}_4(\mathrm{s}) + 5 \mathrm{O}_2(\mathrm{g}) \rightleftharpoons \mathrm{P}_4 \mathrm{O}_{10}(\mathrm{s}) \]In the above reaction, chemical equilibrium is achieved when the rate of formation of \( \mathrm{P}_4 \mathrm{O}_{10} \) from \( \mathrm{P}_4 \) and \( \mathrm{O}_2 \) is equal to the rate at which \( \mathrm{P}_4 \) and \( \mathrm{O}_2 \) are reformed from \( \mathrm{P}_4 \mathrm{O}_{10} \). The reaction appears static, but on a molecular level, the reactants and products are continuously being formed and consumed.

This balance displays a dynamic feature, where molecules of reactants and products continue to exchange, but with no overall changing concentrations. It is crucial for students to understand that at equilibrium, reactions are not halted; they are dynamic and constantly in action.

Equilibrium Constant

The equilibrium constant, often represented as \( K_c \), is a numerical value that quantifies the position of equilibrium for a given chemical reaction at a certain temperature. It is formulated based on the concentrations of the reactants and products at equilibrium.For a general reaction: \[ aA + bB \rightleftharpoons cC + dD \]the expression for the equilibrium constant \( K_c \) is:\[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]This expression illustrates the ratio of the concentration of products to reactants, each raised to the power of their respective stoichiometric coefficients from the balanced chemical equation.

In the exercise's reaction:\[ \mathrm{P}_4(\mathrm{s}) + 5 \mathrm{O}_2(\mathrm{g}) \rightleftharpoons \mathrm{P}_4 \mathrm{O}_{10}(\mathrm{s}) \]We do not include solids like \( \mathrm{P}_4 \) and \( \mathrm{P}_4 \mathrm{O}_{10} \) in the \( K_c \) expression because their concentrations do not change. Instead, the equilibrium constant focuses on the gaseous component:\[ K_c = \frac{1}{[\mathrm{O}_2]^5} \]Here, \( [\mathrm{O}_2] \) is the concentration of oxygen gas at equilibrium, raised to the power indicated by its coefficient in the balanced equation. This highlights the dominance of gases in affecting the equilibrium state.

Phases of Matter

In the context of chemical reactions, the phases of matter (solid, liquid, gas) play a vital role in determining which substances are included in the equilibrium expression. This aspect was clearly illustrated in the original exercise.Matter can exist in different phases:

• Solids: Have a definite shape and volume, with particles closely packed in a rigid structure.
• Liquids: Have a definite volume but take the shape of their container, with particles more loosely packed than solids.
• Gases: Have no definite shape or volume, and their particles move freely and occupy the available space.

A key understanding in equilibrium calculations is that solids and liquids have constant concentrations under constant conditions of temperature and pressure. Therefore, they are not included in the equilibrium expression since their concentration does not change and thus does not affect the equilibrium state.For the exercise reaction:\[ \mathrm{P}_4(\mathrm{s}) + 5 \mathrm{O}_2(\mathrm{g}) \rightleftharpoons \mathrm{P}_4 \mathrm{O}_{10}(\mathrm{s}) \]Only the gaseous \( \mathrm{O}_2 \) is considered in writing the equilibrium expression. Understanding the phases helps to determine which substances impact the reaction's equilibrium and how they should be treated in calculations.