Problem 3 - A Complete Resource Book in Chemistry for JEE Main

Question

The value of \(K_{p}\) in the reaction \(\mathrm{MgCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) is (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{P}\left(\mathrm{CO}_{2}\right)\) (b) \(\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{P}\left(\mathrm{MgCO}_{3}\right)}{\mathrm{P}\left(\mathrm{CO}_{2}\right) \times \mathrm{P}(\mathrm{MgO})}\) (c) \(\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{P}\left(\mathrm{CO}_{2}\right) \times \mathrm{P}\left(\mathrm{CO}_{2}\right) \times \mathrm{P}(\mathrm{MgO})}{\mathrm{P}(\mathrm{MgCO})_{3}}\) (d) \(\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{P}\left(\mathrm{CO}_{2}\right) \times \mathrm{P}(\mathrm{Mg} \mathrm{O})}{\mathrm{P}(\mathrm{MgCO})_{3}}\)

Step-by-Step Solution

Answer

The equilibrium constant \(K_p\) for the reaction is \(K_p = P(\text{CO}_{2})\). Option (a) is correct.

Step 1: Understand the equilibrium expression

The reaction is given as \( ext{MgCO}_{3}( ext{s}) \rightleftharpoons \text{MgO}( ext{s}) + \text{CO}_{2}( ext{g})\). In equilibrium expressions involving gases, only the partial pressures of gaseous species are included. Therefore, only the gaseous carbon dioxide \(\text{CO}_{2}\) is considered when writing the \(K_p\) expression.

Step 2: Recognize solid states

In chemical equilibrium reactions, substances in pure solid state do not appear in the equilibrium expression. Therefore, \(\text{MgCO}_{3}(\text{s})\) and \(\text{MgO}(\text{s})\) are not included in the \(K_p\) expression.

Step 3: Writing the correct equilibrium expression

Since only \(\text{CO}_{2}(\text{g})\) is involved as a gas in the equilibrium expression, the equilibrium constant \(K_p\) is expressed by the partial pressure of \(\text{CO}_{2}\). Thus, \(K_p = P(\text{CO}_{2})\).

Step 4: Select the correct option

From the options given, \(K_p = P(\text{CO}_{2})\) is given in option (a). Therefore, the correct statement for the equilibrium expression is option (a).

Key Concepts

Equilibrium Constant

When you study chemical equilibria, the equilibrium constant is a fundamental idea. It tells you the ratio of the products to the reactants when a reaction is at a point where things have balanced out and no longer change overall. There are different ways to write this constant, depending on whether the substances are in gas form or solution form. In reactions involving gases, we use \(K_{p}\), which is the equilibrium constant expressed using partial pressures of the gaseous components.

• For reactions in equilibrium, only gases and aqueous solutions appear in the equilibrium constant expression. Solids and liquids are omitted because their concentrations remain constant.
• In the given exercise, the reaction is \(\mathrm{MgCO}_{3}(\text{s}) \rightleftharpoons \mathrm{MgO}(\text{s})+\mathrm{CO}_{2}(\text{g})\).
• Since \(\mathrm{CO}_{2}\) is the only gas, the equilibrium constant \(K_{p}\) is expressed solely by the partial pressure of \(\mathrm{CO}_{2}\).

Partial Pressure

Partial pressure is a unique concept you encounter with gases in equilibrium. It represents the pressure exerted by a particular gas in a mixture if it alone occupied the entire volume. In our equilibrium problem, the focus is on carbon dioxide, \(\mathrm{CO}_{2}\).

• Partial pressures add up to give the total pressure of the system.
• Using partial pressures in equilibrium calculations allows chemists to predict how a system will react to changes.

In this specific case, the equilibrium expression deduces that the equilibrium constant is directly related to the partial pressure of \(\mathrm{CO}_{2}\). The mathematical representation is \(K_{p} = P(\mathrm{CO}_{2})\), meaning the equilibrium ratio is directly tied to how much pressure \(\mathrm{CO}_{2}\) alone is exerting.

Gaseous Equilibrium

Chemical equilibrium in a gaseous system is about the balance between forward and reverse reactions equally occurring over time. Reaction mixtures contain both reactants and products that are in constant motion, colliding, and converting between forms at a consistent pace.

• The forward reaction for our exercise turns \(\mathrm{MgCO}_{3}\) into \(\mathrm{MgO}\) and \(\mathrm{CO}_{2}\).
• The process is considered balanced when the partial pressures of reactants and products are stable over time, showing no net change.

The concept of gaseous equilibrium is especially powerful in predicting outcomes when changes in conditions (like pressure or volume) occur. Le Chatelier's Principle guides us in predicting shifts in equilibrium that result from such changes.

Solid State in Equilibrium

At equilibrium, substances in the solid state have a simpler role compared to gases or aqueous solutions. Their effective concentrations don't change, which is why solids like \(\mathrm{MgCO}_{3}\) and \(\mathrm{MgO}\) do not appear in the equilibrium constant expressions.

• The concentration (or "active mass") of a pure solid is considered constant because any added mass doesn’t change its concentration in the mix.
• In the exercise, only the gaseous component, \(\mathrm{CO}_{2}\), affects the equilibrium equation.

This simplification is helpful in reducing the complexity of calculating equilibrium constant expressions. It shows that even though solids participate in the reaction, they do not affect the balance expressed in \(K_{p}\), especially in a gaseous equilibrium setting.