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) \(\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{P}\left(\mathrm{CO}_{2}\right) \times \mathrm{P}(\mathrm{MgO})}{\mathrm{P}(\mathrm{MgCO})_{3}}\)

Step-by-Step Solution

Answer

(a) \( \mathrm{K}_p = \mathrm{P}(\mathrm{CO}_{2}) \)

Step 1: Understand the Reaction

The chemical reaction given is: \( \mathrm{MgCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{MgO}(\mathrm{s}) + \mathrm{CO}_{2}(\mathrm{g}) \). It involves the decomposition of magnesium carbonate into magnesium oxide and carbon dioxide gas.

Step 2: Identify Components Involved in Kp

In the equilibrium expression for \( K_p \), only species in the gaseous state are included. Solids and liquids do not appear in the expression for the equilibrium constant in this context.

Step 3: Write the Expression for Kp

Since only \( \mathrm{CO}_{2}(\mathrm{g}) \) is in the gaseous state, \( K_p \) for this reaction will only involve the pressure of \( \mathrm{CO}_{2}(\mathrm{g}) \). Thus, \( K_p = \mathrm{P}(\mathrm{CO}_{2}) \).

Step 4: Match with Provided Options

From the options given, option (a) matches the expression for \( K_p \): \( \mathrm{K}_p = \mathrm{P}(\mathrm{CO}_{2}) \). Therefore, option (a) is the correct one.

Key Concepts

Partial Pressure

In chemical thermodynamics, the concept of partial pressure is crucial to understanding reactions involving gases, such as the decomposition of magnesium carbonate (\(\mathrm{MgCO}_3\)). Partial pressure refers to the pressure exerted by a single type of gas in a mixture of gases. Each gas in the mixture behaves as if it occupies the entire volume on its own.

In our exercise, the decomposition of \(\mathrm{MgCO}_3\) involves the formation of \(\mathrm{CO}_2\), which is a gas. The equilibrium constant \(K_p\) is based on the partial pressures of gases involved in the equilibrium state. Solids, like \(\mathrm{MgCO}_3\) and \(\mathrm{MgO}\), do not contribute to \(K_p\) because they do not exert pressure.

The partial pressure of \(\mathrm{CO}_2\) is critical in determining the equilibrium position in this reaction. At equilibrium, \(K_p\) is determined only by the pressure of \(\mathrm{CO}_2\), which is why the equilibrium constant is expressed as \(K_p = \mathrm{P}(\mathrm{CO}_2)\). This simplification emerges from the fact that the only reactive gaseous component under examination is \(\mathrm{CO}_2\). Thus, understanding partial pressures simplifies solving equilibrium problems such as this one.

Gaseous State

The gaseous state of matter plays a significant role in this reaction where \(\mathrm{MgCO}_3\) decomposes into \(\mathrm{MgO}\) and \(\mathrm{CO}_2\). Gases are distinct because they are highly compressible, expand to fill their containers, and their molecules have significant freedom of movement.

In the context of equilibrium expressions, gases are represented by their pressures rather than concentrations due to their variable nature. The ideal gas law connects pressure, volume, and temperature, making pressure a convenient and meaningful measure in equilibrium expressions. For the decomposition of \(\mathrm{MgCO}_3\), we focus on the gaseous product, \(\mathrm{CO}_2\), to formulate our equilibrium constant \(K_p\).

It's important to remember that only gases and aqueous solutions appear in expressions for equilibrium constants. Solids and liquids like \(\mathrm{MgCO}_3\) and \(\mathrm{MgO}\) remain constant in concentration and hence do not appear in \(K_p\). Thus, understanding the role of gases is crucial in predicting and calculating chemical equilibria involving gaseous products.

Chemical Decomposition

Chemical decomposition involves breaking down a compound into simpler substances. The reaction \(\mathrm{MgCO}_3(\mathrm{s}) \rightleftharpoons \mathrm{MgO}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g})\) is a classic example of thermal decomposition, where heat triggers the breakdown of a solid compound into a gas and another solid.

The decomposition reaction of \(\mathrm{MgCO}_3\) yields \(\mathrm{CO}_2\), which transitions from a bound state in a solid to a free-moving gaseous state. In equilibrium scenarios, we are particularly interested in how the decomposition affects the concentrations—or in the case of gases, the pressures—of the resulting components.

In terms of the equilibrium constant, solids like \(\mathrm{MgCO}_3\) and \(\mathrm{MgO}\) are not included in the \(K_p\) expression because their activity is essentially constant. Instead, the focus is on the change in pressure of \(\mathrm{CO}_2\) as it decomposes and reaches an equilibrium state. Therefore, appreciating the basics of chemical decomposition is essential for mastering subjects like equilibrium constants and partial pressures in gas reactions.