Problem 167 - A Complete Resource Book in Chemistry for JEE Main

Question

Match the following Column-I (a) \(\mathrm{N}_{2} \mathrm{O}_{4(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NO}_{2(8)}\) (b) \(\mathrm{PCl}_{\text {s(g) }} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}\) (c) \(\mathrm{NH}_{2} \mathrm{COONH}_{4(\mathrm{~s})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}+\mathrm{CO}_{2(\mathrm{~g})}\) (d) \(\mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(8)}\) Column-II (p) \(\mathrm{K}_{\mathrm{p}}=\frac{4 \mathrm{x}^{2} \mathrm{P}}{\mathrm{a}^{2}-\mathrm{x}^{2}}\) (q) \(\mathrm{K}_{\mathrm{p}}=\frac{4 \mathrm{P}^{3}}{27}\) (r) \(K_{p}=K_{c}\) (s) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{c^{\circ}} \mathrm{RT}\) (t) \(\Delta n_{g}>0\)

Step-by-Step Solution

Answer

(a) matches (t), (b) matches (t), (c) matches (t), (d) matches (r).

Step 1: Analyze Reaction (a)

For reaction (a) \( \mathrm{N}_{2} \mathrm{O}_{4(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NO}_{2(8)} \), we need to determine the change in moles of gases, \( \Delta n_{g} \). The initial moles of gases are 1 (since it starts with \( \mathrm{N}_{2} \mathrm{O}_{4} \)), and the final moles are 2 (from \( 2 \mathrm{NO}_{2} \)). Thus, \( \Delta n_{g} = 2 - 1 = 1 \). Since \( \Delta n_{g} > 0 \), this corresponds to Column-II element (t).

Step 2: Analyze Reaction (b)

For reaction (b) \( \mathrm{PCl}_{\text {s(g) }} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})} \), \( \Delta n_{g} = n_{products} - n_{reactants} = (1+1)-1 = 1 \), which is also \( \Delta n_{g} > 0 \). Hence, this reaction also matches Column-II element (t).

Step 3: Analyze Reaction (c)

For reaction (c) \( \mathrm{NH}_{2} \mathrm{COONH}_{4(\mathrm{~s})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}+\mathrm{CO}_{2(\mathrm{~g})} \), only gases are considered in \( \Delta n_{g} \). So, \( \Delta n_{g} = 3 - 0 = 3 \), which is clearly more than zero, \( \Delta n_{g} > 0 \). Thus, this matches with Column-II element (t).

Step 4: Analyze Reaction (d)

For reaction (d) \( \mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(8)} \), the initial and final moles of gases are both 2. Thus, \( \Delta n_{g} = 0 \). According to the thermodynamic relation, this means \( K_{p} = K_{c} \). Therefore, this corresponds with Column-II element (r).

Step 5: Match with provided Column-II

Now, based on the above analysis, we can match the reactions with Column-II options: (a) matches (t), (b) matches (t), (c) matches (t), (d) matches (r).

Key Concepts

Gaseous Equilibrium

Chemical equilibria in gases, or gaseous equilibrium, are fundamental concepts in chemistry. In a gaseous equilibrium, the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products. Gases participating in a chemical reaction are usually enclosed in a container, which helps maintain constant conditions such as temperature and pressure.
In practice, this means that despite continuous molecular activity, the macroscopic properties remain constant. This dynamic balance is crucial in chemical systems like ammonia synthesis or the dissociation of nitrogen dioxide. It ensures that despite random molecular collisions, an exact ratio of products and reactants is achieved and maintained over time.

• Temperature, pressure, and concentration affect the position of equilibrium according to Le Chatelier's principle.
• Changing conditions can favor the formation of either reactants or products.

The concept of gaseous equilibrium forms the backbone of many industrial chemical processes, where shifting the equilibrium position can maximize yield.

Equilibrium Constant (Kp, Kc)

In chemical reactions, the equilibrium constant (K) is a crucial value that expresses the ratio of the concentration of products to reactants at equilibrium. There are two main types of equilibrium constants used in gaseous systems: \(K_p\) and \(K_c\).
\(K_p\) is the equilibrium constant for reactions measured in terms of gas partial pressures, specific to gaseous equilibria, while \(K_c\) relates to concentrations. Calculating \(K_p\) involves placing the partial pressures of the products over those of the reactants, raised to the power of their stoichiometric coefficients.
The relationship between \(K_p\) and \(K_c\) is given by the expression:\[ K_p = K_c (RT)^{\Delta n} \]where \(R\) is the ideal gas constant, \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in moles of gas. To understand better, consider temperature and pressure effects:

• If a system is at constant temperature, but the number of gaseous moles changes, \(K_p\) will differ from \(K_c\).
• For reactions involving only gaseous products and reactants, \(\Delta n\) often determines the relative magnitude of these constants.

Understanding these constants is essential for making predictions about the behavior of a chemical system at equilibrium.

Reaction Stoichiometry

Stoichiometry essentially refers to the calculation of reactants and products in chemical reactions. It is defined by the stoichiometric coefficients, which denote the ratio in which different reactants and products participate in the reaction.
In the context of equilibrium, understanding stoichiometry is crucial as it directly influences the computation of equilibrium constants like \(K_c\) and \(K_p\). For instance, in practice:

• The reaction \(\mathrm{N}_{2} \mathrm{O}_{4(g)} \rightleftharpoons 2 \mathrm{NO}_{2(g)}\) showcases how the coefficients of 1:2 influence the calculation of equilibrium constants.
• Balancing stoichiometric equations accurately is necessary for determining \(\Delta n_g\), the change in moles of gas during the reaction.

Reactions in gaseous state are particularly sensitive to stoichiometric factors because they influence properties like pressure and volume, further affecting equilibrium computation. This makes stoichiometry not just a requirement for chemical balance, but also a tool for predicting reaction outcomes.
Mastering stoichiometry enables chemists to confidently manipulate reactions, understand gas behaviors, and predict how modifying conditions will shift the equilibrium.

Phase Equilibria

Phase equilibria deals with the equilibrium between different phases (solid, liquid, and gas) in a chemical system. It is integral to understanding systems where multiple phases coexist at equilibrium. In reactions involving gases, recognizing the role of solids and liquids is vital, especially when it comes to simplifying calculations.
Since the concentration of a pure solid or liquid is considered constant, it is typically omitted in the expression for equilibrium constants. For example, in the decomposition of solid ammonium carbonate:

• \(\mathrm{NH}_{2} \mathrm{COONH}_{4(s)} \rightleftharpoons 2 \mathrm{NH}_{3(g)} + \mathrm{CO}_{2(g)}\)
• Only the gaseous products are accounted for when evaluating changes in moles (\(\Delta n_g\)).

This simplifies calculations and highlights why the phase of each component in a chemical equilibrium must be clearly understood. Moreover, in heterogeneous equilibria, the effect of changes in pressure and temperature can be different from those in homogeneous systems, often because of these omissions of solid and liquids.
Clear comprehension of phase equilibria allows chemists to efficiently address reaction dynamics in industrial and laboratory settings, ensuring precise control over products formed.