Problem 170 - A Complete Resource Book in Chemistry for JEE Main

Question

Match the following Column-I (a) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (b) \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\) (c) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) (d) \(\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons(\mathrm{g}) \mathrm{SO}_{3}(\mathrm{~g})\) Column-II (p) \(\mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{c}}\) (q) \(\Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ln \mathrm{k}\) (r) Addition of He at constant pressure shifts the equilibrium to right hand side (s) \(\mathrm{K}_{\mathrm{p}}>\mathrm{K}_{\mathrm{c}}\) (t) Increase of pressure favours forward reaction.

Step-by-Step Solution

Answer

(a) with (s) (r), (b) with (s) (r), (c) with (p) (t), (d) with (p) (t).

Step 1: Identify Reaction Type

Look at each reaction in Column-I and identify whether it involves a change in the number of moles of gas from reactants to products. This will help determine which equilibrium constant, whether \(K_p\) or \(K_c\), is greater, as well as how changes in pressure affect the equilibrium.- (a) PCl extsubscript{5} ⇌ PCl extsubscript{3} + Cl extsubscript{2}: Total moles increase from 1 to 2.- (b) N extsubscript{2}O extsubscript{4} ⇌ 2 NO extsubscript{2}: Total moles increase from 1 to 2.- (c) N extsubscript{2} + 3 H extsubscript{2} ⇌ 2 NH extsubscript{3}: Total moles decrease from 4 to 2.- (d) SO extsubscript{2} + 0.5 O extsubscript{2} ⇌ SO extsubscript{3}: Total moles decrease from 1.5 to 1.

Step 2: Match with Column-II (Sample Matches)

For this step, you check if the change in moles results in \(K_p\) being greater or less than \(K_c\) (because of the formula relating these two constants and the presence of the gas constant \(R\), typically \(K_p = K_c(RT)^{\Delta n}\)). Also, consider the effect of pressure changes on equilibrium.- For reactions with increasing moles (a and b), \( K_p > K_c \). Hence, they correspond to option (s) for \( \mathrm{K}_{\mathrm{p}} > \mathrm{K}_{\mathrm{c}} \).- For reactions with decreasing moles (c and d), \( K_p < K_c \). Hence, they correspond to option (p) for \( \mathrm{K}_{\mathrm{p}} < \mathrm{K}_{\mathrm{c}} \).- For pressure effect, increase in pressure favors the side with fewer moles of gas. Hence, reaction (c) and (d) would move forward with increased pressure, matching option (t).

Step 3: Confirm Conditions with General Thermodynamic Equilibrium

Relate the reactions to the general Gibbs free energy equation \(\Delta G^\circ = -RT \ln K\).- Each of these reactions at standard conditions can be described using the equation for Gibbs Free Energy \((q)\), as all standard equilibria are described this way, so they are correctly matched to this condition.

Step 4: Apply External Condition Effect

Consider the effect of adding helium (as an inert gas) at constant pressure. This action will decrease the equilibrium concentration of reactants/support forward reactions if they result in an increase in total moles of gas, hence advancing reactions (a) and (b), due to increased moles of products under constant pressure. - Match to option (r) for reactions (a) and (b).

Key Concepts

Equilibrium Constant

In chemistry, the equilibrium constant provides valuable insight into the position of equilibrium in reversible reactions. It's denoted by either \( K_c \) or \( K_p \), depending on the states of the substances involved. \( K_c \) applies to concentrations, while \( K_p \) relates to partial pressures of gases. The equilibrium constant is crucial because it tells us about the relative concentrations of reactants and products at equilibrium. For the reaction \( \text{PCl}_5 (g) \rightleftharpoons \text{PCl}_3 (g) + \text{Cl}_2 (g) \), the total moles increase from 1 to 2.

• This would actually mean that for such cases, \( K_p \) is greater than \( K_c \), since there is an increase in the number of gas moles, matching the scenario where \( K_p > K_c \).
• Similarly, if the total moles decrease, say in the reaction \( \text{N}_2 (g) + 3\text{H}_2 (g) \rightleftharpoons 2\text{NH}_3 (g) \), \( K_p < K_c \) because the reaction consumes more moles than it produces.

The equilibrium constant varies with temperature, adjusting according to how the system's energy changes. Understanding \( K_c \) and \( K_p \) is pivotal for solving various chemical equilibrium problems.

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept in chemical equilibrium, explaining how systems at equilibrium respond to external changes. This principle states that if a dynamic equilibrium is disturbed, the system will adjust to counteract the disturbance and re-establish equilibrium.One significant application of Le Chatelier's Principle is reflected in pressure changes. When pressure is increased, the equilibrium shifts towards the side with fewer gas moles to alleviate the pressure.
Conversely, if pressure is decreased, the system favors the side with more gas molecules.

• In the context of the reactions discussed, for instance, \( SO_2 (g) + \frac{1}{2} \text{O}_2 (g) \rightleftharpoons \text{SO}_3 (g) \), the reaction produces fewer moles, so increasing pressure favors the forward reaction.
• This principle also helps in understanding why adding an inert gas, like helium, at constant pressure might shift the equilibrium in favor of more expansive reactions (i.e., ones producing more moles of gas).

Thus, Le Chatelier's Principle is invaluable for predicting the behavior of chemical systems undergoing changes in conditions like pressure, concentration, and temperature.

Gibbs Free Energy

Gibbs Free Energy, \( \Delta G \), is a thermodynamic potential that measures the maximum reversible work that may be performed by a thermodynamic system at constant temperature and pressure.The core principle is that a system will tend towards a state that minimizes its Gibbs Free Energy. At equilibrium, the Gibbs Free Energy has reached its minimum and any deviation from this position would increase \( \Delta G \), thus being less favorable.The relationship between Gibbs Free Energy and the equilibrium constant is expressed mathematically as:\[ \Delta G^{\circ} = -RT \ln K \]Where:- \( \Delta G^{\circ} \) is the standard Gibbs Free Energy change.- \( R \) is the universal gas constant.- \( T \) is the temperature in Kelvin.- \( K \) is the equilibrium constant.

• This equation indicates that if \( \Delta G^{\circ} \) is negative, the reaction will proceed spontaneously towards the formation of products, a concept crucial when predicting reaction feasibility.
• For each reaction type you're studying, understanding \( \Delta G^{\circ} \) in relation to its equilibrium constant \( K \) offers insights into why certain reactions are favorable under given conditions, bridging thermodynamics and kinetics beautifully.

Appreciating Gibbs Free Energy's role in chemical equilibria aids in crafting more effective problem-solving approaches in thermochemistry.