Problem 168 - A Complete Resource Book in Chemistry for JEE Main

Question

Match the following Column-I (a) \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\) (b) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) \(\Delta \mathrm{H}=-\mathrm{Q} \mathrm{cal}\) (c) \(2 \mathrm{CO}_{2}(\mathrm{~g})=2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) \(\Delta \mathrm{H}=+\mathrm{Q} \mathrm{cal}\) (d) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\) \(\Delta \mathrm{H}=-\mathrm{Q} \mathrm{cal}\) Column-II (p) Low temperature; high pressure (q) High temperature (r) \(\mathrm{K}_{\mathrm{P}}=\frac{4 \mathrm{x}^{2} \mathrm{P}}{\mathrm{a}^{2}-\mathrm{x}^{2}}\) (s) \(K_{p}=K_{c}\) (t) \(\frac{K_{\text {P }}}{K_{c}}=(R T)^{-2}\)

Step-by-Step Solution

Answer

(a) - (t), (b) - (p), (c) - (q), (d) - (s)

Step 1: Analyzing Reaction (a)

The reaction for (a) is \( \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) \). Since there is a change in the number of moles (1 mole to 2 moles of gas), this indicates that \( K_{p} eq K_{c} \). The relationship \( \frac{K_{p}}{K_{c}} = (RT)^{\Delta n} \equiv (RT)^{1} \) applies. Therefore, this reaction matches with option (t).

Step 2: Analyzing Reaction (b)

The reaction for (b) is \( \mathrm{N}_{2}(\mathrm{~g}) + 3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \), with \( \Delta H = -Q \text{ cal} \). It is exothermic, favoring formation at lower temperatures and higher pressures. Hence, it matches with option (p).

Step 3: Analyzing Reaction (c)

For reaction (c), \( 2 \mathrm{CO}_{2}(\mathrm{~g}) = 2 \mathrm{CO}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{~g}) \), \( \Delta H = +Q \text{ cal} \) indicates endothermic nature, favoring higher temperatures. This reaction matches with option (q) due to the heat absorption.

Step 4: Analyzing Reaction (d)

In the reaction \( \mathrm{H}_{2}(\mathrm{~g}) + \mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{~g}) \), \( \Delta H = -Q \text{ cal} \) indicates an exothermic reaction. The change in moles is zero, making \( \Delta n = 0 \), resulting in \( K_{p} = K_{c} \). This matches with option (s).

Key Concepts

Le Chatelier's Principle

Le Chatelier's Principle helps us predict how equilibrium will shift when we change conditions such as temperature or pressure. The principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change and restore balance.

When we change temperature:- **Exothermic Reactions**: Decrease in temperature shifts equilibrium to the right, favoring products because it removes heat from the system.- **Endothermic Reactions**: Increase in temperature shifts equilibrium to the right, as the process absorbs heat. - For example, the reaction \(\mathrm{2\ CO_2(g)} \rightarrow \mathrm{2\ CO(g) + O_2(g)}\) is endothermic (\(\Delta H = +Q\)), so it favors higher temperatures.

When we change pressure:- Increasing pressure favors the side with fewer moles of gas.- Decreasing pressure favors the side with more moles of gas.- In the case of \(\mathrm{N_2(g) + 3\ H_2(g) \rightarrow 2\ NH_3(g)}\), higher pressure shifts the equilibrium towards forming more \(\mathrm{NH_3}\) due to fewer gas moles.

Equilibrium Constant

The Equilibrium Constant ( heoric, who need to helpedianspeople and oppuutit-)) is a number that shows the ratio of the concentration of products to reactants at equilibrium. It is important because:- It provides a measure of the extent of a reaction.- A large \(K_c\) means more products are formed.- A small \(K_c\) suggests the reaction favors reactants.

For reactions involving gases, two constants can be considered:- \(K_p\): Equilibrium constant with partial pressures.- \(K_c\): Equilibrium constant with concentrations.- For the reaction \(\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2\ HI(g)}\), \(\Delta n = 0\), indicating that \(K_p = K_c\).

The relationship \( \frac{K_p}{K_c} = (RT)^{\Delta n} \) helps us understand how the equilibrium constant changes with respect to temperature and pressure.

Enthalpy Change

Enthalpy change ( heoric, who need to helpedianspeople and oppuutit-)"), denoted as \(\Delta H\), represents the heat absorbed or released during a reaction.- **Exothermic Reactions**: Heat is released, \(\Delta H < 0\).- **Endothermic Reactions**: Heat is absorbed, \(\Delta H > 0\).

The sign of \(\Delta H\) affects the equilibrium direction:- **Negative \(\Delta H\)**: Favorable at low temperatures. Example: \(\mathrm{N_2(g) + 3\ H_2(g) \rightarrow 2\ NH_3(g)}\).- **Positive \(\Delta H\)**: Favorable at high temperatures. Example: \(\mathrm{2\ CO_2(g) = 2\ CO(g) + O_2(g)}\).

Understanding \(\Delta H\) helps anticipate how a reaction responds to thermal energy changes.

Gas Laws

Gas laws are essential tools for understanding how gases behave under different conditions. Here are the core gas laws:- **Boyle's Law**: Pressure and volume are inversely proportional at constant temperature. \[ PV = k \], where \(k\) is a constant.- **Charles's Law**: Volume is directly proportional to temperature at constant pressure. \[ \frac{V}{T} = k \], where \(T\) is in Kelvin.- **Avogadro's Law**: Volume is proportional to the number of moles at constant temperature and pressure. \[ V \propto n \] (where \( n \) is moles).

For reactions like \(\mathrm{N_2O_4(g) \rightleftharpoons 2\ NO_2(g)}\), changes in gas quantities require using the Ideal Gas Law: \[ PV = nRT \].- This reaction shows an increase in moles, thus affecting pressure and volume according to gas law principles.- Understanding these relationships helps predict equilibrium shifts when exploiting Le Chatelier's Principle.