Problem 23 - A Complete Resource Book in Chemistry for JEE Main

Question

Which one of the following has \(\Delta \mathrm{S}^{\circ}\) greater than zero? (a) \(\mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CaCO}_{3}(\mathrm{~g})\) (b) \(\mathrm{NaCl}(\mathrm{aq}) \rightleftharpoons \mathrm{NaCl}(\mathrm{s})\) (c) \(\mathrm{NaNO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{NO}_{3}^{-}(\mathrm{aq})\) (d) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})=2 \mathrm{NH}_{3}(\mathrm{~g})\)

Step-by-Step Solution

Answer

Reaction (c) \( \mathrm{NaNO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \) has \( \Delta S^{\circ} > 0 \).

Step 1: Understanding Entropy Change

Entropy, represented as \( \Delta S^{\circ} \), is a measure of randomness or disorder in a system. An increase in disorder corresponds to a positive \( \Delta S^{\circ} \), while an increase in order corresponds to a negative \( \Delta S^{\circ} \). Generally, when gases are produced in a reaction, entropy increases.

Step 2: Analyzing Reaction (a)

For reaction \( \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CaCO}_{3}(\mathrm{~g}) \), a gas and a solid react to form one solid. As this reaction results in the consumption of a gas and formation of a solid, entropy decreases. Thus, \( \Delta S^{\circ} < 0 \).

Step 3: Analyzing Reaction (b)

For reaction \( \mathrm{NaCl}(\mathrm{aq}) \rightleftharpoons \mathrm{NaCl}(\mathrm{s}) \), an aqueous solution forms a solid, indicating a decrease in disorder, or entropy. Therefore, \( \Delta S^{\circ} < 0 \).

Step 4: Analyzing Reaction (c)

For reaction \( \mathrm{NaNO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \), a solid dissolves into ions in solution, increasing disorder. Thus, \( \Delta S^{\circ} > 0 \).

Step 5: Analyzing Reaction (d)

For reaction \( \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) =2 \mathrm{NH}_{3}(\mathrm{~g}) \), reactants are converted from one mole of gas to two moles. This conversion typically results in an increase in order through reduction of total gas volume, leading to \( \Delta S^{\circ} < 0 \).

Step 6: Determining the Answer

Only reaction (c) results in increased entropy as it transforms a solid into aqueous ions, increasing randomness. Thus, it has \( \Delta S^{\circ} > 0 \).

Key Concepts

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions in a chemical process are equal, stabilizing the concentrations of products and reactants. This balance is dynamic, meaning reactions continue to occur, but there is no net change in concentrations.
Key features of chemical equilibrium include:

• It can be established in both closed and open systems, depending on the nature of the reaction.
• Equilibrium is reached when the Gibbs free energy of the system, in a given state, is minimized.
• At equilibrium, the ratio of products to reactants remains constant, indicated by the equilibrium constant, \( K \).

Chemical equilibrium is vital because it helps predict how changes in conditions, such as temperature and pressure, will affect the composition of a chemical mixture.

Gibbs Free Energy

Gibbs free energy, denoted as \( G \), is a thermodynamic potential that combines enthalpy and entropy into a single value. It is used to predict the direction of chemical reactions and determine whether they will be spontaneous or non-spontaneous. The change in Gibbs free energy (9G) is crucial in understanding reaction favorability:

• If \( \Delta G < 0 \), the reaction is spontaneous under constant temperature and pressure.
• If \( \Delta G > 0 \), the reaction is non-spontaneous as it requires energy input to proceed.
• If \( \Delta G = 0 \), the system is at equilibrium.

The equation that links these concepts is:\[ \Delta G = \Delta H - T \Delta S \]where \( \Delta H \) is the change in enthalpy, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the change in entropy. Gibbs free energy is a critical tool for chemists in understanding and optimizing chemical processes.

Dissolution Process

The dissolution process involves a solid substance dissolving into a solvent to form a homogeneous solution. This transformation often leads to an increase in disorder or randomness, which results in a positive change in entropy (9S), as seen in reaction (c) from the original exercise.
Key aspects of dissolution include:

• The process can be endothermic or exothermic, depending on the nature of solute and solvent interactions.
• As solute particles disperse, they interact with solvent molecules, affecting the overall energy and entropy of the system.
• Entropy plays a significant role—since dissolved particles can move more freely, the levels of disorder increase compared to the solid-state.

Overall, understanding the dissolution process is important for predicting solubility and designing effective chemical mixtures.