Problem 69 - A Complete Resource Book in Chemistry for JEE Main

Question

The standard entropy change for the reaction \(\mathrm{SO}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{SO}_{3}(\mathrm{~g})\) is (where \(\mathrm{S}^{\circ}\) for \(\mathrm{SO}_{2}(\mathrm{~g}), \mathrm{O}_{2}(\mathrm{~g})\) and \(\mathrm{SO}_{3}(\mathrm{~g})\) are \(248.5,205\) and \(256.2\) \(\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\) respectively) (a) \(198.2 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) (b) \(-192.8 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) (c) \(-94.8 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) (d) \(94.8 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The standard entropy change is \(-94.8 \ \text{J} \ \text{K}^{-1} \ \text{mol}^{-1}\), option (c).

Step 1: Identify the Given Values

The standard molar entropy values are given as \( S^\circ(\mathrm{SO}_2(g)) = 248.5 \ \text{J} \ \text{K}^{-1} \ \text{mol}^{-1} \), \( S^\circ(\mathrm{O}_2(g)) = 205 \ \text{J} \ \text{K}^{-1} \ \text{mol}^{-1} \), and \( S^\circ(\mathrm{SO}_3(g)) = 256.2 \ \text{J} \ \text{K}^{-1} \ \text{mol}^{-1} \). We need to find the standard entropy change (\( \Delta S^\circ \)) for the reaction: \( \mathrm{SO}_2(g) + \frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{SO}_3(g) \).

Step 2: Write the Formula for Entropy Change

The formula for the standard entropy change of a reaction is given by: \[ \Delta S^\circ = S^\circ(\text{products}) - S^\circ(\text{reactants}) \]

Step 3: Calculate the Entropy of the Products

There is one mole of \( \mathrm{SO}_3(g) \) as a product. Thus, \( S^\circ(\text{products}) = S^\circ(\mathrm{SO}_3(g)) = 256.2 \ \text{J} \ \text{K}^{-1} \ \text{mol}^{-1} \).

Step 4: Calculate the Entropy of the Reactants

The reactants are one mole of \( \mathrm{SO}_2(g) \) and half a mole of \( \mathrm{O}_2(g) \). Calculate their total entropy: \[ S^\circ(\text{reactants}) = S^\circ(\mathrm{SO}_2(g)) + \frac{1}{2} S^\circ(\mathrm{O}_2(g)) \] Substituting the values: \[ S^\circ(\text{reactants}) = 248.5 + \frac{1}{2} \times 205 = 248.5 + 102.5 = 351 \ \text{J} \ \text{K}^{-1} \ \text{mol}^{-1} \]

Step 5: Compute the Standard Entropy Change

Use the formula from Step 2 to determine \( \Delta S^\circ \): \[ \Delta S^\circ = 256.2 - 351 = -94.8 \ \text{J} \ \text{K}^{-1} \ \text{mol}^{-1} \]

Step 6: Choose the Correct Answer

The calculated standard entropy change is \( -94.8 \ \text{J} \ \text{K}^{-1} \ \text{mol}^{-1} \). This matches option (c), \(-94.8 \ \text{J} \ \text{K}^{-1} \ \text{mol}^{-1} \).

Key Concepts

Standard Entropy Change

In thermodynamics, the standard entropy change (\( \Delta S^\circ \)) is a measure of the disorder or randomness in a chemical reaction's products compared to its reactants under standard conditions (usually 1 atm pressure and 298 K temperature). It quantifies how the entropy of a system changes during a reaction.
The formula to calculate the standard entropy change is:

• \( \Delta S^\circ = S^\circ(\text{products}) - S^\circ(\text{reactants}) \)

The standard entropy values for different substances are often determined empirically and recorded in tables. This helps in calculating \( \Delta S^\circ \) for reactions easily.
A positive \( \Delta S^\circ \) would indicate an increase in disorder, while a negative value signifies a decrease in entropy, meaning the products are more ordered than the reactants.

Chemical Reaction Entropy

Chemical reaction entropy is crucial in understanding how various states of matter interact during a reaction. The entropy values reflect the level of disorder and are essential in predicting reaction spontaneity alongside enthalpy. In general, solid products tend to decrease entropy, while gaseous products can increase it due to their higher freedom of movement.
For any chemical reaction, keeping track of the standard molar entropy (\( S^\circ \)) values for reactants and products is necessary. These values provide a benchmark for calculating how much the entropy changes:

• Reactions resulting in more gaseous molecules typically increase entropy.
• Reactions forming fewer gaseous molecules may decrease entropy, as seen in \( \mathrm{SO}_2(g) + \frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{SO}_3(g) \).

Understanding the concept of entropy changes allows chemists to predict whether a reaction is feasible under standard conditions by considering both \( \Delta H \) and \( \Delta S \) values.

Gaseous Reactions

Gaseous reactions involve reactants and products in the gas phase, and their reactions often result in significant entropy changes. Gases have particles that move freely and occupy a larger volume compared to solids or liquids. Therefore, reactions involving gases tend to involve more pronounced changes in entropy.
In the given example, \( \mathrm{SO}_2(g) + \frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{SO}_3(g) \), there is a reduction in the number of gas molecules from reactants to products, leading to a decrease in entropy. This example highlights that:

• The reduction in the number of gas molecules reduces disorder.
• Such reactions, with a decrease in moles of gas, often result in negative entropy changes.

Predicting entropy changes in gaseous reactions is essential for understanding reaction dynamics, especially when evaluating the feasibility and spontaneity of processes in the gas phase.