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 \(-94.8 \, \mathrm{J} \, \mathrm{K}^{-1} \mathrm{~mol}^{-1}\) (option c).

Step 1: Write the Reaction Formula

The reaction given is \( \mathrm{SO}_{2}(\mathrm{~g}) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{SO}_{3}(\mathrm{~g}) \). This represents the formation of sulfur trioxide from sulfur dioxide and oxygen.

Step 2: Identify Given Entropies

The standard entropies (S^{\circ}) of the reactants and products are given as follows: \( S^{\circ}~\mathrm{SO}_{2}(\mathrm{~g}) = 248.5~\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\), \( S^{\circ}~\mathrm{O}_{2}(\mathrm{~g}) = 205~\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\), and \( S^{\circ}~\mathrm{SO}_{3}(\mathrm{~g}) = 256.2~\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\).

Step 3: Use the Standard Entropy Change Formula

The standard entropy change for a reaction, \( \Delta S^{\circ} \), is calculated as: \[ \Delta S^{\circ} = \sum S^{\circ}_{\text{products}} - \sum S^{\circ}_{\text{reactants}} \] Apply this formula to the reaction.

Step 4: Compute the Total Product and Reactant Entropies

Calculate the total entropy for products: \( S^{\circ}_{\text{product}} = S^{\circ}_{\mathrm{SO}_{3}} = 256.2~\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \). Calculate the total entropy for reactants: \[ S^{\circ}_{\text{reactants}} = S^{\circ}_{\mathrm{SO}_{2}} + \frac{1}{2} S^{\circ}_{\mathrm{O}_{2}} = 248.5 + \frac{1}{2} \times 205 \] \[ S^{\circ}_{\text{reactants}} = 248.5 + 102.5 = 351~\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \]

Step 5: Calculate the Standard Entropy Change

Subtract the total entropy of reactants from products: \[ \Delta S^{\circ} = S^{\circ}_{\text{product}} - S^{\circ}_{\text{reactants}} = 256.2 - 351 = -94.8~\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \]

Step 6: Select the Correct Answer

From the calculation, the standard entropy change \( \Delta S^{\circ} \) for the reaction is \(-94.8 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \). Therefore, the correct option is (c).

Key Concepts

Standard Entropy

In thermodynamics, standard entropy refers to the absolute entropy content of substances in their standard states, which are typically 1 atm pressure and a specific temperature, often 298 K. Entropy itself is a measure of the disorder or randomness within a system. In this exercise, the standard entropies of different chemical species are given: sulfur dioxide

• (\( SO_2(g)\)) with \( 248.5\text{ J K}^{-1} \text{mol}^{-1}\),
• oxygen (\( O_2(g)\)) with \( 205 \text{ J K}^{-1} \text{mol}^{-1}\),
• and sulfur trioxide (\( SO_3(g)\)) with \( 256.2 \text{ J K}^{-1} \text{mol}^{-1}\).

The standard entropy values tell us how much energy at a given temperature is unavailable for doing work in a chemical process. These figures alone, however, do not give a measure of the direction or spontaneity of a reaction, which is why changes in entropy are essential for understanding reactions.

Reaction Entropy Calculation

Calculating the entropy change of a reaction, often labeled \( \Delta S^{\circ}\), involves determining the difference between the total entropy of products and reactants. This is where the standard entropy values come into play.

The formula used is:

• \[\Delta S^{\circ} = \sum S^{\circ}_{\text{products}} - \sum S^{\circ}_{\text{reactants}}\]

In our context, involving the conversion of \( SO_2\) and \( O_2\) to \( SO_3\), we use the entropies previously discussed. After calculating, we find:

• Total entropy of the reactants: \(351 \text{ J K}^{-1} \text{mol}^{-1}\)
• Entropy of the product: \(256.2 \text{ J K}^{-1} \text{mol}^{-1}\)

Thus, inserting these into the formula gives a reaction entropy change (\(\Delta S^{\circ}\)) of \(-94.8 \text{ J K}^{-1} \text{mol}^{-1}\). This negative value signifies that the entropy, and therefore the disorder, decreases during the reaction.

Thermodynamics in Chemistry

Thermodynamics in chemistry connects the principles of heat and energy transformations within chemical processes. It explores energy changes that occur due to chemical reactions and how they influence reaction direction, extent, and spontaneity.

One central role is understanding entropy and its influence on spontaneous reactions. A spontaneous process can be identified when a system moves to a state with higher entropy. However, reactions with negative entropy change (\(\Delta S^{\circ} < 0\)), like the one in our exercise, can still occur spontaneously if they are exothermic, releasing heat that offsets this negative change through Gibbs free energy considerations.

In addition to entropy, thermodynamics looks at enthalpy changes and free energy changes. These three - entropy (\(S\)), enthalpy (\(H\)), and free energy (\(G\)) - are interrelated, providing a fuller picture of reaction feasibility through the equation \(G = H - TS\).