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) \(94.8 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\)

Short answer

The correct answer is (c) \(-94.8 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1}\).

Step-by-step solution

Identify the Formula for Standard Entropy Change

The standard entropy change for a reaction, \( \Delta S^\circ \), can be calculated using the formula: \[ \Delta S^\circ = \sum \text{S}^\circ_{\text{products}} - \sum \text{S}^\circ_{\text{reactants}} \]

List the Standard Entropy Values

For this reaction: \(\text{SO}_2(\text{g}) + \frac{1}{2} \text{O}_2(\text{g}) \rightarrow \text{SO}_3(\text{g})\), the standard entropy values \( \text{S}^\circ \) are: \( \text{SO}_2(\text{g}) = 248.5 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1} \), \( \text{O}_2(\text{g}) = 205 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1} \), \( \text{SO}_3(\text{g}) = 256.2 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1} \).

Calculate \(\sum \text{S}^\circ_{\text{products}}\)

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

Calculate \(\sum \text{S}^\circ_{\text{reactants}}\)

The reactants are one mole of \(\text{SO}_2\) and half a mole of \(\text{O}_2\): \[ \sum \text{S}^\circ_{\text{reactants}} = \text{S}^\circ_{\text{SO}_2} + \frac{1}{2} \times \text{S}^\circ_{\text{O}_2} = 248.5 + \frac{1}{2} \times 205 \] \[ \sum \text{S}^\circ_{\text{reactants}} = 248.5 + 102.5 = 351 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1} \]

Calculate the Standard Entropy Change \(\Delta S^\circ\)

Now calculate \(\Delta S^\circ\) using the formula from Step 1: \[ \Delta S^\circ = 256.2 - 351 = -94.8 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1} \]

Select the Correct Option

The calculated standard entropy change \(\Delta S^\circ\) is \(-94.8 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1}\). Therefore, the correct answer is option (c).

Key concepts

Entropy Calculation

Entropy is a thermodynamic quantity that measures the amount of disorder or randomness in a system. The standard entropy change of a reaction, denoted as \( \Delta S^\circ \), quantifies the difference in entropy between the products and reactants of a chemical reaction under standard conditions.

To calculate \( \Delta S^\circ \) for a reaction, you use the formula:

\[ \Delta S^\circ = \sum \text{S}^\circ_{\text{products}} - \sum \text{S}^\circ_{\text{reactants}} \]
Where \( \text{S}^\circ \) represents the standard entropy, usually given in \( \text{J} \text{K}^{-1} \text{mol}^{-1} \). This calculation requires you to sum up the entropies of the products and subtract the sum of the entropies of the reactants. All values should be standard entropies at the same temperature and pressure, typically 298 K and 1 atm.

Thermodynamics

Thermodynamics is a branch of physics that deals with the relationships between heat, temperature, and other forms of energy. It is concerned with energy transformations in chemical processes and the properties of substances involved.

Entropy plays a crucial role in thermodynamics. It helps us understand the directionality of chemical reactions and processes. A reaction is more likely to occur if the entropy of the system increases, as systems naturally tend to move towards states of higher entropy.

In the context of our exercise, the negative value of \( \Delta S^\circ \) for the formation of \( \text{SO}_3 \) suggests that the disorder of the system decreases as the reaction proceeds. Entropy considerations, combined with enthalpy and the free energy changes, provide insights into the feasibility and spontaneity of reactions.

Chemical Reactions

A chemical reaction involves the transformation of reactants into products, often accompanied by changes in energy and entropy. In any reaction, bonds are broken, and new bonds are formed, leading to the redistribution of energy.

For the reaction given in the exercise, \( \text{SO}_2(\text{g}) + \frac{1}{2} \text{O}_2(\text{g}) \rightarrow \text{SO}_3(\text{g}) \), the reactants \( \text{SO}_2 \) and \( \frac{1}{2} \text{O}_2 \) combine to form \( \text{SO}_3 \). This transformation impacts the entropy of the system as described by the change in standard entropy \( \Delta S^\circ \).

Understanding how different reactions affect entropy helps chemists and engineers to predict reaction behavior and to design processes that align with desired outcomes, such as maximizing product yield or minimizing energy consumption.

Entropy of Products and Reactants

In a chemical reaction, both the products and the reactants have associated entropy values, which represent their degree of disorder. The standard entropy values \( \text{S}^\circ \) for these substances provide a snapshot of their partial arrangement and energy distribution.

In our exercise, the entropy values given for \( \text{SO}_2(\text{g}) \), \( \text{O}_2(\text{g}) \), and \( \text{SO}_3(\text{g}) \) reflect their relative order. The reactants \( \text{SO}_2 \) and \( \frac{1}{2} \text{O}_2 \) have higher total entropy compared to the product \( \text{SO}_3 \). This indicates a more ordered product, resulting in a negative \( \Delta S^\circ \).

It's crucial to accurately use these values when calculating entropy changes, as errors can lead to incorrect predictions about a reaction's direction and spontaneity. The careful calculation of entropy change is fundamental in studying and applying chemical thermodynamics.