Question

In the reaction: \(\mathrm{CO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})\) the change in \(\Delta \mathrm{S}^{\circ}\) is (given \(\mathrm{S}^{\circ}\) for \(\mathrm{CO}, \mathrm{O}_{2}\) and \(\mathrm{CO}_{2}\) are \(197.6,205.3\) and \(213.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) respectively) (a) \(-78.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (b) \(-50 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (c) \(-86.5 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (d) \(-30 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)

Short answer

The change in \( \Delta S^{\circ} \) is \(-86.5 \, \text{J K}^{-1} \text{mol}^{-1}\). The correct answer is (c).

Step-by-step solution

Write Down Given Values

Identify and write down the standard entropy values provided in the problem statement: \[ S^{\circ}(\text{CO}) = 197.6 \, \text{J K}^{-1} \text{mol}^{-1}, \quad S^{\circ}(\text{O}_2) = 205.3 \, \text{J K}^{-1} \text{mol}^{-1}, \quad S^{\circ}(\text{CO}_2) = 213.6 \, \text{J K}^{-1} \text{mol}^{-1} \]

Recall Formula for Entropy Change

Use the formula for calculating the change in standard entropy \( \Delta S^{\circ} \) for a chemical reaction: \[ \Delta S^{\circ} = \sum S^{\circ}_{\text{products}} - \sum S^{\circ}_{\text{reactants}} \] In this reaction: \( \text{Reactants: CO and } \frac{1}{2} \text{O}_2 \) \( \text{Product: CO}_2 \)

Calculate Entropy of Products

Calculate the total entropy of the products. For the product \( \text{CO}_2 \): \[ \sum S^{\circ}_{\text{products}} = S^{\circ}(\text{CO}_2) = 213.6 \, \text{J K}^{-1} \text{mol}^{-1} \]

Calculate Entropy of Reactants

Calculate the total entropy of the reactants. For the reactants \(\text{CO}\) and \(\frac{1}{2} \text{O}_2\): \[ \sum S^{\circ}_{\text{reactants}} = S^{\circ}(\text{CO}) + \frac{1}{2} S^{\circ}(\text{O}_2) = 197.6 + \frac{1}{2} \times 205.3 = 197.6 + 102.65 = 300.25 \, \text{J K}^{-1} \text{mol}^{-1} \]

Calculate Change in Entropy

Now, calculate \( \Delta S^{\circ} \) using the values obtained: \[ \Delta S^{\circ} = 213.6 - 300.25 = -86.65 \, \text{J K}^{-1} \text{mol}^{-1} \] This can be rounded to \(-86.5 \, \text{J K}^{-1} \text{mol}^{-1}\).

Select the Correct Option

Compare the calculated \( \Delta S^{\circ} = -86.5 \, \text{J K}^{-1} \text{mol}^{-1} \) with the given options. The correct answer is option (c) \(-86.5 \, \text{J K}^{-1} \text{mol}^{-1}\).

Key concepts

Standard Entropy

Standard entropy is a measure of disorder or randomness in a substance and is expressed as the amount of energy per temperature that is not available to do work. Each substance has its characteristic standard entropy value under standard conditions, which are 1 atmosphere of pressure and a specified temperature, usually 298 K (25°C). These values are critical when assessing reactions and predicting their spontaneity. Understanding standard entropy values helps in determining how molecules are arranged and move. A higher standard entropy indicates more disorder and a greater number of arrangements possible for the substance. In our exercise:

• CO has a standard entropy of 197.6 J K⁻¹ mol⁻¹
• O₂ has a standard entropy of 205.3 J K⁻¹ mol⁻¹
• CO₂ has a standard entropy of 213.6 J K⁻¹ mol⁻¹

These values serve as foundational components in calculating the change in entropy for the chemical reaction, guiding us to evaluate whether a reaction will be more or less ordered as it progresses.

Chemical Reactions

In the context of chemical reactions, understanding how substances transform and interact is crucial for predicting changes in entropy. A chemical reaction involves the breaking and forming of bonds, which can lead to significant changes in the disorder of the system.Our example reaction is \[ \text{CO(g)} + \frac{1}{2} \text{O}_2(\text{g}) \longrightarrow \text{CO}_2(\text{g}) \]This reaction entails combining carbon monoxide (CO) and oxygen (O₂) to form carbon dioxide (CO₂). When products have different standard entropy than reactants, there is a change in entropy:

• Reactants (CO and ½ O₂) possess a combined standard entropy of 300.25 J K⁻¹ mol⁻¹.
• The product (CO₂) has a lower standard entropy of 213.6 J K⁻¹ mol⁻¹.

This decrease in standard entropy indicates that the products are more ordered than the reactants, leading to a negative change in entropy for the reaction.

Thermodynamics Calculations

Thermodynamics encompasses the study of energy transformations and how these relate to energy and work within a system. Calculating changes in entropy during a reaction is a part of this domain. The change in standard entropy, denoted as \( \Delta S^{\circ} \), can be calculated using:\[ \Delta S^{\circ} = \sum S^{\circ}_{\text{products}} - \sum S^{\circ}_{\text{reactants}} \]For our chemical reaction, this means taking the standard entropy of the products and subtracting the standard entropy of the reactants:

• Total entropy of the products: 213.6 J K⁻¹ mol⁻¹ (from CO₂)
• Total entropy of the reactants: 300.25 J K⁻¹ mol⁻¹ (from CO and ½ O₂)
• Thus, \( \Delta S^{\circ} = 213.6 - 300.25 = -86.65 \) J K⁻¹ mol⁻¹, rounded to -86.5 J K⁻¹ mol⁻¹

A negative \( \Delta S^{\circ} \) suggests a decrease in disorder or that the system becomes more ordered after the reaction. This helps in predicting the feasibility of reactions and understanding energy distribution in the system.