Question

Use Table \(17.1\) to calculate \(\Delta S^{\circ}\) for each of the following reactions. (a) \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(l)\) (b) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)\) (c) \(\mathrm{BaCO}_{3}(s) \longrightarrow \mathrm{BaO}(s)+\mathrm{CO}_{2}(g)\) (d) \(2 \mathrm{NaCl}(s)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{NaF}(s)+\mathrm{Cl}_{2}(g)\)

Short answer

Question: Calculate the standard entropy change (\(\Delta S^{\circ}\)) for the following reactions: (a) CO(g) + 2H2(g) -> CH3OH(l) (b) N2(g) + O2(g) -> 2 NO(g) (c) BaCO3(s) -> BaO(s) + CO2(g) (d) 2 NaCl(s) + F2(g) -> 2 NaF(s) + Cl2(g) Provide your answers in J/mol·K. Answer: (a) \(\Delta S^{\circ} = -331\: J\: mol^{-1} K^{-1}\) (b) \(\Delta S^{\circ} = 25.4\: J\: mol^{-1} K^{-1}\) (c) \(\Delta S^{\circ} = 155.7\: J\: mol^{-1} K^{-1}\) (d) \(\Delta S^{\circ} = -60.2\: J\: mol^{-1} K^{-1}\)

Step-by-step solution

(a) Calculate \(\Delta S^{\circ}\) for the given reaction.

Reaction (a): $\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(l)$ Follow these steps to calculate \(\Delta S^{\circ}\): 1. Find the standard entropies for the reactants and products in Table 17.1: $S^{\circ}_{CO} = 197.6\: J\: mol^{-1} K^{-1} \\ S^{\circ}_{H2} = 130.6\: J\: mol^{-1} K^{-1} \\ S^{\circ}_{CH3OH} = 126.8\: J\: mol^{-1} K^{-1}$ 2. Apply the formula to calculate \(\Delta S^{\circ}\): \(\Delta S^{\circ} = S^{\circ}_{CH3OH} - (S^{\circ}_{CO} + 2S^{\circ}_{H2}) = 126.8 - (197.6 + 2(130.6))\) 3. Calculate the result: \(\Delta S^{\circ} = -331\: J\: mol^{-1} K^{-1}\)

(b) Calculate \(\Delta S^{\circ}\) for the given reaction.

Reaction (b): \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)\) 1. Find the standard entropies for the reactants and products in Table 17.1: $S^{\circ}_{N2} = 191.5\: J\: mol^{-1} K^{-1} \\ S^{\circ}_{O2} = 205.0\: J\: mol^{-1} K^{-1} \\ S^{\circ}_{NO} = 210.7\: J\: mol^{-1} K^{-1}$ 2. Apply the formula to calculate \(\Delta S^{\circ}\): \(\Delta S^{\circ} = 2S^{\circ}_{NO} - (S^{\circ}_{N2} + S^{\circ}_{O2}) = 2(210.7) - (191.5 + 205.0)\) 3. Calculate the result: \(\Delta S^{\circ} = 25.4\: J\: mol^{-1} K^{-1}\)

(c) Calculate \(\Delta S^{\circ}\) for the given reaction.

Reaction (c): \(\mathrm{BaCO}_{3}(s) \longrightarrow \mathrm{BaO}(s)+\mathrm{CO}_{2}(g)\) 1. Find the standard entropies for the reactants and products in Table 17.1: $S^{\circ}_{BaCO3} = 112.2\: J\: mol^{-1} K^{-1} \\ S^{\circ}_{BaO} = 54.3\: J\: mol^{-1} K^{-1} \\ S^{\circ}_{CO2} = 213.6\: J\: mol^{-1} K^{-1}$ 2. Apply the formula to calculate \(\Delta S^{\circ}\): \(\Delta S^{\circ} = (S^{\circ}_{BaO} + S^{\circ}_{CO2}) - S^{\circ}_{BaCO3} = (54.3 + 213.6) - 112.2\) 3. Calculate the result: \(\Delta S^{\circ} = 155.7\: J\: mol^{-1} K^{-1}\)

(d) Calculate \(\Delta S^{\circ}\) for the given reaction.

Reaction (d): \(2 \mathrm{NaCl}(s)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{NaF}(s)+\mathrm{Cl}_{2}(g)\) 1. Find the standard entropies for the reactants and products in Table 17.1: $S^{\circ}_{NaCl} = 72.1\: J\: mol^{-1} K^{-1} \\ S^{\circ}_{F2} = 202.7\: J\: mol^{-1} K^{-1} \\ S^{\circ}_{NaF} = 51.4\: J\: mol^{-1} K^{-1} \\ S^{\circ}_{Cl2} = 223.0\: J\: mol^{-1} K^{-1}$ 2. Apply the formula to calculate \(\Delta S^{\circ}\): \(\Delta S^{\circ} = (2S^{\circ}_{NaF} + S^{\circ}_{Cl2}) - (2S^{\circ}_{NaCl} + S^{\circ}_{F2}) = (2(51.4) + 223.0) - (2(72.1) + 202.7)\) 3. Calculate the result: \(\Delta S^{\circ} = -60.2\: J\: mol^{-1} K^{-1}\) The standard entropies for the given reactions are: (a) \(\Delta S^{\circ} = -331\: J\: mol^{-1} K^{-1}\) (b) \(\Delta S^{\circ} = 25.4\: J\: mol^{-1} K^{-1}\) (c) \(\Delta S^{\circ} = 155.7\: J\: mol^{-1} K^{-1}\) (d) \(\Delta S^{\circ} = -60.2\: J\: mol^{-1} K^{-1}\)

Key concepts

Entropy

Entropy is a key concept in thermodynamics that describes the degree of disorder or randomness in a system. It essentially provides a way to measure the number of ways energy can be distributed within a system. Entropy helps predict the direction in which processes will naturally occur. For example, if a process increases the disorder or randomness, it usually results in a positive entropy change.

Here are some noteworthy points about entropy:

• It is a state function, meaning its value depends only on the state of the system, not on the path taken to reach that state.
• Entropy is denoted by the symbol \(S\) and is typically expressed in units of \(J\,mol^{-1}\,K^{-1}\).
• Systems tend to move towards a state of higher entropy or disorder.

Standard Entropy

Standard entropy refers to the entropy content of a substance under standard state conditions, typically at a pressure of 1 bar and a specified temperature, often 25°C (298 K). This standard measurement allows chemists to tabulate entropies that can be used to compare different substances.

Key characteristics of standard entropy include:

• Denoted as \(S^{ eflectbox{c}}\),
• The standard entropy values are tabulated in reference books, making it easier to calculate entropy change in chemical reactions.
• Substances in different physical states (solid, liquid, gas) have different standard entropies.

These values are crucial when predicting reaction spontaneity and are used in the calculation of the change in entropy for reactions.

Chemical Reactions

Chemical reactions are processes in which substances, known as reactants, are converted into different substances, known as products. These reactions can involve the breaking and formation of chemical bonds and typically result in energy changes, sometimes influencing the entropy of the system.
Common types of chemical reactions include:

• Synthesis reactions: Two or more substances combine to form a single compound.
• Decomposition reactions: A compound breaks down into two or more simpler substances.
• Redox reactions: Involve the transfer of electrons between reactants.

Entropy plays a significant role in these processes by helping to predict the direction and spontaneity of reactions. Generally, reactions that lead to an increase in the overall entropy of the system, often also aided by energy release, are more likely to occur spontaneously.

Entropy Change

Entropy change, denoted by \(\Delta S\), measures how the entropy of a system varies as it undergoes a chemical or physical transformation. Calculating \(\Delta S\) for reactions involves comparing the standard entropy values of products and reactants. The formula is given by:

\[ \Delta S^{\circ} = \sum S^{\circ}_{\text{products}} - \sum S^{\circ}_{\text{reactants}} \]
Where, each \(S^{\circ}\) represents the standard entropy of substances involved under standard conditions.

Important aspects of entropy change include:

• A positive \(\Delta S\) indicates an increase in disorder, favoring spontaneity.
• A negative \(\Delta S\) shows a decrease in disorder, generally requiring input of energy to proceed.
• This concept assists in understanding reaction feasibility and is fundamental in Gibbs Free Energy calculations, which combine both enthalpy and entropy changes to predict spontaneity.

Understanding \(\Delta S\) is pivotal as it provides insight into how a reaction progresses under particular conditions.