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Predict the sign of \(\Delta S^{\circ}\) for each of the following reactions. (a) \(\mathrm{O}_{3}(g) \longrightarrow \mathrm{O}_{2}(g)+\mathrm{O}(g)\) (b) \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(g)\) (c) \(\mathrm{CuSO}_{4}(s)+5 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(s)\)

Short Answer

Expert verified
Question: For each of the following reactions, predict the sign of the change in entropy (∆S°): (a) O3(g) → O2(g) + O(g) (b) PCl3(g) + Cl2(g) → PCl5(g) (c) CuSO4(s) + 5 H2O(l) → CuSO4·5H2O(s) Answer: (a) The sign of ∆S° is positive. (b) The sign of ∆S° is negative. (c) The sign of ∆S° is negative.

Step by step solution

01

Reaction (a): O3(g) → O2(g) + O(g) #

In this reaction, one Ozone (O3) molecule decomposes into one Oxygen (O2) molecule and one separate Oxygen (O) atom. Since the number of particles in the gaseous state increases from 1 to 2, the overall disorder of the system increases. Therefore, the entropy increases, and the sign of \(\Delta S^{\circ}\) will be positive.
02

Reaction (b): PCl3(g) + Cl2(g) → PCl5(g) #

In this reaction, phosphoryl chloride (PCl3) reacts with chlorine gas (Cl2), forming phosphorus pentachloride (PCl5). Here, two gaseous molecules are combining to form one gaseous molecule. This means the number of particles in the gaseous state decreases from 2 to 1, leading to a decrease in the disorder of the system. Thus, the entropy decreases, and the sign of \(\Delta S^{\circ}\) will be negative.
03

Reaction (c): CuSO4(s) + 5 H2O(l) → CuSO4·5H2O(s) #

In this reaction, solid copper sulfate (CuSO4) and liquid water (H2O) form a copper sulfate pentahydrate (CuSO4·5H2O) crystal. Since the number of particles in the solid and liquid states doesn't change and both reactants and products are present in similar phases (solid and liquid), there is no significant change in the disorder of the system. However, the process forms a more ordered structure (hydrated crystal) due to the bonding of water molecules to the copper sulfate. The formation of this structured crystal suggests a slight decrease in entropy. Consequently, the sign of \(\Delta S^{\circ}\) will be negative.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermodynamics
When we discuss thermodynamics in the context of chemical reactions, we are looking at the energy changes associated with these reactions. A fundamental concept within this field is entropy (\( S \)), which is a measure of the disorder or randomness in a system.

The change in entropy, denoted as \((\Delta S^\circ)\), tells us how the disorder of a system changes as a reaction proceeds. In simple terms, if a reaction results in more dispersed energy and a higher level of disorder, we say that entropy increases, and \((\Delta S^\circ)\) is positive. Conversely, if the reaction leads to a more ordered system with less randomness, entropy decreases, and \((\Delta S^\circ)\) is negative.

For instance, in a reaction where a gas forms from a solid, the gas particles are more spread out than the particles in the solid, leading to an increase in entropy. This concept is crucial for predicting the spontaneity of reactions; typically, processes that involve an increase in entropy tend to be naturally favorable. Entropy, alongside enthalpy (\( H \)), allows for the calculation of a reaction's Gibbs free energy (\( G \)), which predicts the spontaneity of a process. The higher the entropy change \((\Delta S^\circ)\) and the lower the enthalpy change \((\Delta H^\circ)\), the more negative \((\Delta G^\circ)\), and the more likely a reaction will occur spontaneously.
Chemical Kinetics
Chemical kinetics deals with the speed, or rate, of a chemical reaction and the factors which affect this rate. It's another facet of physical chemistry that complements studies on thermodynamics by not just determining whether a reaction is possible but also how fast it will go.

Unlike thermodynamics, which is concerned with the initial and final states of a chemical process, kinetics looks at the path taken from reactants to products. This involves analyzing reaction mechanisms, energy profiles, and transition states to understand how and why certain reactions are fast or slow. While entropy change can give us insights into the energy distribution of a system, kinetics provides a dynamic view of how those energy changes manifest in the form of reaction rates.

For example, in a scenario where we have a high \((\Delta S^\circ)\) that suggests a reaction should happen spontaneously, chemical kinetics might reveal that the reaction is actually very slow due to a high activation energy barrier. Therefore, to fully grasp a chemical process, it's important to consider both thermodynamic properties, like entropy, and kinetic aspects, such as the energy required to initiate the reaction, to predict the behavior of a system.
Physical Chemistry
At the crossroads of physics and chemistry lies the specialty of physical chemistry, which is concerned with the physical properties of molecules, the forces that act upon them, and the energy changes that accompany chemical reactions.

Under the umbrella of physical chemistry, thermodynamics and chemical kinetics are just two of several topics being explored. This branch of science seeks a deeper understanding of the principles governing the behavior of matter and energy on an atomic and molecular level. By studying aspects like reaction equilibria, electrochemistry, and quantum chemistry, physical chemists can explain why and how chemical reactions occur.

For instance, in the exercises mentioned, the prediction of entropy change \((\Delta S^\circ)\) relies on physical chemistry concepts, such as the nature of substances (solid, liquid, or gas) involved in reactions and the molecular structure's impact on the system's disorder. Understanding the principles of physical chemistry thus provides students with the tools to interpret more complex phenomena, like reaction dynamics and equilibrium, in a nuanced and comprehensive manner.

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Most popular questions from this chapter

Show by calculation whether the reaction $$ \mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q) \quad \Delta G^{\circ}=18.0 \mathrm{~kJ} $$ is spontaneous at \(25^{\circ} \mathrm{C}\) when (a) \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=0.78 \mathrm{M}\) and \([\mathrm{HF}]=0.24 \mathrm{M}\) (b) \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=0.0030 \mathrm{M}\) and \([\mathrm{HF}]=1.85 \mathrm{M}\)

The overall reaction that occurs when sugar is metabolized is $$ \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)+12 \mathrm{O}_{2}(g) \longrightarrow 12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) $$ For this reaction, \(\Delta H^{\circ}\) is \(-5650 \mathrm{~kJ}\) and \(\Delta G^{\circ}\) is \(-5790 \mathrm{~kJ}\) $$ \text { at } 25^{\circ} \mathrm{C} $$ (a) If \(25 \%\) of the free energy change is actually converted to useful work, how many kilojoules of work are obtained when one gram of sugar is metabolized at body temperature, \(37^{\circ} \mathrm{C}\) ? (b) How many grams of sugar would a 120 -lb woman have to eat to get the energy to climb the Jungfrau in the Alps, which is \(4158 \mathrm{~m}\) high? \(\left(w=9.79 \times 10^{-3} \mathrm{mb},\right.\) where \(w=\) work in kilojoules, \(m\) is body mass in kilograms, and \(h\) is height in meters.)

Given that \(\Delta H_{\mathrm{t}}^{\circ}\) for \(\mathrm{HF}(a q)\) is \(-320.1 \mathrm{~kJ} / \mathrm{mol}\) and \(S^{\circ}\) for \(\mathrm{HF}(a q)\) is \(88.7 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K},\) find \(K_{\mathrm{a}}\) for \(\mathrm{HF}\) at \(25^{\circ} \mathrm{C}\)

Predict the order of the following reactions in terms of increasing \(\Delta S\) : (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{C}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(s)\) (d) \(\mathrm{I}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{ICl}(g)\)

A \(0.250 \mathrm{M}\) solution of a weak base \(\mathrm{R}_{2} \mathrm{NH}\) has a \(\mathrm{pH}\) of 10.60 at \(25^{\circ} \mathrm{C}\). What is \(\Delta G^{\circ}\) for the dissociation of the weak base in water at \(25^{\circ} \mathrm{C}\) ? $$ \mathrm{R}_{2} \mathrm{NH}(a q)+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{R}_{2} \mathrm{NH}_{2}^{+}(a q)+\mathrm{OH}^{-}(a q) $$

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