The reaction, \(2 \mathrm{Cl}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow 2 \mathrm{Cl}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \Delta H=\) \(-161 \mathrm{kJ},\) is expected to be (a) spontaneous at all temperatures; (b) spontaneous at low temperatures, but nonspontaneous at high temperatures; (c) nonspontaneous at all temperatures; (d) spontaneous at high temperatures only.

Short Answer

Expert verified
The reaction is spontaneous at high temperatures only.

Step by step solution

01

Determine the sign of entropy change

Entropy (S) is a measure of randomness or disorder in a system. Considering the reaction \(2Cl2O(g) \rightarrow 2Cl2(g) + O2(g)\), there are 2 moles of gaseous reactants producing 3 moles of gaseous products. Because there are more moles of gas formed than were used up, the reaction results in a higher degree of disorder, and therefore, an increase in entropy (\(\Delta S > 0\)).
02

Use the Gibb’s free energy equation

The Gibbs free energy change (\(\Delta G\)) can be calculated by Gibbs free energy equation: \(\Delta G = \Delta H - T\Delta S\), where \(\Delta H\) is the change in enthalpy, \(\Delta S\) is the change in entropy and T is the temperature in Kelvin. We know that a reaction is spontaneous if \(\Delta G <0\), and non-spontaneous if \(\Delta G > 0\). In this reaction, since both \(\Delta H\) and \(\Delta S\) are positive, the sign of \(\Delta G\) depends on the term \(T\Delta S\). If \(T\Delta S > \Delta H\), then \(\Delta G < 0\), and the reaction is spontaneous. If \(T\Delta S < \Delta H\), then \(\Delta G > 0\), and the reaction is non-spontaneous.
03

Determine the condition for spontaneity

Considering the sign of \(\Delta H\) and \(\Delta S\), it can be seen that the spontaneity change takes place at a certain temperature. When the temperature increases, \(T\Delta S\) also increases. At low temperatures, \(T\Delta S\) is less than \(\Delta H\), resulting in a positive \(\Delta G\), so the reaction is non-spontaneous. As the temperature increases, there will come a point where \(T\Delta S\) becomes greater than \(\Delta H\), resulting in a negative \(\Delta G\), thus making the reaction spontaneous. Therefore, the reaction is non-spontaneous at low temperatures and becomes spontaneous at high temperatures (or when \(T\Delta S\) exceeds \(\Delta H\)).

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Key Concepts

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

Spontaneity of Reactions
When studying chemical reactions, it's essential to understand what determines whether a reaction will occur on its own, without any external energy input. This concept is known as the spontaneity of reactions. A spontaneous reaction is one that can proceed in a given direction without continuous outside assistance.

The key to determining the spontaneity of a reaction lies in the Gibbs free energy change, denoted as \(\Delta G\). The sign of \(\Delta G\) dictates whether a reaction is spontaneous or non-spontaneous at a given temperature. If \(\Delta G < 0\), the reaction is spontaneous, and if \(\Delta G > 0\), it is non-spontaneous. It's important to note that \(\Delta G\) is influenced by both enthalpy change (\(\Delta H\)) and entropy change (\(\Delta S\)) in the reaction, as well as the temperature at which the reaction is taking place (T). When both \(\Delta H\) and \(\Delta S\) are positive, as in our example exercise, the term \(T\Delta S\) becomes crucial in determining the spontaneity at various temperatures.

Additionally, a reaction may also be at equilibrium (\(\Delta G = 0\)), where there is no net change over time, although reactions at equilibrium can still have dynamic exchanges between reactants and products. Understanding how these factors interplay can help students predict whether a reaction will occur and under which conditions.
Entropy Change
Entropy, symbolized as S, is a fundamental concept in chemistry and thermodynamics that deals with disorder within a system. An increase in entropy (\(\Delta S > 0\)) means that the system has become more disordered or random. In chemical reactions, entropy often plays a significant role in deciding the direction and spontaneity of the reaction.

In the example reaction, moving from two moles of \(Cl_2O\) gas to form three moles of gaseous products (\(2Cl_2\) and \(O_2\)) leads to an increase in the number of gas particles. This increase causes more possible ways to arrange the molecules, hence raising the system's disorder and the entropy. It's crucial for students to envision the microscopic view of a reaction to grasp the concept of entropy change. More molecules moving freely equate to higher entropy.

Furthermore, conditions involving temperature can greatly affect entropy. Generally, as temperature increases, so does the kinetic energy of particles, contributing to higher entropy. Analyzing entropy change helps to comprehend why certain reactions favor particular conditions, enhancing the student's capability to predict chemical behavior.
Enthalpy Change
Enthalpy change, represented by \(\Delta H\), is a measure of the total energy change that occurs during a chemical reaction. It indicates whether a reaction is exothermic (releasing heat, \(\Delta H < 0\)) or endothermic (absorbing heat, \(\Delta H > 0\)). In the context of spontaneous reactions, enthalpy change combines with entropy change to determine Gibbs free energy.

For example, the textbook problem shows that the enthalpy change for the given reaction is negative (\(-161 kJ\)), which means it is exothermic. Exothermic reactions tend to be favorable for spontaneity since they release energy to the surroundings, which could instinctively seem to make a reaction spontaneous. However, it is not the only factor. The complete picture of spontaneity involves evaluating \(\Delta H\) alongside \(\Delta S\) and temperature, as reflected in the Gibbs free energy equation (\(\Delta G = \Delta H - T\Delta S\)).

When teaching enthalpy, it's also helpful to use real-world analogies or engaging lab demonstrations to cement the concept. The goal is to ensure students can connect the abstractness of enthalpy with tangible examples, enhancing their comprehension of how energy exchange influences chemical processes.

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Most popular questions from this chapter

The following data are given for the two solid forms of \(\mathrm{HgI}_{2}\) at \(298 \mathrm{K}\). $$\begin{array}{llll} \hline & \Delta H_{f}^{\circ} & \Delta G_{f,}^{\circ} & S^{\circ} \\ & \text { kJ mol }^{-1} & \text {kJ mol }^{-1} & \text {J mol }^{-1} \text {K }^{-1} \\ \hline \mathrm{HgI}_{2} \text { (red) } & -105.4 & -101.7 & 180 \\ \mathrm{Hg} \mathrm{I}_{2} \text { (yellow) } & -102.9 & (?) & (?) \\ \hline \end{array}$$ Estimate values for the two missing entries. To do this, assume that for the transition \(\mathrm{HgI}_{2}(\mathrm{red}) \longrightarrow\) \(\mathrm{HgI}_{2}(\text { yellow }),\) the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) at \(25^{\circ} \mathrm{C}\) have the same values that they do at the equilibrium temperature of \(127^{\circ} \mathrm{C}\).

For the dissociation of \(\mathrm{CaCO}_{3}(\mathrm{s})\) at \(25^{\circ} \mathrm{C}, \mathrm{CaCO}_{3}(\mathrm{s})\) \(\rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \Delta G^{\circ}=+131 \mathrm{kJ} \mathrm{mol}^{-1} .\) A sample of pure \(\mathrm{CaCO}_{3}(\mathrm{s})\) is placed in a flask and connected to an ultrahigh vacuum system capable of reducing the pressure to \(10^{-9} \mathrm{mmHg}\) (a) Would \(\mathrm{CO}_{2}(\mathrm{g})\) produced by the decomposition of \(\mathrm{CaCO}_{3}(\mathrm{s})\) at \(25^{\circ} \mathrm{C}\) be detectable in the vacuum system at \(25^{\circ} \mathrm{C} ?\) (b) What additional information do you need to determine \(P_{\mathrm{CO}_{2}}\) as a function of temperature? (c) With necessary data from Appendix D, determine the minimum temperature to which \(\mathrm{CaCO}_{3}(\mathrm{s})\) would have to be heated for \(\mathrm{CO}_{2}(\mathrm{g})\) to become detectable in the vacuum system.

If a reaction can be carried out only by electrolysis, which of the following changes in a thermodynamic property must apply: (a) \(\Delta H>0 ;\) (b) \(\Delta S>0\) (c) \(\Delta G=\Delta H ;\) (d) \(\Delta G>0 ?\) Explain.

From the data given in the following table, determine \(\Delta S^{\circ} \quad\) for the reaction \(\quad \mathrm{NH}_{3}(\mathrm{g})+\mathrm{HCl}(\mathrm{g}) \longrightarrow\) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) .\) All data are at \(298 \mathrm{K}\) $$\begin{array}{lcc} \hline & \Delta H_{f}^{\circ}, \mathrm{kJ} \mathrm{mol}^{-1} & \Delta G_{f,}^{\circ} \mathrm{kJ} \mathrm{mol}^{-1} \\ \hline \mathrm{NH}_{3}(\mathrm{g}) & -46.11 & -16.48 \\ \mathrm{HCl}(\mathrm{g}) & -92.31 & -95.30 \\ \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) & -314.4 & -202.9 \\ \hline \end{array}$$

Consider the vaporization of water: \(\mathrm{H}_{2} \mathrm{O}(1) \longrightarrow\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(100^{\circ} \mathrm{C},\) with \(\mathrm{H}_{2} \mathrm{O}(1)\) in its standard state, but with the partial pressure of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(2.0 \mathrm{atm}\) Which of the following statements about this vaporization at \(100^{\circ} \mathrm{C}\) are true? (a) \(\Delta G^{\circ}=0,\) (b) \(\Delta G=0\) (c) \(\Delta G^{\circ}>0,\) (d) \(\Delta G>0 ?\) Explain.

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