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Explain why (a) some exothermic reactions do not occur spontaneously, and (b) some reactions in which the entropy of the system increases do not occur spontaneously.

Short Answer

Expert verified
Some exothermic reactions don't occur spontaneously due to a decrease in entropy or low temperature, resulting in a positive Gibbs free energy. Reactions in which the entropy of the system increases do not occur spontaneously if the enthalpy change is positive and large enough over the entropy term or the temperature is low, thus making the Gibbs free energy positive.

Step by step solution

01

Part A: Understanding Exothermic Reactions

Exothermic reactions release energy (usually in the form of heat), and thus, the enthalpy (\( \Delta H \)) of the reaction is negative. However, spontaneity of a reaction is determined not only by the enthalpy but also by the entropy (\( \Delta S \)) and absolute temperature (T), as formulated in the Gibbs free energy (\( \Delta G \)). The equation is \( \Delta G = \Delta H - T\Delta S \). Even if \( \Delta H \) is negative, for a reaction to be spontaneous, \( \Delta G \) should be negative. It means that an increase in \( \Delta S \) (change in entropy) or T can still yield a positive \( \Delta G \) thus making the reaction non-spontaneous, even if it is exothermic.
02

Part B: Understanding Entropy Increase

An increase in entropy (\( \Delta S \)) usually suggests a more disordered system, which is a favored condition for a spontaneous reaction. However, as again explained by the Gibbs free energy equation (\( \Delta G = \Delta H - T\Delta S \)), a reaction to be spontaneous, \( \Delta G \) should be negative. Even if \( \Delta S \) is positive, if the \( \Delta H \) is positive (endothermic reaction) and large enough to overcompensate the \( T\Delta S \) term or if the temperature (T) is very low, the \( \Delta G \) can be positive, thus making the reaction non-spontaneous.

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Key Concepts

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

Exothermic Reactions
In chemistry, exothermic reactions are well known for their energy release, which usually occurs as heat. During these reactions, the system's enthalpy, or heat content (\( \Delta H \)), decreases. Consequently, the enthalpy change is negative, indicating energy is emitted into the environment. However, this energy release does not automatically mean the reaction will occur spontaneously.

The critical factor determining spontaneity is the Gibbs Free Energy (\( \Delta G \)). The Gibbs Free Energy equation is represented as:\[ \Delta G = \Delta H - T\Delta S \]where:
  • \( \Delta H \) = Change in enthalpy
  • \( T \) = Absolute temperature
  • \( \Delta S \) = Change in entropy
Even if a reaction is exothermic, \( \Delta G \) must also be negative for spontaneity. A high entropy decrease or low temperature can counteract the negative enthalpy, resulting in a reaction that is non-spontaneous regardless of its exothermic nature. This highlights the interplay between heat and disorder in determining whether a reaction will proceed on its own.
Entropy
Entropy, denoted as\( \Delta S \), is a measure of disorder or randomness in a system. Increasing entropy often signifies a move toward disorder. Systems naturally tend towards higher entropy as it is statistically favored. Thus, reactions with positive entropy change (\( \Delta S \) > 0) might seem inclined to be spontaneous. However, the bigger picture involves examining the effects on Gibbs Free Energy (\( \Delta G \)):
\[ \Delta G = \Delta H - T\Delta S \]

When entropy increases in a reaction, it may contribute negatively to \( \Delta G \), pushing it towards spontaneity. Nevertheless, if the reaction is endothermic (requiring energy), the positive \( \Delta H \) might nullify the entropy benefit, especially if the temperature is not sufficiently high. Therefore, even with disorder rising, a reaction could remain constrained without releasing energy or occurring spontaneously.

To sum up, entropy's role in spontaneity is nuanced. While higher disorder promotes spontaneity, the context of the reaction's enthalpy and temperature is crucial.
Spontaneity
Spontaneity in chemical reactions refers to the ability of a process to occur on its own, without any external input. It's not solely about an energy release but also about the conditions of the system and how they interact with each other through the Gibbs Free Energy (\( \Delta G \)). The condition for spontaneity (\( \Delta G \)<0) is crucial here:
  • A negative \( \Delta H \) (favorable energy release) can drive spontaneity.
  • A positive \( \Delta S \) (increased entropy) supports spontaneity, especially at higher temperatures.
  • However, not all reactions with favorable energy or entropy changes are spontaneous due to the delicate interplay of factors.

For example, a reaction might release energy (exothermic) but still require an entropy increase to be spontaneous. Similarly, an endothermic reaction might proceed spontaneously if the entropy boost and temperature are exceedingly high. Spontaneity is about the balance of energy, order, and the environment, making it a more sophisticated aspect of understanding reactions in general.

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Most popular questions from this chapter

A tabulation of more precise thermodynamic data than are presented in Appendix D lists the following values for \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(298.15 \mathrm{K},\) at a standard state pressure of 1 bar. $$\begin{array}{llll} \hline & \Delta H_{f}^{\circ}, & \Delta G_{f,}^{\circ} & S_{\prime}^{\circ} \\\ & \text { kJ mol }^{-1} & \text {kJ mol }^{-1} & \text {J mol }^{-1} \text {K }^{-1} \\ \hline \mathrm{H}_{2} \mathrm{O}(1) & -285.830 & -237.129 & 69.91 \\ \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) & -241.818 & -228.572 & 188.825 \\ \hline \end{array}$$ (a) Use these data to determine, in two different ways, \(\Delta G^{\circ}\) at \(298.15 \mathrm{K}\) for the vaporization: \(\mathrm{H}_{2} \mathrm{O}(1,1 \mathrm{bar}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g}, 1 \mathrm{bar}) .\) The value you obtain will differ slightly from that on page 838 because here, the standard state pressure is 1 bar, and there, it is 1 atm. (b) Use the result of part (a) to obtain the value of \(K\) for this vaporization and, hence, the vapor pressure of water at \(298.15 \mathrm{K}\) (c) The vapor pressure in part (b) is in the unit bar. Convert the pressure to millimeters of mercury. (d) Start with the value \(\Delta G^{\circ}=8.590 \mathrm{kJ}\), given on page 838 and calculate the vapor pressure of water at 298.15 K in a fashion similar to that in parts (b) and (c). In this way, demonstrate that the results obtained in a thermodynamic calculation do not depend on the convention we choose for the standard state pressure, as long as we use standard state thermodynamic data consistent with that choice.

A handbook lists the following standard a handbook lists the following standard enthalpies of formation at \(298 \mathrm{K}\) for cyclopentane, \(\mathrm{C}_{5} \mathrm{H}_{10}: \quad \Delta H_{\mathrm{f}}^{\mathrm{g}}\left[\mathrm{C}_{5} \mathrm{H}_{10}(1)\right]=-105.9 \mathrm{kJ} / \mathrm{mol} \quad\) and \(\Delta H_{\mathrm{f}}^{\mathrm{o}}\left[\mathrm{C}_{5} \mathrm{H}_{10}(\mathrm{g})\right]=-77.2 \mathrm{kJ} / \mathrm{mol}\) (a) Estimate the normal boiling point of cyclopentane. (b) Estimate \(\Delta G^{\circ}\) for the vaporization of cyclopentane at \(298 \mathrm{K}\). (c) Comment on the significance of the sign of \(\Delta G^{\circ}\) at \(298 \mathrm{K}\)

\(\mathrm{H}_{2}(\mathrm{g})\) can be prepared by passing steam over hot iron: \(3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})+4 \mathrm{H}_{2}(\mathrm{g})\) (a) Write an expression for the thermodynamic equilibrium constant for this reaction. (b) Explain why the partial pressure of \(\mathrm{H}_{2}(\mathrm{g})\) is independent of the amounts of \(\mathrm{Fe}(\mathrm{s})\) and \(\mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})\) present. (c) Can we conclude that the production of \(\mathrm{H}_{2}(\mathrm{g})\) from \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) could be accomplished regardless of the proportions of \(\mathrm{Fe}(\mathrm{s})\) and \(\mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})\) present? Explain.

Two correct statements about the reversible reaction \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) are \((\mathrm{a}) K=K_{\mathrm{p}}\) (b) the equilibrium amount of NO increases with an increased total gas pressure; (c) the equilibrium amount of NO increases if an equilibrium mixture is transferred from a \(10.0 \mathrm{L}\) container to a \(20.0 \mathrm{L}\) container; (d) \(K=K_{c} ;\) (e) the composition of an equilibrium mixture of the gases is independent of the temperature.

For each of the following reactions, indicate whether \(\Delta S\) for the reaction should be positive or negative. If it is not possible to determine the sign of \(\Delta S\) from the information given, indicate why. (a) \(\mathrm{CaO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})\) (b) \(2 \mathrm{HgO}(\mathrm{s}) \longrightarrow 2 \mathrm{Hg}(1)+\mathrm{O}_{2}(\mathrm{g})\) (c) \(2 \mathrm{NaCl}(1) \longrightarrow 2 \mathrm{Na}(1)+\mathrm{Cl}_{2}(\mathrm{g})\) (d) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{g})\) (e) \(\operatorname{Si}\left(\text { s) }+2 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{SiCl}_{4}(\mathrm{g})\right.\)

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