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Chapter 19: Problem 21

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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Most popular questions from this chapter

Use the following data to estimate the standard molar entropy of gaseous benzene at \(298.15 \mathrm{K} ;\) that is, \(S^{\circ}\left[\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g}, 1 \mathrm{atm})\right] .\) For \(\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{s}, 1 \mathrm{atm})\) at its melting point of \(5.53^{\circ} \mathrm{C}, S^{\circ}\) is \(128.82 \mathrm{Jmol}^{-1} \mathrm{K}^{-1}\). The enthalpy of fusion is \(9.866 \mathrm{kJ} \mathrm{mol}^{-1} .\) From the melting point to 298.15 K, the average heat capacity of liquid benzene is \(134.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1} .\) The enthalpy of vaporization of \(\mathrm{C}_{6} \mathrm{H}_{6}(1)\) at \(298.15 \mathrm{K}\) is \(33.85 \mathrm{kJ} \mathrm{mol}^{-1},\) and in the vapor- ization, \(\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g})\) is produced at a pressure of 95.13 Torr. Imagine that this vapor could be compressed to 1 atm pressure without condensing and while behaving as an ideal gas. Calculate \(S^{\circ}\left[\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g}, 1 \text { atm) }] .[ \text { Hint: Refer to }\right.\) Exercise \(88,\) and note the following: For infinitesimal quantities, \(d S=d q / d T ;\) for the compression of an ideal gas, \(d q=-d w ;\) and for pressure-volume work, \(d w=-P d V\).

The Gibbs energy change of a reaction can be used to assess (a) how much heat is absorbed from the surroundings; (b) how much work the system does on the surroundings; (c) the net direction in which the reaction occurs to reach equilibrium; (d) the proportion of the heat evolved in an exothermic reaction that can be converted to various forms of work.

At \(298 \mathrm{K}, \Delta G_{\mathrm{f}}^{\mathrm{p}}[\mathrm{CO}(\mathrm{g})]=-137.2 \mathrm{kJ} / \mathrm{mol}\) and \(K_{\mathrm{p}}=\) \(6.5 \times 10^{11}\) for the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons\) \(\mathrm{COCl}_{2}(\mathrm{g}) . \quad\) Use these data to determine \(\Delta G_{f}\left[\mathrm{COCl}_{2}(\mathrm{g})\right],\) and compare your result with the value in Appendix D.

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 \(298 \mathrm{K}\) $$\begin{array}{lcc} \hline & \Delta H_{f}^{\circ}, \mathrm{kJ} \mathrm{mol}^{-1} & \Delta G_{f,}^{\circ} \mathrm{kJ} \mathrm{mol}^{-1} \\ \hline \mathrm{NH}_{3}(\mathrm{g}) & -46.11 & -16.48 \\ \mathrm{HCl}(\mathrm{g}) & -92.31 & -95.30 \\ \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) & -314.4 & -202.9 \\ \hline \end{array}$$

On page 822 the terms states and microstates were introduced. Consider a system that has four states (i.e., energy levels), with energy \(\varepsilon=0,1,2,\) and 3 energy units, and three particles labeled \(A, B,\) and \(C .\) The total energy of the system, in energy units, is 3 . How many microstates can be generated?
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