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Chapter 16: Problem 25

Which of the following processes are spontaneous? a. Salt dissolves in \(\mathrm{H}_{2} \mathrm{O}\). b. A clear solution becomes a uniform color after a few drops of dye are added. c. Iron rusts. d. You clean your bedroom.

Short Answer

Expert verified
Processes a, b, and c (salt dissolving in \(\mathrm{H}_{2} \mathrm{O}\), a clear solution becoming uniform in color, and iron rusting) are spontaneous as they result in an increase in entropy and involve energy changes favorable for the process to occur. Process d (cleaning your bedroom) is non-spontaneous as it requires manual effort and results in a decrease in entropy.

Step by step solution

01

Process a: Salt dissolves in \(\mathrm{H}_{2} \mathrm{O}\).

Dissolving salt in water is a spontaneous process because it causes an increase in entropy (disorder) by mixing the salt and water molecules. Moreover, during this process, the energy is released (exothermic), making it favorable.
02

Process b: A clear solution becomes a uniform color after a few drops of dye are added.

This process is also spontaneous, as the dye molecules spread throughout the solution, increasing its entropy. Diffusion is a natural process, and the dye particles move randomly from a high concentration to a low concentration region, resulting in a uniform color without any external assistance.
03

Process c: Iron rusts.

Rusting of iron is a spontaneous process as it is an oxidation reaction. When iron is exposed to oxygen and moisture, it reacts to form rust (iron oxide), which is a more stable compound. This reaction releases energy, and the entropy increases due to the formation of more moles from the reactants, which makes this process spontaneous.
04

Process d: You clean your bedroom.

Cleaning your bedroom is not a spontaneous process, as it requires your manual effort to organize the items and clean the space. It also involves a decrease in entropy (disorder) as the bedroom becomes more organized after cleaning. Therefore, this process is non-spontaneous. From the above analysis, processes a, b, and c are spontaneous, while process d is not spontaneous.

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Most popular questions from this chapter

Consider the reaction $$\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)$$ a. Use \(\Delta G_{\mathrm{f}}^{\circ}\) values in Appendix 4 to calculate \(\Delta G^{\circ}\) for this reaction. b. Is this reaction spontaneous under standard conditions at \(298 \mathrm{K} ?\) c. The value of \(\Delta H^{\circ}\) for this reaction is \(100 .\) kJ. At what temperatures is this reaction spontaneous at standard conditions? Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

What types of experiments can be carried out to determine whether a reaction is spontaneous? Does spontaneity have any relationship to the final equilibrium position of a reaction? Explain.

Given the values of \(\Delta H\) and \(\Delta S,\) which of the following changes will be spontaneous at constant \(T\) and \(P ?\) a. \(\Delta H=+25 \mathrm{kJ}, \Delta S=+5.0 \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}\) b. \(\Delta H=+25 \mathrm{kJ}, \Delta S=+100 . \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}\) c. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=+5.0 \mathrm{J} / \mathrm{K}, T=298 \mathrm{K}\) d. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=-40 . \mathrm{J} / \mathrm{K}, T=200 . \mathrm{K}\)

Hydrogen sulfide can be removed from natural gas by the reaction $$2 \mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \rightleftharpoons 3 \mathrm{S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ Calculate \(\Delta G^{\circ}\) and \(K\) (at 298 K) for this reaction. Would this reaction be favored at a high or low temperature?

Consider the system $$\mathrm{A}(g) \longrightarrow \mathrm{B}(g)$$ at $$25^{\circ} \mathrm{C}$$ a. Assuming that \(G_{A}^{\circ}=8996 \mathrm{J} / \mathrm{mol}\) and \(G_{\mathrm{B}}^{\circ}=11,718 \mathrm{J} / \mathrm{mol}\) calculate the value of the equilibrium constant for this reaction. b. Calculate the equilibrium pressures that result if 1.00 mole of \(\mathrm{A}(g)\) at 1.00 atm and 1.00 mole of \(\mathrm{B}(g)\) at 1.00 atm are mixed at \(25^{\circ} \mathrm{C}.\) c. Show by calculations that \(\Delta G=0\) at equilibrium.
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