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Chapter 17: Problem 23

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 dissolves in \(\mathrm{H}_2 \mathrm{O}\), a clear solution becomes a uniform color after adding dye, and iron rusts) are spontaneous as they occur naturally without external intervention and increase the entropy of the system. Process d (cleaning your bedroom) is non-spontaneous, as it requires external intervention and decreases entropy.

Step by step solution

01

Process a: Salt dissolves in \(\mathrm{H}_2 \mathrm{O}\)

Salt dissolving in water is a spontaneous process because it occurs naturally without any external intervention. In this process, the salt ions interact with water molecules, and the dissolution takes place based on the electrostatic attraction between them. This process increases the entropy (disorder) of the system, making it spontaneous.
02

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

When a few drops of dye are added to a clear solution, the dye molecules spread throughout the solution to give a uniform color. This process is also a spontaneous process, as it occurs naturally without external intervention. The molecules of dye are distributed evenly in the solution by diffusion, and this process increases the overall entropy of the system, contributing to its spontaneity.
03

Process c: Iron rusts

Iron rusting is a spontaneous process because it occurs naturally in the presence of air and moisture. When iron comes in contact with air and water, it forms iron oxide or rust. This process is a slow chemical reaction, known as oxidation, and it leads to an increase in entropy. This increase in entropy makes the process of iron rusting spontaneous.
04

Process d: You clean your bedroom

Cleaning your bedroom is a non-spontaneous process because it requires external intervention (in this case, your effort) to organize and decrease the overall entropy of the system (your bedroom). The natural tendency of things is towards an increase in entropy, which is opposite to the result of cleaning your bedroom. As the process relies on external intervention and decreases entropy, it is considered non-spontaneous.

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Most popular questions from this chapter

Consider the following reaction: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$Calculate \(\Delta G\) for this reaction under the following conditions (assume an uncertainty of \(\pm 1\) in all quantities): a. \(T=298 \mathrm{~K}, P_{\mathrm{N}_{2}}=P_{\mathrm{H}_{2}}=200 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=50 \mathrm{~atm}\) b. \(T=298 \mathrm{~K}, P_{\mathrm{N}_{2}}=200 \mathrm{~atm}, P_{\mathrm{H}_{2}}=600 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=200 \mathrm{~atm}\)

Consider a weak acid, HX. If a \(0.10 M\) solution of HX has a pH of \(5.83\) at \(25^{\circ} \mathrm{C}\), what is \(\Delta G^{\circ}\) for the acid's dissociation reaction at \(25^{\circ} \mathrm{C}\) ?

For mercury, the enthalpy of vaporization is \(58.51 \mathrm{~kJ} / \mathrm{mol}\) and the entropy of vaporization is \(92.92 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). What is the normal boiling point of mercury?

Given the following data: $$\begin{array}{lr}2 \mathrm{C}_{6} \mathrm{H}_{6}(l)+15 \mathrm{O}_{2}(g) \longrightarrow 12 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta G^{\circ}=-6399 \mathrm{~kJ} \\\\\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \Delta G^{\circ}=-394 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \Delta G^{\circ}=-237 \mathrm{~kJ} \end{array}$$ calculate \(\Delta G^{\circ}\) for the reaction $$6 \mathrm{C}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)$$

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