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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

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 \(298 \mathrm{~K}\) ) for this reaction. Would this reaction be favored at a high or low temperature?

The third law of thermodynamics states that the entropy of a perfect crystal at \(0 \mathrm{~K}\) is zero. In Appendix \(4, \mathrm{~F}^{-}(a q), \mathrm{OH}^{-}(a q)\), and \(\mathrm{S}^{2-}(a q)\) all have negative standard entropy values. How can \(S^{\circ}\) values be less than zero?

Impure nickel, refined by smelting sulfide ores in a blast furnace, can be converted into metal from \(99.90 \%\) to \(99.99 \%\) purity by the Mond process. The primary reaction involved in the Mond process is $$\mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g)$$ a. Without referring to Appendix 4, predict the sign of \(\Delta S^{\circ}\) for the above reaction. Explain. b. The spontaneity of the above reaction is temperature dependent. Predict the sign of \(\Delta S_{\text {sarr }}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{~kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) at \(298 \mathrm{~K}\). Using these values and data in Appendix 4, calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the above reaction. d. Calculate the temperature at which \(\Delta G^{\circ}=0(K=1)\) for the above reaction, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. e. The first step of the Mond process involves equilibrating impure nickel with \(\mathrm{CO}(\mathrm{g})\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) at about \(50^{\circ} \mathrm{C}\). The purpose of this step is to convert as much nickel as possible into the gas phase. Calculate the equilibrium constant for the preceding reaction at \(50 .{ }^{\circ} \mathrm{C}\). f. In the second step of the Mond process, the gaseous \(\mathrm{Ni}(\mathrm{CO})_{4}\) is isolated and heated to \(227^{\circ} \mathrm{C}\). The purpose of this step is to deposit as much nickel as possible as pure solid (the reverse of the preceding reaction). Calculate the equilibrium constant for the preceding reaction at \(227^{\circ} \mathrm{C}\). g. Why is temperature increased for the second step of the Mond process? h. The Mond process relies on the volatility of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for its success. Only pressures and temperatures at which \(\mathrm{Ni}(\mathrm{CO})_{4}\) is a gas are useful. A recently developed variation of the Mond process carries out the first step at higher pressures and a temperature of \(152^{\circ} \mathrm{C}\). Estimate the maximum pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) that can be attained before the gas will liquefy at \(152^{\circ} \mathrm{C}\). The boiling point for \(\mathrm{Ni}(\mathrm{CO})_{4}\) is \(42^{\circ} \mathrm{C}\) and the enthalpy of vaporization is \(29.0 \mathrm{~kJ} / \mathrm{mol}\).

At 1 atm, liquid water is heated above \(100^{\circ} \mathrm{C}\). For this process, which of the following choices (i-iv) is correct for \(\Delta S_{\text {sur }}\) ? \(\Delta S\) ? \(\Delta S_{\text {univ }} ?\) Explain each answer. i. greater than zero ii. less than zero iii. equal to zero iv. cannot be determined

The enthalpy of vaporization of chloroform \(\left(\mathrm{CHCl}_{3}\right)\) is \(31.4\) \(\mathrm{kJ} / \mathrm{mol}\) at its boiling point \(\left(61.7^{\circ} \mathrm{C}\right) .\) Determine \(\Delta S_{\mathrm{sys}}, \Delta S_{\mathrm{sur}}\), and \(\Delta S_{\text {univ }}\) when \(1.00 \mathrm{~mol}\) chloroform is vaporized at \(61.7^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm} .\)

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