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On the basis of your experience, predict which of the following reactions are spontaneous. (a) \(\mathrm{CO}_{2}(s) \longrightarrow \mathrm{CO}_{2}(g)\) at \(25^{\circ} \mathrm{C}\) (b) \(\mathrm{NaCl}(s) \longrightarrow \mathrm{NaCl}(l)\) at \(25^{\circ} \mathrm{C}\) (c) \(2 \mathrm{NaCl}(s) \longrightarrow 2 \mathrm{Na}(s)+\mathrm{Cl}_{2}(g)\) (d) \(\mathrm{CO}_{2}(g) \longrightarrow \mathrm{C}(s)+\mathrm{O}_{2}(g)\)

Short Answer

Expert verified
(a) CO2(s) -> CO2(g) (b) NaCl(s) -> NaCl(l) (c) 2 NaCl(s) -> 2 Na(s) + Cl2(g) (d) CO2(g) -> C(s) + O2(g) Answer: (a) CO2(s) -> CO2(g)

Step by step solution

01

(a) CO2(s) -> CO2(g) at 25°C

Consider the sublimation of solid carbon dioxide (CO2) to gaseous CO2. At 25°C (298 K), the reaction involves a phase change from solid to gas. CO2 has a sublimation temperature of -78°C. Since 25°C is higher than the sublimation temperature of CO2, the natural tendency would be for solid CO2 to go into the gaseous phase, making this reaction spontaneous at this temperature.
02

(b) NaCl(s) -> NaCl(l) at 25°C

In this reaction, solid sodium chloride (NaCl) is melting to form liquid NaCl. The melting point of NaCl is 801°C, which is much higher than the given temperature of 25°C. Hence, at this temperature, it would not be favorable for solid NaCl to melt into liquid NaCl, meaning this reaction is not spontaneous at 25°C.
03

(c) 2 NaCl(s) -> 2 Na(s) + Cl2(g)

The reaction involves the decomposition of solid sodium chloride (NaCl) into solid sodium (Na) and gaseous chlorine (Cl2). In general, ionic compounds such as NaCl are very stable and are unlikely to decompose into their constituent elements spontaneously, as there is a significant energy cost associated with breaking the ionic bonds. Moreover, sodium and chlorine both prefer to exist in their respective ionic states rather than as neutral atoms. Hence, overall, this reaction is not spontaneous.
04

(d) CO2(g) -> C(s) + O2(g)

This reaction involves the decomposition of gaseous carbon dioxide (CO2) into solid carbon (C) and gaseous oxygen (O2). As gaseous CO2 is a very stable molecule and has a lower energy state than the combination of solid carbon and gaseous oxygen, it is energetically unfavorable for CO2 to decompose into its constituent elements. Therefore, this reaction is not spontaneous at the given conditions.

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Most popular questions from this chapter

For the decomposition of \(\mathrm{Ag}_{2} \mathrm{O}\) : $$ 2 \mathrm{Ag}_{2} \mathrm{O}(s) \longrightarrow 4 \mathrm{Ag}(s)+\mathrm{O}_{2}(g) $$ (a) Obtain an expression for \(\Delta G^{\circ}\) as a function of temperature. Prepare a table of \(\Delta G^{\circ}\) values at \(100 \mathrm{~K}\) intervals between \(100 \mathrm{~K}\) and \(500 \mathrm{~K}\). (b) Calculate the temperature at which \(\Delta G^{\circ}=0\).

Hydrogen has been suggested as the fuel of the future. One way to store it is to convert it to a compound that can be heated to release the hydrogen. One such compound is calcium hydride, \(\mathrm{CaH}_{2}\). This compound has a heat of formation of \(-186.2 \mathrm{~kJ} / \mathrm{mol}\) and a standard entropy of \(42.0 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\). What is the minimum temperature to which calcium hydride would have to be heated to produce hydrogen at one atmosphere pressure?

Consider the reaction $$ \mathrm{NH}_{4}^{+}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{NH}_{3}(a q) $$ Use \(\Delta G_{f}^{\circ}\) for \(\mathrm{NH}_{3}(a q)\) at \(25^{\circ} \mathrm{C}=-26.7 \mathrm{~kJ} / \mathrm{mol}\) and the appropriate tables to calculate (a) \(\Delta G^{\circ}\) at \(25^{\circ} \mathrm{C}\) (b) \(K_{\mathrm{a}}\) at \(25^{\circ} \mathrm{C}\)

Use Table 16.1 to calculate \(\Delta S^{\circ}\) for each of the following reactions: (a) \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) (b) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l)+\mathrm{N}_{2} \mathrm{H}_{4}(l) \longrightarrow \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)\) (d) \(\mathrm{CH}_{4}(g)+3 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CHCl}_{3}(l)+3 \mathrm{HCl}(g)\)

It has been proposed that wood alcohol, \(\mathrm{CH}_{3} \mathrm{OH}\), a relatively inexpensive fuel to produce, be decomposed to produce methane. Methane is a natural gas commonly used for heating homes. Is the decomposition of wood alcohol to methane and oxygen thermodynamically feasible at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm} ?\)

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