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Chapter 19: Problem 93

Ammonium nitrate dissolves spontaneously and endothermally in water at room temperature. What can you deduce about the sign of \(\Delta S\) for this solution process?

Short Answer

Expert verified
The sign of \(\Delta S\) for the dissolution of ammonium nitrate in water at room temperature is positive. This conclusion is based on the given information that the process is spontaneous and endothermic, and by analyzing the Gibbs Free Energy equation, which states that for a spontaneous process, \(\Delta G < 0\). A positive \(\Delta S\) indicates an increase in the disorder of the system during the dissolution process.

Step by step solution

01

Understand the given information

First, let's understand what is given in the exercise: 1. The dissolution of ammonium nitrate in water is spontaneous - this means the process occurs naturally without any external intervention. 2. The dissolution is endothermic - this implies that heat is absorbed by the system, and its surroundings experience a decrease in temperature.
02

Examine the Second Law of Thermodynamics

According to the Second Law of Thermodynamics, the total entropy change of a spontaneous process should be positive (\(\Delta S_{total} > 0\)). For an endothermic reaction, the change in enthalpy (\(\Delta H\)) is positive.
03

Apply the Gibbs Free Energy equation

The Gibbs Free Energy equation can be used to determine the spontaneity of a process: \[\Delta G = \Delta H - T\Delta S\] Where: - \(\Delta G\) is the change in Gibbs Free Energy - \(\Delta H\) is the change in enthalpy (positive for an endothermic reaction) - \(T\) is the temperature in Kelvin - \(\Delta S\) is the change in entropy (what we're trying to deduce) For a spontaneous process, \(\Delta G < 0\).
04

Calculate the sign of \(\Delta S\)

Since \(\Delta G < 0\) for a spontaneous process, and we know that the dissolution of ammonium nitrate in water is a spontaneous endothermic process, we can deduce the sign of \(\Delta S\). Given that \(\Delta H\) is positive (endothermic process), in order for the Gibbs Free Energy equation (\(\Delta G = \Delta H - T\Delta S\)) to be less than 0 (spontaneous), the term \(-T\Delta S\) must be negative and its magnitude should be greater than that of \(\Delta H\). Thus, for the equation to hold true, \(\Delta S\) must be positive. A positive change in entropy suggests an increase in the disorder of the system during the dissolution process. In conclusion, the sign of \(\Delta S\) for the dissolution of ammonium nitrate in water at room temperature is positive.

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Most popular questions from this chapter

Using \(S^{\circ}\) values from Appendix C, calculate \(\Delta S^{\circ}\) values for the following reactions. In each case account for the sign of \(\Delta S^{\circ} .\) (a) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)\) (b) \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) (c) \(\mathrm{Be}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{BeO}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (d) \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\)

(a) Which of the thermodynamic quantities \(T, E, q, w,\) and \(S\) are state functions? (b) Which depend on the path taken from one state to another? (c) How many reversible paths are there between two states of a system? (d) For a reversible isothermal process, write an expression for \(\Delta E\) in terms of \(q\) and \(w\) and an expression for \(\Delta S\) in terms of \(q\) and \(T\).

Use data from Appendix \(\mathrm{C}\) to calculate the equilibrium constant, \(K,\) at \(298 \mathrm{~K}\) for each of the following reactions: (a) \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{Hl}(g)\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6}(g)\)

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be made by the controlled oxidation of methane: $$ \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g) $$ (a) Use data in Appendix C to calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. (b) How is \(\Delta G^{\circ}\) for the reaction expected to vary with increasing temperature? (c) Calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). Under standard conditions, is the reaction spontaneous at this temperature? (d) Is there a temperature at which the reaction would be at equilibrium under standard conditions and that is low enough so that the compounds involved are likely to be stable?

The conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\), is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2}(g) $$ In practice, this reaction is carried out in the presence of oxygen: $$ 2 \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) Using the data in Appendix \(C,\) calculate \(K\) for these reactions at \(25^{\circ} \mathrm{C}\) and \(500{ }^{\circ} \mathrm{C}\). (b) Is the difference in \(\Delta G^{\circ}\) for the two reactions due primarily to the enthalpy term \((\Delta H)\) or the entropy term \((-T \Delta S) ?(\mathbf{c})\) Explain how the preceding reactions are an example of driving a nonspontaneous reaction, as discussed in the "Chemistry and Life" box in Section 19.7 . (d) The reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\) to form \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{H}_{2} \mathrm{O}\) must be carried out carefully to avoid a competing reaction. What is the most likely competing reaction?
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