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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

Use data in Appendix \(\mathrm{C}\) to calculate \(\Delta H^{\circ}, \Delta S^{\circ},\) and \(\Delta G^{\circ}\) at \(25^{\circ} \mathrm{C}\) for each of the following reactions. In each case show that \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}\) (a) \(2 \mathrm{Cr}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CrO}_{3}(s)\) (b) \(\mathrm{BaCO}_{3}(s) \longrightarrow \mathrm{BaO}(s)+\mathrm{CO}_{2}(g)\) (c) \(2 \mathrm{P}(s)+10 \mathrm{HF}(g) \longrightarrow 2 \mathrm{PF}_{5}(g)+5 \mathrm{H}_{2}(g)\) (d) \(\mathrm{K}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{KO}_{2}(s)\)

The fuel in high-efficiency natural gas vehicles consists primarily of methane \(\left(\mathrm{CH}_{4}\right) .\) (a) How much heat is produced in burning 1 mol of \(\mathrm{CH}_{4}(g)\) under standard conditions if reactants and products are brought to \(298 \mathrm{~K}\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) is formed? (b) What is the maximum amount of useful work that can be accomplished under standard conditions by this system?

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) \(3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6}(g)\)

Using data in Appendix C, calculate \(\Delta H^{\circ}, \Delta S^{\circ},\) and \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) for each of the following reactions. In each case show that \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} .\) (a) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)\) (b) \(\mathrm{C}(s,\) graphite \()+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(g)\) (c) \(2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{POCl}_{3}(g)\) (d) \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\)

The element gallium (Ga) freezes at \(29.8^{\circ} \mathrm{C},\) and its molar enthalpy of fusion is \(\Delta H_{\text {fus }}=5.59 \mathrm{~kJ} / \mathrm{mol}\). (a) When molten gallium solidifies to \(\mathrm{Ga}(s)\) at its normal melting point, is \(\Delta S\) positive or negative? (b) Calculate the value of \(\Delta S\) when \(60.0 \mathrm{~g}\) of \(\mathrm{Ga}(l)\) solidifies at \(29.8^{\circ} \mathrm{C}\)

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