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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
In this exercise, we are asked to deduce the sign of the change in entropy, \(\Delta S\), for the dissolution of ammonium nitrate in water. The dissolution is spontaneous and endothermic at room temperature. We know that for a process to be spontaneous, the Gibbs Free Energy change, \(\Delta G\), must be negative. Furthermore, the enthalpy change, \(\Delta H\), is positive since the process is endothermic. Analyzing the \(\Delta G\) equation, we can deduce that the sign of \(\Delta S\) must be positive, as this allows \(\Delta G\) to be negative, which is necessary for a spontaneous process.

Step by step solution

01

Identify the given information

We know that the dissolution of ammonium nitrate in water is spontaneous and endothermic at room temperature. Spontaneous means that the process occurs without any external influence, and endothermic means that it absorbs heat from its surroundings.
02

Use the Gibbs Free Energy formula

The Gibbs Free Energy change, \(\Delta G\), can be used to determine the spontaneity of a process. The formula for Gibbs Free Energy change is: \[ \Delta G = \Delta H - T\Delta S \] where \(\Delta G\) is the change in Gibbs Free Energy, \(\Delta H\) is the change in enthalpy (heat), \(\Delta S\) is the change in entropy, and \(T\) is the temperature in Kelvin.
03

Determine the sign of the enthalpy change

Since we are given that the dissolution is endothermic, this means that the process absorbs heat from its surroundings. Therefore, the enthalpy change, \(\Delta H\), is positive.
04

Analyze the spontaneity

For a process to be spontaneous, the Gibbs Free Energy change, \(\Delta G\), must be negative. Since we know that \(\Delta H\) is positive in this case, we can deduce the sign of \(\Delta S\) by analyzing the \(\Delta G\) equation: - If \(\Delta S\) is negative, the \(-T\Delta S\) term would be positive, making \(\Delta G\) positive and the process non-spontaneous. - If \(\Delta S\) is positive, the \(-T\Delta S\) term would be negative, and since \(\Delta H\) is also positive, the overall \(\Delta G\) could be negative and the process can be spontaneous.
05

Conclusion

Based on the given information, we can deduce that the sign of the change in entropy, \(\Delta S\), for the dissolution of ammonium nitrate in water must be positive. This is because it allows the Gibbs Free Energy change, \(\Delta G\), to be negative, which is necessary for a spontaneous process.

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Most popular questions from this chapter

Using the data in Appendix \(\mathrm{C}\) and given the pressures listed, calculate \(K_{\rho}\) and \(\Delta G\) for each of the following reactions: (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) \(R_{\mathrm{N}_{2}}=2.6 \mathrm{~atm}, P_{\mathrm{H}_{2}}=5.9 \mathrm{~atm}, P_{\mathrm{NH}_{5}}=1.2 \mathrm{~atm}\) (b) \(2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) \(P_{\mathrm{N}_{2} \mathrm{H}_{4}}=\mathrm{PNo}_{1}=5.0 \times 10^{-2} \mathrm{~atm}\). \(P_{\mathrm{N}_{2}}=0.5 \mathrm{~atm}, P_{\mathrm{H}_{2} O}=0.3 \mathrm{~atm}\) (c) \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{H}_{2}(\mathrm{~g})\) \(P_{\mathrm{N}_{2} \mathrm{H}_{4}}=0.5 \mathrm{~atm}, P_{\mathrm{N}_{2}}=1.5 \mathrm{~atm}, P_{\mathrm{H}_{1}}=2.5 \mathrm{~atm}\)

(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 \(T\).

Indicate whether each statement is true or false. (a) \(\Delta S\) for an isothermal process depends on both the temperature and the amount of heat reversibly transferred. (b) \(\Delta S\) is a state function. (c) The second law of thermodynamics says that the entropy of the system increases for all spontaneous processes.

(a) For each of the following reactions, predict the sign of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) without doing any calculations. (b) Based on your general chemical knowledge, predict which of these reactions will have \(K>1\). (c) In each case indicate whether \(K\) should increase or decrease with increasing temperature. (i) \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{MgO}(s)\) (ii) \(2 \mathrm{KI}(s) \rightleftharpoons 2 \mathrm{~K}(g)+\mathrm{I}_{2}(g)\) (iii) \(\mathrm{Na}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Na}(\mathrm{g})\) (iv) \(2 \mathrm{~V}_{2} \mathrm{O}_{5}(s) \rightleftharpoons 4 \mathrm{~V}(\mathrm{~s})+5 \mathrm{O}_{2}(\mathrm{~g})\)

For each of the following processes, indicate whether the signs of \(\Delta S\) and \(\Delta H\) are expected to be positive, negative, or about zero. (a) A solid sublimes. (b) The temperature of a sample of
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