/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 86 Ammonium nitrate dissolves spont... [FREE SOLUTION] | 91Ó°ÊÓ

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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, because the process is spontaneous and endothermic.

Step by step solution

01

Understand Gibbs Free Energy Formula

To find the sign of \(\Delta S\), we can use the Gibbs free energy formula, which relates the change in entropy and the change in enthalpy of a process to its spontaneity. The formula 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, \(T\) is the temperature in Kelvin, and \(\Delta S\) is the change in entropy. A process is spontaneous if \(\Delta G<0\), nonspontaneous if \(\Delta G>0\), and at equilibrium if \(\Delta G=0\).
02

Identify the given data

The problem gives us the following information: - The dissolution of ammonium nitrate in water is spontaneous. - This process is endothermic. From this, we can deduce that: - \(\Delta G < 0\) (since the process is spontaneous) - \(\Delta H > 0\) (since the process is endothermic)
03

Use the data to determine the sign of delta_S

We know that \(\Delta G = \Delta H - T\Delta S\) and we need to determine the sign of \(\Delta S\). We can rearrange the formula to solve for \(\Delta S\): \[ \Delta S = \frac{\Delta H - \Delta G}{T} \] Since the process is spontaneous, \(\Delta G < 0\). And because the process is endothermic, \(\Delta H > 0\). So, the difference between \(\Delta H\) and \(\Delta G\) will also be positive: \[ \Delta H - \Delta G > 0 \] As the temperature is always positive (in Kelvin scale), the sign of \(\Delta S\) will be determined by the numerator \((\Delta H - \Delta G)\). Because \(\Delta H - \Delta G > 0\), \(\Delta S\) must be positive for this process to be spontaneous at room temperature.
04

Conclusion

The sign of \(\Delta S\) for the dissolution of ammonium nitrate in water at room temperature is positive, because the process is spontaneous and endothermic.

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Most popular questions from this chapter

Using data from Appendix \(C\), calculate the change in Gibbs free energy for each of the following reactions. In each case indicate whether the reaction is spontaneous under standard conditions. (a) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{HCl}(g)\) (b) \(\mathrm{MgCl}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{MgO}(s)+2 \mathrm{HCl}(g)\) (c) \(2 \mathrm{NH}_{3}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g)\) (d) \(2 \mathrm{NOCl}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)\)

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 \(\mathrm{Co}(s)\) is lowered from \(60^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\) (c) Ethyl alcohol evaporates from a beaker. (d) \(\mathrm{A}\) diatomic molecule dissociates into atoms. (e) A piece of charcoal is combusted to form \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).

A system goes from state 1 to state 2 and back to state 1 . (a) What is the relationship between the value of \(\Delta E\) for going from state 1 to state 2 to that for going from state 2 back to state \(1 ?\) (b) Without further information, can you conclude anything about the amount of heat transferred to the system as it goes from state 1 to state 2 as compared to that upon going from state 2 back to state \(1 ?\) (c) Suppose the changes in state are reversible processes. Can you conclude anything about the work done by the system upon going from state 1 to state 2 as compared to that upon going from state 2 back to state \(1 ?\)

Consider what happens when a sample of the explosive TNT (Section 8.8: "Chemistry Put to Work: Explosives and Alfred Nobel") is detonated. (a) Is the detonation a spontaneous process? (b) What is the sign of \(q\) for this process? (c) Can you determine whether \(w\) is positive, negative, or zero for the process? Explain. (d) Can you determine the sign of \(\Delta E\) for the process? Explain.

How would each of the following changes affect the number of microstates available to a system: (a) increase in temperature, (b) decrease in volume, (c) change of state from liquid to gas?

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