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 dissolution of ammonium nitrate in water is an endothermic and spontaneous process at room temperature. Based on the Gibbs free energy equation (\(\Delta G = \Delta H - T\Delta S\)), and given that \(\Delta G < 0\) and \(\Delta H > 0\), we can deduce that the change in entropy (ΔS) for this dissolution process is positive (ΔS > 0).

Step by step solution

01

Understanding the relationship between Gibbs free energy, enthalpy, and entropy

The Gibbs free energy change (ΔG) is related to the changes in enthalpy (ΔH) and entropy (ΔS) by the following equation: \[ \Delta G = \Delta H - T \Delta S\] where T is the temperature in Kelvin. For a spontaneous process, ΔG must be negative.
02

Analyzing given information

We are given that the dissolution of ammonium nitrate in water is endothermic, which means that heat is absorbed in the process, and thus ΔH > 0. The process is also spontaneous at room temperature, so ΔG < 0.
03

Deduce the sign of the change in entropy (ΔS)

As per the given information, ΔH > 0 and ΔG < 0. Now, let's substitute these conditions in the Gibbs free energy equation: \[\Delta G = \Delta H - T \Delta S < 0\] Since ΔH > 0, we need ΔS to be positive in order for ΔG to be negative. Therefore, we can deduce that the change in entropy (ΔS) for this dissolution process is positive (ΔS > 0).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Indicate whether \(\Delta G\) increases, decreases, or stays the same for each of the following reactions as the partial pressure of \(\mathrm{O}_{2}\) is increased: (a) \(\mathrm{HgO}(s) \longrightarrow \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) (b) $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)$ (c)

(a) Write the chemical equations that correspond to \(\Delta G_{i}^{9}\) for \(\mathrm{CH}_{4}(g)\) and for \(\mathrm{NaCl}(s) .\) (b) For these formation reactions, compare \(\Delta G_{f}^{\circ}\) and \(\Delta H_{f}\). (c) In general, under which condition is \(\Delta G\), more negative (less positive) than \(\Delta H_{f}\) ? (i) When the temperature is high, (ii) when \(\Delta S_{f}^{\circ}\) is positive, (iii) when the reaction is reversible.

In each of the following pairs, which compound would you expect to have the higher standard molar entropy: (a) \(\mathrm{C}_{3} \mathrm{H}_{\mathrm{s}}(g)\) or $\mathrm{C}_{4} \mathrm{H}_{10}(g)$, (b) \(\mathrm{C}_{4} \mathrm{H}_{10}(l)\) or \(\mathrm{C}_{4} \mathrm{H}_{10}(g)\)

Which of the following processes are spontaneous and which are nonspontaneous: (a) mixing of water and ethanol, \((\mathbf{b})\) dissolution of sugar in a cup of hot coffee, (c) formation of oxygen atoms from \(\mathrm{O}_{2}\) molecules at \(\mathrm{STP}\), (d) rusting of iron, (e) formation of glucose from \(\mathrm{CO}_{2}\) and $\mathrm{H}_{2} \mathrm{O}\( at \)\mathrm{STP} ?$

The normal boiling point of the element mercury (Hg) is $356.7{ }^{\circ} \mathrm{C},\( and its molar enthalpy of vaporization is \)\Delta H_{\text {vap }}=59.11 \mathrm{~kJ} / \mathrm{mol} .$ (a) When Hg boils at its nor- mal boiling point, does its entropy increase or decrease? (b) Calculate the value of \(\Delta S\) when \(2.00 \mathrm{~mol}\) of \(\mathrm{Hg}\) is vaporized at \(356.7^{\circ} \mathrm{C}\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free