Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Problem 6

Which substance in each of the following pairs would have the greater entropy? Explain. (a) at \(75^{\circ} \mathrm{C}\) and 1 atm: \(1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}(1)\) or \(1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (b) at \(5^{\circ} \mathrm{C}\) and 1 atm: \(50.0 \mathrm{g} \mathrm{Fe}(\mathrm{s})\) or \(0.80 \mathrm{mol} \mathrm{Fe}(\mathrm{s})\) (c) 1 mol \(\mathrm{Br}_{2}\left(1,1 \text { atm }, 8^{\circ} \mathrm{C}\right)\) or \(1 \mathrm{mol} \mathrm{Br}_{2}(\mathrm{s}, 1 \mathrm{atm},\) \(\left.-8^{\circ} \mathrm{C}\right)\) (d) \(0.312 \mathrm{mol} \mathrm{SO}_{2}\left(\mathrm{g}, 0.110 \mathrm{atm}, 32.5^{\circ} \mathrm{C}\right)\) or \(0.284 \mathrm{mol}\) \(\mathrm{O}_{2}\left(\mathrm{g}, 15.0 \mathrm{atm}, 22.3^{\circ} \mathrm{C}\right)\)

Short Answer

Expert verified
For each pair, the substance with greater entropy is: (a) 1 mol H2O(g), (b) 0.80 mol Fe(s), (c) 1 mol Br2(l), (d) 0.312 mol SO2(g).

Step by step solution

01

Understand entropy and factors

Entropy is essentially the measure of the disorder or randomness of a system. The greater the disorder, the greater the entropy. The factors that increase entropy include increasing temperature, the number of particles and transitioning from a solid to a liquid or a gas.
02

Analyze Part (a)

In part (a), we have 1 mol H2O in liquid (l) state and 1 mol H2O in gaseous (g) state at the same temperature and pressure. It's important to know that at the same temperature and pressure, gases have more entropy than liquids due to higher disorder. Therefore, 1 mol H2O(g) will have more entropy.
03

Analyze Part (b)

In part (b), the comparisons are between 50.0 g Fe(s) and 0.80 mol Fe(s). The one with more moles or particles will have more entropy as the increase in the number of particles increases disorder. Therefore, 0.80 mol Fe(s), which is approximately 44.8 grams, will have more entropy than 50.0 g Fe(s).
04

Analyze Part (c)

In part (c), we compare the entropy of 1 mol Br2 in the liquid state at 8°C and 1 mol Br2 in the solid state at -8°C. As a general rule, transitioning from solid to liquid or gas increases the entropy. Therefore, 1 mol Br2 (l) has greater entropy than 1 mol Br2(s).
05

Analyze Part (d)

In part (d), we compare 0.312 mol SO2(g) at 0.110 atm and 32.5°C and 0.284 mol O2(g) at 15.0 atm and 22.3°C. The gas with the higher number of moles and the higher temperature will have more entropy, which is 0.312 mol SO2(g).

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.

Start your free trial

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

The standard molar entropy of solid hydrazine at its melting point of \(1.53^{\circ} \mathrm{C}\) is \(67.15 \mathrm{Jmol}^{-1} \mathrm{K}^{-1}\). The enthalpy of fusion is \(12.66 \mathrm{kJmol}^{-1} .\) For \(\mathrm{N}_{2} \mathrm{H}_{4}(1)\) in the interval from \(1.53^{\circ} \mathrm{C}\) to \(298.15 \mathrm{K}\), the molar heat capacity at constant pressure is given by the expression \(C_{p}=97.78+0.0586(T-280) .\) Determine the standard molar entropy of \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l})\) at \(298.15 \mathrm{K}\). [Hint: The heat absorbed to produce an infinitesimal change in the temperature of a substance is \(d q_{\mathrm{rev}}=C_{p} d T\).

To establish the law of conservation of mass, Lavoisier carefully studied the decomposition of mercury(II) oxide: $$\mathrm{HgO}(\mathrm{s}) \longrightarrow \mathrm{Hg}(1)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ At \(25^{\circ} \mathrm{C}, \Delta H^{\circ}=+90.83 \mathrm{kJ}\) and \(\Delta G^{\circ}=+58.54 \mathrm{kJ}\) (a) Show that the partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) in equilibrium with \(\mathrm{HgO}(\mathrm{s})\) and \(\mathrm{Hg}(\mathrm{l})\) at \(25^{\circ} \mathrm{C}\) is extremely low. (b) What conditions do you suppose Lavoisier used to obtain significant quantities of oxygen?

A tabulation of more precise thermodynamic data than are presented in Appendix D lists the following values for \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(298.15 \mathrm{K},\) at a standard state pressure of 1 bar. $$\begin{array}{llll} \hline & \Delta H_{f}^{\circ}, & \Delta G_{f,}^{\circ} & S_{\prime}^{\circ} \\\ & \text { kJ mol }^{-1} & \text {kJ mol }^{-1} & \text {J mol }^{-1} \text {K }^{-1} \\ \hline \mathrm{H}_{2} \mathrm{O}(1) & -285.830 & -237.129 & 69.91 \\ \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) & -241.818 & -228.572 & 188.825 \\ \hline \end{array}$$ (a) Use these data to determine, in two different ways, \(\Delta G^{\circ}\) at \(298.15 \mathrm{K}\) for the vaporization: \(\mathrm{H}_{2} \mathrm{O}(1,1 \mathrm{bar}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g}, 1 \mathrm{bar}) .\) The value you obtain will differ slightly from that on page 838 because here, the standard state pressure is 1 bar, and there, it is 1 atm. (b) Use the result of part (a) to obtain the value of \(K\) for this vaporization and, hence, the vapor pressure of water at \(298.15 \mathrm{K}\) (c) The vapor pressure in part (b) is in the unit bar. Convert the pressure to millimeters of mercury. (d) Start with the value \(\Delta G^{\circ}=8.590 \mathrm{kJ}\), given on page 838 and calculate the vapor pressure of water at 298.15 K in a fashion similar to that in parts (b) and (c). In this way, demonstrate that the results obtained in a thermodynamic calculation do not depend on the convention we choose for the standard state pressure, as long as we use standard state thermodynamic data consistent with that choice.

For each of the following reactions, indicate whether \(\Delta S\) for the reaction should be positive or negative. If it is not possible to determine the sign of \(\Delta S\) from the information given, indicate why. (a) \(\mathrm{CaO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})\) (b) \(2 \mathrm{HgO}(\mathrm{s}) \longrightarrow 2 \mathrm{Hg}(1)+\mathrm{O}_{2}(\mathrm{g})\) (c) \(2 \mathrm{NaCl}(1) \longrightarrow 2 \mathrm{Na}(1)+\mathrm{Cl}_{2}(\mathrm{g})\) (d) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{g})\) (e) \(\operatorname{Si}\left(\text { s) }+2 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{SiCl}_{4}(\mathrm{g})\right.\)

Consider the vaporization of water: \(\mathrm{H}_{2} \mathrm{O}(1) \longrightarrow\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(100^{\circ} \mathrm{C},\) with \(\mathrm{H}_{2} \mathrm{O}(1)\) in its standard state, but with the partial pressure of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(2.0 \mathrm{atm}\) Which of the following statements about this vaporization at \(100^{\circ} \mathrm{C}\) are true? (a) \(\Delta G^{\circ}=0,\) (b) \(\Delta G=0\) (c) \(\Delta G^{\circ}>0,\) (d) \(\Delta G>0 ?\) Explain.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation
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