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Chapter 15: Problem 56

We can represent the freezing of \(\mathrm{H}_{2} \mathrm{O}(1)\) at \(0^{\circ} \mathrm{C}\ \mathrm{as} \mathrm{H}_{2} \mathrm{O}\) \(\left(1, d=1.00 \mathrm{g} / \mathrm{cm}^{3}\right) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}\left(\mathrm{s}, d=0.92 \mathrm{g} / \mathrm{cm}^{3}\right) . \quad \mathrm{Ex}\) plain why increasing the pressure on ice causes it to melt. Is this the behavior you expect for solids in general? Explain.

Short Answer

Expert verified
Increasing the pressure on ice causes it to melt because the increased pressure favors the phase with the higher density. Water is unique in that its solid form (ice) is less dense than its liquid form because of the way its molecules are arranged. This characteristic causes it to behave differently than most solid substances under pressure.

Step by step solution

01

Understanding Phase Transitions

First, consider the phase diagram of water which shows the changes in states of matter (solid, liquid, gas) under varying temperature and pressure. At 0°C, where the freezing point of water is, we can observe a transition from liquid water (1.00 g/cm³) to ice (0.92 g/cm³). This transition is unique due to the density difference between water and ice.
02

Understanding the Density Difference

The density of liquid water is higher than that of ice. This is because the molecules in ice are arranged in a crystal lattice that leaves relatively large gaps between them. These gaps are not present in the structure of liquid water, which allows more molecules to occupy a given volume, increasing its density.
03

Understanding the Effect of Pressure

If pressure is applied to this system, it will favour the phase with the higher density. As liquid water has a higher density than ice, increasing the pressure will cause the ice to transition back into a liquid hence melting the ice. This is due to Le Chatelier's Principle, which states that a system in equilibrium will adjust to counteract the change in conditions (in this case, the increase in pressure).
04

Comparing to Other Solids

This behavior is not common in most solids. Most solids have a higher density than their liquid forms, so increasing pressure typically favours the solid phase. Water is a unique exception due to its lower density when solid versus liquid.

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Most popular questions from this chapter

A mixture consisting of \(0.150 \mathrm{mol} \mathrm{H}_{2}\) and \(0.150 \mathrm{mol} \mathrm{I}_{2}\) is brought to equilibrium at \(445^{\circ} \mathrm{C},\) in a 3.25 L flask. What are the equilibrium amounts of \(\mathrm{H}_{2}, \mathrm{I}_{2},\) and HI? $$\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) \quad K_{\mathrm{c}}=50.2\ \mathrm\ {at}\ 445^{\circ} \mathrm{C}$$

In the reaction \(2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}), 0.455\) \(\mathrm{mol} \mathrm{SO}_{2}, 0.183 \mathrm{mol} \mathrm{O}_{2},\) and \(0.568 \mathrm{mol} \mathrm{SO}_{3}\) are introduced simultaneously into a 1.90 L vessel at \(1000 \mathrm{K}\). (a) If \(K_{c}=2.8 \times 10^{2},\) is this mixture at equilibrium? (b) If not, in which direction will a net change occur?

Based on these descriptions, write a balanced equation and the corresponding \(K_{c}\) expression for each reversible reaction. (a) Carbonyl fluoride, \(\mathrm{COF}_{2}(\mathrm{g}),\) decomposes into gaseous carbon dioxide and gaseous carbon tetrafluoride. (b) Copper metal displaces silver(I) ion from aqueous solution, producing silver metal and an aqueous solution of copper(II) ion. (c) Peroxodisulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\), oxidizes iron(II) ion to iron(III) ion in aqueous solution and is itself reduced to sulfate ion.

The following is an approach to establishing a relationship between the equilibrium constant and rate constants mentioned in the section on page 660 \(\bullet\)Work with the detailed mechanism for the reaction. \(\bullet\) Use the principle of microscopic reversibility, the idea that every step in a reaction mechanism is reversible. (In the presentation of elementary reactions in Chapter \(14,\) we treated some reaction steps as reversible and others as going to completion. However, as noted in Table \(15.3,\) every reaction has an equilibrium constant even though a reaction is generally considered to go to completion if its equilibrium constant is very large.) \(\bullet\) Use the idea that when equilibrium is attained in an overall reaction, it is also attained in each step of its mechanism. Moreover, we can write an equilibrium constant expression for each step in the mechanism, similar to what we did with the steady-state assumption in describing reaction mechanisms. \(\bullet\)Combine the \(K_{\mathrm{c}}\) expressions for the elementary steps into a \(K_{\mathrm{c}}\) expression for the overall reaction. The numerical value of the overall \(K_{c}\) can thereby be expressed as a ratio of rate constants, \(k\) Use this approach to establish the equilibrium constant expression for the overall reaction, $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) $$ The mechanism of the reaction appears to be the following: Fast: \(\quad \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\) Slow: \(\quad 2 \mathrm{I}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)

Determine values of \(K_{c}\) from the \(K_{p}\) values given. (a) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) \(K_{\mathrm{p}}=2.9 \times 10^{-2} \mathrm{at} 303 \mathrm{K}\) (b) \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g})\) \(K_{\mathrm{p}}=1.48 \times 10^{4} \mathrm{at} 184^{\circ} \mathrm{C}\) (c) \(\mathrm{Sb}_{2} \mathrm{S}_{3}(\mathrm{s})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Sb}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) \(K_{\mathrm{p}}=0.429\) at \(713 \mathrm{K}\)
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