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Chapter 15: Problem 40

Use Le Ch芒telier's principle to explain why the equilibrium vapor pressure of a liquid increases with increasing temperature.

Short Answer

Expert verified
According to Le Ch芒telier's principle, a system at equilibrium will adjust to counteract any changes. When the temperature increases, the equilibrium of a liquid-vapor system shifts towards the vapor phase to counteract the added energy. This results in more molecules in the vapor phase, and thus an increase in the equilibrium vapor pressure.

Step by step solution

01

Understanding Le Ch芒telier's Principle

Le Ch芒telier's principle states that if a system in equilibrium is subjected to a change, the system will adjust itself in a way that counteracts the change. In the context of phase transitions, if the equilibrium between liquid and vapor phases is disturbed by a change in temperature, the system will counteract this by shifting either towards the liquid or vapor phase to re-establish equilibrium.
02

Equilibrium and Temperature

An increase in temperature adds energy to the system. In the context of a liquid-vapor equilibrium, this additional energy allows more liquid particles to overcome the intermolecular forces that keep them in the liquid phase, resulting in evaporation. Hence, more particles move to the gas phase.
03

Applying Le Ch芒telier's principle to Increasing Temperature

According to Le Ch芒telier's principle, the system will react to this increase in temperature by shifting the equilibrium to counteract the change. Since more energy results in more evaporation, the system counters this change by shifting the equilibrium towards the vapor phase. This shift implies an increase in the number of molecules in the vapor phase, which is reflected as an increase in the vapor pressure.

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Most popular questions from this chapter

At \(1024^{\circ} \mathrm{C}\), the pressure of oxygen gas from the decomposition of copper(II) oxide \((\mathrm{CuO})\) is \(0.49 \mathrm{~atm}\) $$ 4 \mathrm{CuO}(s) \rightleftharpoons 2 \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{O}_{2}(g) $$ (a) What is the \(K_{P}\) for the reaction? (b) Calculate the fraction of \(\mathrm{CuO}\) decomposed if 0.16 mole of it is placed in a 2.0 - \(L\) flask at \(1024^{\circ} \mathrm{C}\). (c) What would be the fraction if a 1.0 -mole sample of \(\mathrm{CuO}\) were used? (d) What is the smallest amount of \(\mathrm{CuO}\) (in moles) that would establish the equilibrium?

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ is \(3.8 \times 10^{-5}\) at \(727^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{c}}\) and \(K_{P}\) for the equilibrium $$ 2 \mathrm{I}(g) \rightleftharpoons \mathrm{I}_{2}(g) $$ at the same temperature.

For the reaction $$ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) $$ at \(700^{\circ} \mathrm{C}, K_{\mathrm{c}}=0.534 .\) Calculate the number of moles of \(\mathrm{H}_{2}\) formed at equilibrium if a mixture of 0.300 mole of \(\mathrm{CO}\) and 0.300 mole of \(\mathrm{H}_{2} \mathrm{O}\) is heated to \(700^{\circ} \mathrm{C}\) in a 10.0 - \(\mathrm{L}\) container.

The equilibrium constant \(K_{P}\) for the reaction $$ 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) $$ is found to be \(2 \times 10^{-42}\) at \(25^{\circ} \mathrm{C}\). (a) What is \(K_{\mathrm{c}}\) for the reaction at the same temperature? (b) The very small value of \(K_{P}\left(\right.\) and \(\left.K_{\mathrm{c}}\right)\) indicates that the reaction overwhelmingly favors the formation of water molecules. Explain why, despite this fact, a mixture of hydrogen and oxygen gases can be kept at room temperature without any change.

In 1899 the German chemist Ludwig Mond developed a process for purifying nickel by converting it to the volatile nickel tetracarbonyl \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) (b.p. \(=\) \(\left.42.2^{\circ} \mathrm{C}\right)\) $$ \mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) $$ (a) Describe how you can separate nickel and its solid impurities. (b) How would you recover nickel? \(\left[\Delta H_{\mathrm{f}}^{\circ}\right.\) for \(\mathrm{Ni}(\mathrm{CO})_{4}\) is \(\left.-602.9 \mathrm{kj} / \mathrm{mol} .\right]\)
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