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

Short Answer

Expert verified
Increasing temperature increases molecular kinetic energy, favoring evaporation and increasing vapor pressure.

Step by step solution

01

Understanding Le Châtelier's Principle

Le Châtelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium will shift to counteract the change. In the context of vapor pressure, the system will try to alleviate any imposed change in temperature.
02

Analyzing the Effect of Temperature on Vapor Pressure

When the temperature of a liquid is increased, the kinetic energy of its molecules also increases. This makes it easier for molecules to escape from the liquid phase into the vapor phase, thereby increasing the vapor pressure.
03

Applying Le Châtelier's Principle

According to Le Châtelier's principle, heating the system encourages it to oppose that change by favoring the endothermic process. Evaporation is an endothermic process (it absorbs heat), so increasing temperature shifts the equilibrium towards more evaporation, raising vapor pressure.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Vapor Pressure
Equilibrium vapor pressure is a fundamental concept that explains the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases. At a given temperature, the vapor pressure is constant because the rate of evaporation of liquid molecules equals the rate of condensation. Essentially, it is the pressure at which a liquid's molecules transition equally between liquid and vapor phases without net change in mass. This aspect of equilibrium indicates that the vapor pressure depends heavily on temperature. As molecules move between phases, the system reaches a point where neither evaporation nor condensation is favored more. This equilibrium state is crucial in understanding how different liquids behave under various temperature conditions.
Temperature Effect
Temperature plays a pivotal role in altering equilibrium vapor pressure. As temperature increases, more heat is provided to the liquid molecules, resulting in increased kinetic energy. This additional energy helps more molecules overcome intermolecular forces, and thus they escape into the vapor phase more readily. In essence, when you heat a liquid, you supply energy that aids in breaking the bonds holding the molecules in the liquid state. The effect of temperature is ultimately reflected in Le Châtelier's principle, as the system attempts to counterbalance the increase in temperature by favoring the evaporation process. Consequently, higher temperatures lead to increased vapor pressures because the system shifts in such a way to minimize this disturbance in equilibrium.
Kinetic Energy
Kinetic energy is the energy of motion, and it significantly impacts the evaporation process. When a liquid is heated, the kinetic energy of its molecules increases, causing them to move more vigorously. This increase in molecular motion facilitates the breaking of intermolecular bonds, allowing more molecules to escape from the liquid phase into the gaseous phase. Higher kinetic energy also means that a greater fraction of the molecules possess the necessary energy threshold to overcome the attractive forces keeping them in the liquid. This not only raises the vapor pressure but also explains why temperature is an essential factor in phase transitions. Understanding kinetic energy's role helps in grasping the dynamic nature of molecular interactions and phase changes.
Evaporation Process
Evaporation is a process where liquid molecules transition into the gaseous phase. It is inherently endothermic, meaning it absorbs energy from its surroundings to occur. During evaporation, only molecules with sufficient kinetic energy can break free from the liquid surface. This process explains why a liquid's surface is often cooler after its molecules evaporate. In the context of temperature and vapor pressure, evaporation accelerates with a rise in temperature due to increased kinetic energy. According to Le Châtelier's principle, increasing temperature shifts equilibrium to favor evaporation, since it is an endothermic reaction that alleviates the added energy. Therefore, understanding evaporation is pivotal when analyzing humidity, boiling points, and overall atmospheric conditions.

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Most popular questions from this chapter

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction: $$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftarrows 2 \mathrm{HI}(g)$$ is 54.3 at \(430^{\circ} \mathrm{C}\). At the start of the reaction, there are \(0.714 \mathrm{~mol}\) of \(\mathrm{H}_{2}, 0.984 \mathrm{~mol}\) of \(\mathrm{I}_{2}\), and \(0.886 \mathrm{~mol}\) of HI in a 2.40-L reaction chamber. Calculate the concentrations of the gases at equilibrium.

The aqueous reaction: L-glutamate \(+\) pyruvate \(\rightleftarrows \alpha\) -ketoglutarate \(+\mathrm{L}\) -alanine is catalyzed by the enzyme \(\mathrm{L}\) -glutamate-pyruvate aminotransferase. At \(300 \mathrm{~K},\) the equilibrium constant for the reaction is 1.11 . Predict whether the forward reaction will occur if the concentrations of the reactants and products are [L-glutamate] \(=3.0 \times 10^{-5} \mathrm{M}\), [pyruvate] \(=3.3 \times 10^{-4} M,[\alpha\) -ketoglutarate \(]=1.6 \times 10^{-2} M\), and \([\mathrm{L}\) -alanine \(]=6.25 \times 10^{-3} \mathrm{M}\)

A mixture containing 3.9 moles of \(\mathrm{NO}\) and 0.88 mole of \(\mathrm{CO}_{2}\) was allowed to react in a flask at a certain temperature according to the equation: $$ \mathrm{NO}(g)+\mathrm{CO}_{2}(g) \rightleftarrows \mathrm{NO}_{2}(g)+\mathrm{CO}(g) $$ At equilibrium, 0.11 mole of \(\mathrm{CO}_{2}\) was present. Calculate the equilibrium constant \(K_{\mathrm{c}}\) of this reaction.

At \(1000 \mathrm{~K},\) a sample of pure \(\mathrm{NO}_{2}\) gas decomposes: $$ 2 \mathrm{NO}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ The equilibrium constant \(K_{P}\) is 158 . Analysis shows that the partial pressure of \(\mathrm{O}_{2}\) is 0.25 atm at equilibrium. Calculate the pressure of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) in the mixture.

For which of the following reactions is \(K_{\mathrm{c}}\) equal to \(K_{P}\) ? For which can we not write a \(K_{P}\) expression? (a) \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftarrows 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) (b) \(\mathrm{CaCO}_{3}(s) \rightleftarrows \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\) (c) \(\mathrm{Zn}(s)+2 \mathrm{H}^{+}(a q) \rightleftarrows \mathrm{Zn}^{2+}(a q)+\mathrm{H}_{2}(g)\) (d) \(\mathrm{PCl}_{3}(g)+3 \mathrm{NH}_{3}(g) \rightleftarrows 3 \mathrm{HCl}(g)+\mathrm{P}\left(\mathrm{NH}_{2}\right)_{3}(g)\) (e) \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \rightleftarrows \mathrm{NH}_{4} \mathrm{Cl}(s)\) (f) \(\mathrm{NaHCO}_{3}(s)+\mathrm{H}^{+}(a q) \rightleftarrows \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)+\) \(\mathrm{Na}^{+}(a q)\) (g) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftarrows 2 \mathrm{HF}(g)\) (h) \(\mathrm{C}(\) graphite \()+\mathrm{CO}_{2}(g) \rightleftarrows 2 \mathrm{CO}(g)\)

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