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Consider the vaporization of liquid water to steam at a pressure of 1 atm. (a) Is this process endothermic or exothermic? (b) In what temperature range is it a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) At what temperature are the two phases in equilibrium?

Short Answer

Expert verified
a) The vaporization of liquid water to steam is an endothermic process. b) It is spontaneous for temperatures higher than \(\frac{\Delta H}{\Delta S}\). c) It is nonspontaneous for temperatures lower than \(\frac{\Delta H}{\Delta S}\). d) The two phases are in equilibrium at 100°C (373.15 K) at 1 atm pressure.

Step by step solution

01

a) Endothermic or Exothermic

To answer this question, we need to think about the process of vaporization. When liquid water turns into steam, it needs to absorb energy in the form of heat to overcome the intermolecular forces holding the water molecules together in the liquid phase. As the water absorbs heat, the process is endothermic.
02

b) Spontaneous Temperature Range

To determine if the process is spontaneous, we will consider the sign of Gibbs free energy change (ΔG). If ΔG is negative, the process is spontaneous. The equation for Gibbs free energy change is: ΔG = ΔH - TΔS, where ΔH is the enthalpy change, T is the temperature in Kelvin, and ΔS is the entropy change. In the case of vaporization, we know that ΔH > 0 (since it's endothermic) and ΔS > 0 (since gases have higher entropy than liquids). So, for the process to be spontaneous (ΔG < 0), we need: 0 > ΔH - TΔS, which implies: T > ΔH/ΔS. Thus, the process is spontaneous for temperatures higher than ΔH/ΔS.
03

c) Nonspontaneous Temperature Range

For nonspontaneous processes, ΔG > 0. Using the same Gibbs free energy change equation (ΔG = ΔH - TΔS), we can write the condition for nonspontaneity as: 0 < ΔH - TΔS, which implies: T < ΔH/ΔS. Thus, the process is nonspontaneous for temperatures lower than ΔH/ΔS.
04

d) Equilibrium Temperature

At equilibrium, the Gibbs free energy change (ΔG) is zero; both phases coexist, and the rate of vaporization equals the rate of condensation. Using the Gibbs free energy change equation (ΔG = ΔH - TΔS), we can write the condition for equilibrium as: 0 = ΔH - TΔS, which implies: T = ΔH/ΔS. The Clapeyron equation is often used to find the equilibrium temperature, but it requires additional information such as the specific enthalpy and entropy changes for the process or the saturation vapor pressure at different temperatures. In the case of water, at 1 atm pressure, the equilibrium temperature between liquid water and steam is 373.15 K (100°C), which is the normal boiling point of water. In summary: a) The vaporization of liquid water to steam is an endothermic process. b) It is spontaneous for temperatures higher than ΔH/ΔS. c) It is nonspontaneous for temperatures lower than ΔH/ΔS. d) The two phases are in equilibrium at 100°C (373.15 K) at 1 atm pressure.

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Key Concepts

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

Endothermic process
When we talk about an endothermic process, we refer to a reaction or transformation that absorbs energy from its surroundings in the form of heat. This concept is quite significant in understanding how substances like water change from one phase to another. During vaporization, when water turns into steam, energy is needed to overcome the intermolecular forces that hold water molecules together in liquid form.
  • The absorbed energy is used to break these bonds, allowing molecules to move more freely in the gaseous state.
  • This absorption of heat means the process is endothermic; it requires an input of energy from the environment.
Thus, when we observe boiling water and the steam rising above it, we are witnessing the classic example of an endothermic process in action. The water absorbs heat from its surroundings to transition from liquid to vapor.
Gibbs free energy
In thermodynamics, Gibbs free energy is a critical concept for predicting whether a process can occur spontaneously. It combines enthalpy, entropy, and temperature to determine the outcome of a reaction. The equation used is:\[\Delta G = \Delta H - T\Delta S\]In this equation:
  • \( \Delta G \) represents the change in Gibbs free energy.
  • \( \Delta H \) is the change in enthalpy — the heat absorbed or released.
  • \( T \) is the temperature in Kelvin.
  • \( \Delta S \) refers to the change in entropy — the measure of disorder or randomness.
For vaporization:
  • \( \Delta H \) is positive as it requires heat absorption.
  • \( \Delta S \) is also positive, since gases exhibit higher entropy than liquids.
The key to spontaneity lies in \( \Delta G \). If it is negative, the process will proceed spontaneously under the given conditions.
Equilibrium temperature
The equilibrium temperature is the point at which two phases coexist in balance. This concept can be illustrated using the vaporization of water. At equilibrium:
  • The speed at which liquid water turns into steam equals the speed at which steam condenses back into water.
  • The Gibbs free energy change \( (\Delta G) \) is zero, meaning the process is perfectly balanced.
To find this temperature, the equation:\[0 = \Delta H - T\Delta S\]rearranges to:\[T = \frac{\Delta H}{\Delta S}\]This equation shows that at equilibrium, temperature results from the balance between enthalpy and entropy changes. For water at 1 atm pressure, this equilibrium temperature is traditionally known as the boiling point — 100°C or 373.15 K.
Spontaneous process
A process is considered spontaneous when it occurs naturally without the need for external energy. Using Gibbs free energy as our guide, we identify spontaneity by a negative \( \Delta G \). For the vaporization of water:
  • Spontaneity occurs at temperatures where the system's entropy increase and absorbed heat lead to a decrease in free energy (\( \Delta G < 0 \)).
  • Above a specific temperature, determined by \( T > \frac{\Delta H}{\Delta S} \), the vaporization becomes spontaneous.
Therefore, when water boils at temperatures above its boiling point, it naturally turns to steam without external influences. This inherent drive towards greater disorder — from more orderly liquid to more chaotic vapor — is a hallmark of spontaneous processes in chemistry.

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Most popular questions from this chapter

Use Appendix \(\mathrm{C}\) to compare the standard entropies at \(25^{\circ} \mathrm{C}\) for the following pairs of substances: (a) \(\mathrm{Sc}(s)\) and \(\mathrm{Sc}(g)\), \(\mathrm{NH}_{3}(g)\) and \(\mathrm{NH}_{3}(a q)\) (c) \(1 \mathrm{~mol} \mathrm{P}_{4}(g)\) and \(2 \mathrm{~mol} \mathrm{P}_{2}(g)\), (d) C(graphite) and C(diamond). In each case explain the difference in the entropy values.

The conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\), is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2}(g) $$ In practice, this reaction is carried out in the presence of oxygen: $$ 2 \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) Using the data in Appendix \(C,\) calculate \(K\) for these reactions at \(25^{\circ} \mathrm{C}\) and \(500{ }^{\circ} \mathrm{C}\). (b) Is the difference in \(\Delta G^{\circ}\) for the two reactions due primarily to the enthalpy term \((\Delta H)\) or the entropy term \((-T \Delta S) ?(\mathbf{c})\) Explain how the preceding reactions are an example of driving a nonspontaneous reaction, as discussed in the "Chemistry and Life" box in Section 19.7 . (d) The reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\) to form \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{H}_{2} \mathrm{O}\) must be carried out carefully to avoid a competing reaction. What is the most likely competing reaction?

(a) What do you expect for the sign of \(\Delta S\) in a chemical reaction in which two moles of gaseous reactants are converted to three moles of gaseous products? (b) For which of the processes in Exercise 19.11 does the entropy of the system increase?

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be made by the controlled oxidation of methane: $$ \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g) $$ (a) Use data in Appendix C to calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. (b) How is \(\Delta G^{\circ}\) for the reaction expected to vary with increasing temperature? (c) Calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). Under standard conditions, is the reaction spontaneous at this temperature? (d) Is there a temperature at which the reaction would be at equilibrium under standard conditions and that is low enough so that the compounds involved are likely to be stable?

Suppose we vaporize a mole of liquid water at \(25^{\circ} \mathrm{C}\) and another mole of water at \(100{ }^{\circ} \mathrm{C}\). (a) Assuming that the enthalpy of vaporization of water does not change much between \(25^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C},\) which process involves the larger change in entropy? (b) Does the entropy change in either process depend on whether we carry out the process reversibly or not? Explain.

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