Chemistry 112 Chapter 18

¡Supera tus tareas y exámenes ahora con Quizwiz!

18.8 What is meant by the standard free-energy change ∆G° for a reaction? What is meant by the standard free energy of formation DGf 8 of a substance?

. The standard free-energy change, ΔG°, equals ΔH° − TΔS°; that is, it is the difference between the standard enthalpy change of a system and the product of temperature and the standard entropy change of a system. The standard free-energy change of formation is the free-energy change when 1 mole of a substance is formed from its elements in their stable states at 1 atm and at a standard temperature, usually 25°C.

18.20 Given the following information at 25°C, calculate ∆G° at 25°C for the reaction 2A(g) + B(g)h = 3C(g) Substance ∆H (kJ/mol) S (J/mol∙K) A(g) 191 244 B(g) 70.8 300 C(g) −197 164 a −956 kJ b 956 kJ c −346 kJ d 346 kJ e −1.03 × 10^3 kJ

A

18.12 How is the concept of coupling of reactions useful in explaining how a nonspontaneous change could be made to occur?

A nonspontaneous reaction can be made to occur by coupling it with a spontaneous reaction having a sufficiently negative ΔG° to furnish the required energy. (The net ΔG° of the coupled reactions must be negative.)

18.1 What is a spontaneous process? Give three examples of spontaneous processes. Give three examples of nonspontaneous processes.

A spontaneous process is a chemical and/or a physical change that occurs by itself without the continuing intervention of an outside agency. Three examples are (1) a rock on a hilltop rolls down, (2) heat flows from a hot object to a cold one, and (3) iron rusts in moist air. Three examples of nonspontaneous processes are (1) a rock rolls uphill by itself, (2) heat flows from a cold object to a hot one, and (3) rust is converted to iron and oxygen.

18.13 Explain how the free energy changes as a spontaneous reaction occurs. Show by means of a diagram how G changes with the extent of reaction.

As a spontaneous reaction proceeds, the free energy decreases until equilibrium is reached at a minimum ΔG

18.2 Which contains greater entropy, a quantity of frozen benzene or the same quantity of liquid benzene at the same temperature? Explain in terms of the dispersal of energy in the substance.

Because the energy is more dispersed in liquids than in solids, liquid benzene contains more entropy than does the same quantity of frozen benzene.

18.14 Explain how an equilibrium constant can be obtained from thermal data alone (that is, from measurements of heat only).

Because the equilibrium constant is related to ΔH° and ΔS° by −RT ln K = ΔH° − TΔS°, heat measurements alone can be used to obtain it. The standard enthalpy, ΔH°, is the heat of reaction measured at constant pressure. The standard entropy change, ΔS°, can be calculated from standard entropies, which are obtained from heat-capacity data.

18.18 You run a reaction that has a negative entropy change and is exothermic. Assuming that the entropy and enthalpy do not change with temperature, you could predict that as you increase the temperature: I. the equilibrium shifts to favor the reaction products. II. the reaction becomes more spontaneous. III. ∆G for the reaction increases (becomes more positive). a I only b II only c III only d I and III only e II and III only

C

18.19 A reaction has a ∆G = −10.0 kJ and a ∆H = −20.0 kJ. If ∆S = −1.82 × 10^3 J/K, what was the temperature at which the reaction occurred? a 6.04 K b 6.04 × 10^−3 K c 5.49 × 10^−3 K d 5.49 K e 8.55 K

D

18.17 Which of the following are true about the process of water making the transition from the liquid to the gaseous state at 110°C? a ∆G > 0, ∆H < 0, and ∆S = 0 b ∆G < 0, ∆H > 0, and ∆S < 0 c ∆G > 0, ∆H > 0, and ∆S > 0 d ∆G < 0, ∆H < 0, and ∆S > 0 e ∆G < 0, ∆H > 0, and ∆S > 0

E

18.7 Define the free energy G. How is ∆G related to ∆H and ∆S?

Free energy, G, equals H − TS; that is, it is the difference between the enthalpy of a system and the product of temperature and entropy. The free-energy change, ΔG, equals ΔH − TΔS

18.9 Explain how ∆G° can be used to decide whether a chemical equation is spontaneous in the direction written.

If ΔG° for a reaction is negative, the equation for the reaction is spontaneous in the direction written; that is, the reactants form the products as written. If it is positive, then the equation as written is nonspontaneous

18.10 What is the useful work obtained in the ideal situation in which a chemical reaction with free-energy change ∆G is run so that it produces no entropy?

In principle, if a reaction is carried out so that no entropy is produced, the useful work obtained is the maximum useful work, wmax, and is equal to ΔG of the reaction

18.15 Discuss the different sign combinations of ∆H° and ∆S° that are possible for a process carried out at constant temperature and pressure. For each combination, state whether the process must be spontaneous or not, or whether both situations are possible. Explain.

The four combinations are as follows: (1) A negative ΔH° and a positive ΔS° always give a negative ΔG° and a spontaneous reaction. (2) A positive ΔH° and a negative ΔS° always give a positive ΔG° and a nonspontaneous reaction. (3) A negative ΔH° and a negative ΔS° may give a negative or a positive ΔG°. At low temperatures, ΔG° will usually be negative and the reaction spontaneous; at high temperatures, ΔG° will usually be positive and the reaction nonspontaneous. (4) A positive ΔH° and a positive ΔS° may give a negative or a positive ΔG°. At low temperatures, ΔG° will usually be positive and the reaction nonspontaneous; at high temperatures, ΔG° will usually be negative and the reaction spontaneous.

Concept Check 18.1 You have a sample of 1.0 mg of solid iodine at room temperature. Later, you notice that the iodine has sublimed (passed into the vapor state). What can you say about the change of entropy of the iodine?

The process is I2(s) → I2(g). The iodine atoms have gone from a state of some order (crystalline iodine) to one that is more disordered (gas). The entropy will have increased.

18.4 The entropy change ∆S for a phase transition equals ∆H/T, where ∆H is the enthalpy change. Why is it that the entropy change for a system in which a chemical reaction occurs spontaneously does not equal ∆H/T?

The relationship between entropy and enthalpy can be expressed in terms of the following equation: ΔS = (Δ H - ΔG)/T At equilibrium, ΔG equals 0, so the equation reduces to ΔH/T, whereas, when not at equilibrium, ΔG ≠ 0, so this is not the case. In contrast to a phase change at equilibrium, the entropy change for a spontaneous chemical reaction (at constant pressure) does not equal ΔH/T because entropy is created by the spontaneous reaction. This can be an increase in the entropy of the surroundings or of the system. An example of the latter is the reaction N2O4(g) → 2NO2(g), where one reactant molecule forms two product molecules, thus increasing the randomness.

18.3 State the second law of thermodynamics.

The second law of thermodynamics states that for a spontaneous process, the total entropy of a system and its surroundings always increases. As stated in Section 18.2, a spontaneous process actually creates energy dispersal, or entropy

18.5 Describe how the standard entropy of hydrogen gas at 25°C can be obtained from heat measurements.

The standard entropy of hydrogen gas at 25°C can be obtained by starting near 0.0 K as a reference point, where the entropy of perfect crystals of hydrogen is almost zero. Then, warm to room temperature in small increments, and calculate ΔS° for each incremental temperature change (say, 2 K) by dividing the heat absorbed by the average temperature (1 K is used as the average for 0 K to 2 K), and also take into account the entropy increases that accompany a phase change

Concept Check 18.2 Consider the reaction of nitrogen, N2, and oxygen, O2, to form nitrogen monoxide, NO: N2(g) + O2(g) —> 2NO(g) From the standard free energy of formation of NO, what can you say about this reaction?

The standard free energy of formation for NO(g) is 86.60 kJ/mol. This rather large positive value means the equilibrium constant is small. At equilibrium, the NO concentration is low

18.6 Describe what you would look for in a reaction involving gases in order to predict the sign of ∆S°. Explain.

To predict the sign of ΔS°, look for a change, Δngas, in the number of moles of gas. If there is an increase in moles of gas in the products (Δngas is positive), then ΔS° should be positive. A decrease in moles of gas in the products suggests ΔS° should be negative

18.11 Give an example of a chemical reaction used to obtain useful work.

When gasoline burns in an automobile engine, the change in free energy shows up as useful work. Gasoline, a mixture of hydrocarbons such as C8H18 or octane, burns to yield energy, gaseous CO2, and gaseous H2O.

18.16 Consider a reaction in which ∆H° and ∆S° are positive. Suppose the reaction is nonspontaneous at room temperature. How would you estimate the temperature at which the reaction becomes spontaneous?

You can estimate the temperature at which a nonspontaneous reaction becomes spontaneous by substituting zero for ΔG° into the equation ΔG° = ΔH° − TΔS° and then solving for T using the form T = ΔH°/ΔS°.

Concept Check 18.4 Consider the decomposition of dinitrogen tetroxide, N2O4, to nitrogen dioxide, NO2: N2O4(g) —> 2NO2(g) How would you expect the spontaneity of the reaction to behave with temperature change?

You must break a chemical bond when N2O4 goes to 2NO2. It will take energy to do that, so you expect ΔH to be positive (the reaction is endothermic). Also, when you break a molecule in two, you generally increase the entropy (ΔS is positive). According to Table 18.3, the reaction should be nonspontaneous at relatively low temperature, but spontaneous at some higher temperature. That is, the concentration of NO2 increases with temperature

Concept Check 18.3 The following reaction is spontaneous in the direction given. A(g) + B(g) —> C(g) + D(g) Suppose you are given a vessel containing an equilibrium mixture of A, B, C, and D, and you increase the concentration of C by increasing its partial pressure. a How is the standard free-energy change, ∆G°, affected by the addition of C to the vessel? b How is the free-energy change, ∆G, affected by the addition of C to the vessel?

a. The standard free-energy change, ΔG°, is independent of concentration, so it will not change. b. The relationship between ΔG and ΔG° is given by ΔG = ΔG° + RT ln Q, where Q is the reaction quotient. If the concentration of C is increased, it will cause the value of Q to increase and, in turn, ΔG to increase.


Conjuntos de estudio relacionados

Assessment of Musculoskeletal Function (CH. 40)

View Set

Monopoly, Antitrust, and Regulation

View Set

CIS Chapter 7 - Computer Hardware

View Set

CHAPTER 16: ILLINOIS REAL ESTATE LICENSING LAW

View Set

Chapter 27 Safety, security, Emergency and Preparedness

View Set

Z for Zachariah test study guide

View Set