Chapter Six Thermochemistry

Copyright © Houghton Mifflin Company. All rights reserved.6 | 2 Question Calculate the work for the expansion of an ideal gas from 2.0 L to 5.0 L against a pressure of 2.0 atm at constant temperature. a)6.0 L  atm b)-6.0 L  atm c)0 d)3.0 L  atm e)-3.0 L  atm

Copyright © Houghton Mifflin Company. All rights reserved.6 | 3 Answer b)–6.0 L  atm Section 6.1, The Nature of Energy To calculate the work: w = –PΔV = –(2.0 atm)(5.0 L – 2.0 L) = –6.0 L · atm

Copyright © Houghton Mifflin Company. All rights reserved.6 | 4 Question Label the following process as exothermic or endothermic: Water freezes. a)Exothermic b)Endothermic

Copyright © Houghton Mifflin Company. All rights reserved.6 | 5 Answer a)Exothermic Section 6.1, The Nature of Energy For water to freeze, heat must be released from the water to the surroundings.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 6 Question Label the following process as exothermic or endothermic: Your hand gets cold when you touch ice. a)Exothermic b)Endothermic

Copyright © Houghton Mifflin Company. All rights reserved.6 | 7 Answer a)Exothermic Section 6.1, The Nature of Energy For your hand to get cold, heat must be transferred from your hand to the ice.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 8 Question Label the following process as exothermic or endothermic: Ice increases in temperature when you touch it. a)Exothermic b)Endothermic

Copyright © Houghton Mifflin Company. All rights reserved.6 | 9 Answer b)Endothermic Section 6.1, The Nature of Energy Heat is transferred from your hand (surroundings) to the ice (system).

Copyright © Houghton Mifflin Company. All rights reserved.6 | 10 Question Label the following process as exothermic or endothermic: Water vapor condenses on a cold pipe. a)Exothermic b)Endothermic

Copyright © Houghton Mifflin Company. All rights reserved.6 | 11 Answer a)Exothermic Section 6.1, The Nature of Energy For water vapor to condense, heat must be transferred from the water vapor to the cold pipe.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 12 Question Gas A 2 reacts with gas B 2 to form gas AB. The bond energy of AB is much greater than the bond energy of either A 2 or B 2. Is the reaction for the formation of AB exothermic or endothermic? a)Exothermic b)Endothermic

Copyright © Houghton Mifflin Company. All rights reserved.6 | 13 Answer a)Exothermic Section 6.1, The Nature of Energy Because the bond energy of AB is much greater, the energy released when AB is formed is greater than the energy required to break apart A 2 and B 2. Thus energy is released overall.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 14 Question Which of the following statements is true? a)Change in enthalpy is a state function. b)In exothermic reactions, the reactants are lower in potential energy than the products. c)A chemist takes the surroundings point of view when determining the sign for work or heat. d)Heat of reaction and change in enthalpy can always be used interchangeably.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 15 Answer a)Change in enthalpy is a state function. Section 6.1, The Nature of Energy; Section 6.2, Enthalpy and Calorimetry Change in enthalpy does not depend on the pathway, making it a state function.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 16 Question For a particular process, q = –10 kJ and w = 25 kJ. Which of the following statements is true? a)Heat flows from the surroundings to the system. b)The surroundings do work on the system. c)  E = –35 kJ.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 17 Answer b)The surroundings do work on the system. Section 6.1, The Nature of Energy Because heat is negative, it flows from the system to the surroundings. The internal energy change  E = q + w = 15 kJ. Because work is positive, the surroundings do work on the system.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 18 Question Which of the following statements is correct? a)The internal energy of a system increases when more work is done by the system than heat was flowing into the system. b)The internal energy of a system decreases when work is done on the system and heat is flowing into the system. c)The system does work on the surroundings when an ideal gas expands against a constant external pressure.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 19 Answer c)The system does work on the surroundings when an ideal gas expands against a constant external pressure. Section 6.1, The Nature of Energy The internal energy of a system increases when heat is flowing into the system and work is done on the system. Work has a negative sign and is done on the surroundings when –P  V is negative. This is the case when an ideal gas expands against a constant external pressure.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 20 Question Two different metals of equal mass with different heat capacities are subjected to the same amount of heat. Which undergoes the smallest change in temperature? a)The metal with the higher heat capacity shows the smallest change in temperature. b)The metal with the lower heat capacity shows the smallest change in temperature. c)Because they have equal mass, both metals undergo the same change in temperature.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 21 Answer a)The metal with the higher heat capacity shows the smallest change in temperature. Section 6.2, Enthalpy and Calorimetry Since each metal is subjected to the same amount of heat: q metal = q metal. Thus (heat capacity metal )(  T) = (heat capacity metal )(  T) Since the heat capacity and the change in temperature are inversely related, for a smaller change in temperature, the heat capacity would have to be higher.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 22 Question If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is a)boiling. b)completely vaporized. c)frozen solid. d)still a liquid.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 23 Answer d)still a liquid. Section 6.2, Enthalpy and Calorimetry To decide on an answer, recognize that 5.0  10 3 J = (4.18 J · g –1 · °C –1 )(15.5 g)(T – 10.°C) T = 87°C, so the water is still a liquid.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 24 Question A 50.0-g sample of water at 80°C is added to a 50.0-g sample of water at 20°C. The final temperature of the water should be a)between 20°C and 50°C. b)50°C. c)between 50°C and 80°C.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 25 Answer b) 50°C. Section 6.2, Enthalpy and Calorimetry In both samples, mass and heat capacity are the same. The final temperature should be the average of the two temperatures.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 26 Question A 50.0-g sample of water at 80°C is added to a g sample of water at 20°C. The final temperature of the water should be a)between 20°C and 50°C. b)50°C. c)between 50°C and 80°C.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 27 Answer a)between 20°C and 50°C. Section 6.2, Enthalpy and Calorimetry The mass of the cooler water is greater, so the final temperature should be lower than the average of the two temperatures.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 28 Question You have a 50.0-g sample of water at 20°C. To it, you add a 50.0-g sample of iron at 80°C. The final temperature of the water should be a)between 20°C and 50°C. b)50°C. c)between 50°C and 80°C.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 29 Answer a) between 20°C and 50°C. Section 6.2, Enthalpy and Calorimetry Iron has a lower heat capacity than water, and the masses of both samples are the same. Therefore, the change in the temperature of the water will be less than the change in the temperature of the iron. The final temperature will be less than the average temperature of the iron and the water.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 30 Question Given the equation S(s) + O 2 (g)  SO 2 (g),  H = –296 kJ, which of the following statement(s) is (are) true? I.The reaction is exothermic. II.When mol sulfur is reacted, 148 kJ of energy is released. III.When 32.0 g of sulfur is burned, 2.96  10 5 J of energy is released. a)All are true. b)None is true. c)I and II are true. d)I and III are true. e)Only II is true.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 31 Answer a)All are true. Section 6.3, Hess’s Law A reaction with a negative change in enthalpy is exothermic. Based on the stoichiometry, if mol of sulfur was consumed,  H = (½)(–296 kJ). If 1.00 mol (32.0 g) of sulfur is burned, 296 kJ (2.96  10 5 J) of energy is released.

Chapter Sixteen Spontaneity, Entropy, and Free Energy

Copyright © Houghton Mifflin Company. All rights reserved.6 | 33 Question Which of the following substances has the higher positional probability per mole? a)Liquid water at 100°C. b)Gaseous water at 100°C. c)Because the temperatures are the same, the positional probabilities are the same.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 34 Answer b) Gaseous water at 100°C. Section 16.1, Spontaneous Processes and Entropy A mole of gaseous water at 100°C has a greater volume than a mole of liquid water at 100°C.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 35 Question Predict the sign of the change in positional probability for the process of dry ice (solid CO 2 ) subliming to form gaseous CO 2. a)The positional probability increases. b)The positional probability decreases. c)The positional probability does not change.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 36 Answer a) The positional probability increases. Section 16.1, Spontaneous Processes and Entropy The process involves a change from a relatively small volume to a much larger volume.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 37 Question For which process is  S positive? a)Water freezing at -20  C b)An ideal gas being compressed reversibly at a constant temperature and pressure c)A precipitation reaction d)A spontaneous endothermic process at a constant temperature and pressure

Copyright © Houghton Mifflin Company. All rights reserved.6 | 38 Answer d)A spontaneous endothermic process at a constant temperature and pressure Section 16.2, Entropy and the Second Law of Thermodynamics For an endothermic process,  S surr is negative. For  S univ to be positive,  S sys (  S) must be positive.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 39 Question For any spontaneous process, a)  S univ,  S sys, and  S surr must all be positive. b)  S univ and  S sys must be positive. c)  S sys and  S surr must be positive. d)Only  S univ must be positive. e)Only  S sys must be positive.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 40 Answer d) Only  S univ must be positive. Section 16.2, Entropy and the Second Law of Thermodynamics The second law of thermodynamics states that for any spontaneous process there is always an increase in the entropy of the universe.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 41 Question (part 1) A mixture of hydrogen and chlorine remains unreacted until it is exposed to ultraviolet light from a burning magnesium strip. Then the following reaction occurs very rapidly: H 2 (g) + Cl 2 (g)  2HCl(g)  G = –45.54 kJ  H = –44.12 kJ  S = –4.76 J/K Select the statement that best explains this behavior.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 42 Question (part 2) a)The reactants are thermodynamically more stable than the products. b)The reaction has a small equilibrium constant. c)The ultraviolet light raises the temperature of the system and makes the reaction more favorable. d)The negative value for  S slows down the reaction. e)The reaction is spontaneous, but the reactants are kinetically stable.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 43 Answer e)The reaction is spontaneous, but the reactants are kinetically stable. Section 16.4, Free Energy Because the free energy change has a negative value, the reaction is spontaneous, which is a thermodynamic consideration. Until the ultraviolet light is brought to bear, the reaction is proceeding very slowly, which is a kinetic consideration.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 44 Question For the vaporization of a liquid at a given pressure, a)  G is positive at all temperatures. b)  G is negative at all temperatures. c)  G is positive at low temperatures, but negative at high temperatures. d)  G is negative at low temperatures, but positive at high temperatures.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 45 Answer c)  G is positive at low temperatures, but negative at high temperatures. Section 16.4, Free Energy Recall that  G =  H – T  S.  H is the heat of vaporization, which is a positive number, and  S is positive for a phase change from liquid to gas.  G will be positive for low temperatures where  H > T  S and negative for high temperatures where  H < T  S.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 46 Question In which case must a reaction be spontaneous at all temperatures? a)  H is positive, and  S is positive. b)  H is negative, and  S is negative. c)  H is positive, and  S is negative. d)  H is negative, and  S is positive. e)  H = 0, and  S = 0.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 47 Answer d)  H is negative, and  S is positive. Section 16.4, Free Energy Recall that  G =  H – T  S. Spontaneous reactions are those with  G < 0, so reactions with  H negative and  S positive will be spontaneous at all temperatures.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 48 Question At constant pressure, the reaction 2NO 2 (g)  N 2 O 4 (g) is exothermic. The reaction (as written) is a)always spontaneous. b)spontaneous at low temperatures, but not at high temperatures. c)spontaneous at high temperatures, but not at low temperatures. d)never spontaneous.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 49 Answer b)spontaneous at low temperatures, but not at high temperatures. Section 16.4, Free Energy Recall that  G =  H – T  S. Spontaneous reactions are those with  G < 0 (constant T, P). The given reaction is exothermic (  H < 0) and proceeding toward more order (  S < 0). Thus the reaction (as written) will be spontaneous at low temperatures.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 50 Question When a stable diatomic molecule spontaneously forms from its atoms at constant pressure and temperature, what is true about  H°? a)  H° is negative. b)  H° is positive. c)  H° is zero.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 51 Answer a)  H° is negative. Section 6.2, Enthalpy and Calorimetry; Section 16.4, Free Energy When a diatomic molecule is formed from atoms, energy is released as heat (the reaction is exothermic). Thus  H° is negative.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 52 Question When a stable diatomic molecule spontaneously forms from its atoms at constant pressure and temperature, what is true about  S°? a)  S° is negative. b)  S° is positive. c)  S° is zero.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 53 Answer a)  S° is negative. Section 16.4, Free Energy Consider the reaction A + A → A 2. When 2 mol of reactant forms 1 mol of product, there is a decrease in entropy; thus the value of  S° is negative.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 54 Question When a stable diatomic molecule spontaneously forms from its atoms at constant pressure and temperature, what is true about  G°? a)  G° is negative. b)  G° is positive. c)  G° is zero.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 55 Answer a)  G° is negative. Section 16.4, Free Energy Since the process is spontaneous, the value of  G° must be negative.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 56 Question A liquid is vaporized at its boiling point at constant pressure. What is true about  H°? a)  H° is negative. b)  H° is positive. c)  H° is zero.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 57 Answer b)  H° is positive. Section 6.2, Enthalpy and Calorimetry; Section 16.4, Free Energy For a liquid to be vaporized, energy must be added. This process is endothermic, so the value of  H° is positive.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 58 Question A liquid is vaporized at its boiling point at constant pressure. What is true about  S°? a)  S° is negative. b)  S° is positive. c)  S° is zero.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 59 Answer b)  S° is positive. Section 16.4, Free Energy When a liquid is vaporized, the volume increases greatly, increasing entropy. Thus the value of  S° is positive.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 60 Question A liquid is vaporized at its boiling point at constant pressure. What is true about  G°? a)  G° is negative. b)  G° is positive. c)  G° is zero.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 61 Answer c)  G° is zero. Section 16.4, Free Energy Since the liquid is vaporized at its boiling point, the system is at equilibrium. Thus the value of  G° is zero.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 62 Question For the dissociation reaction of the acid HF,  S is negative. HF(aq) H + (aq) + F – (aq) Which statement best explains why  S is negative? a)Each HF molecule produces two ions when it dissociates. b)The ions are hydrated. c)The reaction is expected to be exothermic, so  S should be negative. d)The reaction is expected to be endothermic, so  S should be negative.

Copyright © Houghton Mifflin Company. All rights reserved.6 | 63 Answer b)The ions are hydrated. Section 16.5, Entropy Changes in Chemical Reactions When a species such as HF dissociates in water, the ions are surrounded by many water molecules. Initially, many species are free. Later, however, the hydrated ions result in fewer free species and less disorder.