AP* Chapter 6. Thermochemistry

AP* Chapter 6 Thermochemistry

Section 6.1 The Nature of Energy Energy Capacity to do work or to produce heat. Law of conservation of energy energy can be converted from one form to another but can be neither created nor destroyed. The total energy content of the universe is constant. Copyright Cengage Learning. All rights reserved 2

Section 6.1 The Nature of Energy Energy Potential energy energy due to position or composition. Kinetic energy energy due to motion of the object and depends on the mass of the object and its velocity. Copyright Cengage Learning. All rights reserved 3

Section 6.1 The Nature of Energy Final Position After A has rolled down the hill, the potential energy lost by A has been converted to random motions of the components of the hill (frictional heating) and to the increase in the potential energy of B. Copyright Cengage Learning. All rights reserved 5

Section 6.1 The Nature of Energy Energy Heat involves the transfer of energy between two objects due to a temperature difference. Work force acting over a distance. Energy is a state function; work and heat are not State Function property that does not depend in any way on the system s past or future (only depends on present state). Copyright Cengage Learning. All rights reserved 6

Section 6.1 The Nature of Energy Chemical Energy System part of the universe on which we wish to focus attention. Surroundings include everything else in the universe. Copyright Cengage Learning. All rights reserved 7

Section 6.1 The Nature of Energy Chemical Energy Endothermic Reaction: Heat flow is into a system. Absorb energy from the surroundings. Exothermic Reaction: Energy flows out of the system. Energy gained by the surroundings must be equal to the energy lost by the system. Copyright Cengage Learning. All rights reserved 8

Section 6.1 The Nature of Energy Endothermic and Exothermic Systems and Energy

Section 6.1 The Nature of Energy CONCEPT CHECK! Classify each process as exothermicor endothermic. Explain. The system is underlined in each example. Exo Endo Endo Exo Endo a) Your handgets cold when you touch ice. b) The icegets warmer when you touch it. c) Waterboils in a kettle being heated on a stove. d) Water vapor condenses on a cold pipe. e) Ice cream melts. Copyright Cengage Learning. All rights reserved 11

Section 6.1 The Nature of Energy CONCEPT CHECK! For each of the following, define a systemand its surroundingsand give the directionof energy transfer. a) Methane is burning in a Bunsen burner in a laboratory. b) Water drops, sitting on your skin after swimming, evaporate. Copyright Cengage Learning. All rights reserved 12

Section 6.1 The Nature of Energy CONCEPT CHECK! Hydrogen gas and oxygen gas react violently to form water. Explain. Which is lowerin energy: a mixture of hydrogen and oxygen gases, or water? Copyright Cengage Learning. All rights reserved 13

Section 6.1 The Nature of Energy Thermodynamics The study of energy and its interconversions is called thermodynamics. Law of conservation of energy is often called the first law of thermodynamics. Copyright Cengage Learning. All rights reserved 14

Section 6.1 The Nature of Energy Internal Energy Internal energy Eof a system is the sum of the kinetic and potential energies of all the particles in the system. To change the internal energy of a system: ΔE= q+ w q represents heat w represents work Copyright Cengage Learning. All rights reserved 15

Section 6.1 The Nature of Energy Internal Energy Thermodynamic quantities consist of two parts: Number gives the magnitude of the change. Sign indicates the direction of the flow. Copyright Cengage Learning. All rights reserved 16

Section 6.1 The Nature of Energy Internal Energy Sign reflects the system s point of view. Endothermic Process: qis positive Exothermic Process: qis negative Copyright Cengage Learning. All rights reserved 17

Section 6.1 The Nature of Energy Internal Energy Sign reflects the system s point of view. System does work on surroundings: wis negative Surroundings do work on the system: wis positive Copyright Cengage Learning. All rights reserved 18

Section 6.1 The Nature of Energy Work Work = P A Δh= PΔV P is pressure. Ais area. Δhis the piston moving a distance. ΔVis the change in volume. 19

Section 6.1 The Nature of Energy Work For an expanding gas, ΔVis a positive quantity because the volume is increasing. Thus ΔVand w must have opposite signs: w= PΔV To convert between L atm and Joules, use 1 L atm = 101.3 J. Copyright Cengage Learning. All rights reserved 20

Section 6.1 The Nature of Energy EXERCISE! Which of the following performs more work? a) A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L. b) A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L. They perform the same amount of work. Copyright Cengage Learning. All rights reserved 21

Section 6.1 The Nature of Energy Interactive Example 6.3 -Internal Energy, Heat, and Work A balloon is being inflated to its full extent by heating the air inside it In the final stages of this process, the volume of the balloon changes from 4.00 10 6 L to 4.50 10 6 L by the addition of 1.3 10 8 J of energy as heat Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate ΔE for the process To convert between L atm and J, use 1 L atm = 101.3 J

Section 6.1 The Nature of Energy Interactive Example 6.3 - Solution Where are we going? To calculate ΔE What do we know? V 1 = 4.00 10 6 L q = +1.3 10 8 J P = 1.0 atm 1 L atm = 101.3 J V 2 = 4.50 10 6 L

Section 6.1 The Nature of Energy Interactive Example 6.3 - Solution (Continued 1) What do we need? How do we get there? What is the work done on the gas? What is ΔV? E= q+ w w= P V V = V V = = 6 6 5 2 1 4.50 10 L 4.00 10 L 5.0 10 L

Section 6.1 The Nature of Energy Interactive Example 6.3 - Solution (Continued 2) What is the work? 5 5 w= P V = 1.0 atm 5.0 10 L= 5.0 10 L atm The negative sign makes sense because the gas is expanding and doing work on the surroundings To calculate ΔE, we must sum q and w However, since q is given in units of J and w is given in units of L atm, we must change the work to units of joules

Section 6.1 The Nature of Energy Interactive Example 6.3 - Solution (Continued 3) 5 w= 5.0 10 L atm 101.3 J L atm Then, 8 7 7 E= q+ w= +1.3 10 J + 5.1 10 J = 8 10 J Reality check Since more energy is added through heating than the gas expends doing work, there is a net increase in the internal energy of the gas in the balloon Hence ΔE is positive 7 = 5.1 10 J ( ) ( )

Section 6.1 The Nature of Energy Exercise A balloon filled with 39.1 moles of helium has a volume of 876 L at 0.0 C and 1.00 atm pressure The temperature of the balloon is increased to 38.0 C as it expands to a volume of 998 L, the pressure remaining constant Calculate q, w, and ΔE for the helium in the balloon The molar heat capacity for helium gas is 20.8 J/ C mol q = 30.9 kj, w = 12.4 kj, and ΔE = 18.5 kj

Section 6.1 The Nature of Energy CONCEPT CHECK! Determine the sign of ΔEfor each of the following with the listed conditions: a) An endothermic process that performs work. work > heat ΔE= negative work < heat ΔE= positive b) Work is done on a gas and the process is exothermic. work > heat work < heat ΔE= positive ΔE= negative Copyright Cengage Learning. All rights reserved 28

Section 6.2 Enthalpy and Calorimetry EXERCISE! Consider the combustion of propane: C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l) ΔH= 2221 kj Assume that all of the heat comes from the combustion of propane. Calculate ΔHin which 5.00 g of propane is burned in excess oxygen at constant pressure. 252 kj Copyright Cengage Learning. All rights reserved 30

Section 6.2 Enthalpy and Calorimetry Calorimetry Science of measuring heat Specific heat capacity: The energy required to raise the temperature of one gram of a substance by one degree Celsius. Molar heat capacity: The energy required to raise the temperature of one mole of substance by one degree Celsius. Copyright Cengage Learning. All rights reserved 31

Section 6.2 Enthalpy and Calorimetry Calorimetry If two reactants at the same temperature are mixed and the resulting solution gets warmer, this means the reaction taking place is exothermic. An endothermic reaction cools the solution. Copyright Cengage Learning. All rights reserved 32

Section 6.2 Enthalpy and Calorimetry Calorimetry Energy released (heat) = s m ΔT s = specific heat capacity (J/ C g) m= mass of solution (g) ΔT = change in temperature ( C) Copyright Cengage Learning. All rights reserved 34

Section 6.2 Enthalpy and Calorimetry CONCEPT CHECK! A 100.0 g sample of water at 90 C is added to a 100.0 g sample of water at 10 C. The final temperature of the water is: a) Between 50 C and 90 C b) 50 C c) Between 10 C and 50 C Copyright Cengage Learning. All rights reserved 35

Section 6.2 Enthalpy and Calorimetry CONCEPT CHECK! A 100.0 g sample of water at 90. C is added to a 500.0 g sample of water at 10. C. The final temperature of the water is: a) Between 50 C and 90 C b) 50 C c) Between 10 C and 50 C Calculate the final temperature of the water. 23 C Copyright Cengage Learning. All rights reserved 36

Section 6.2 Enthalpy and Calorimetry CONCEPT CHECK! You have a Styrofoam cup with 50.0 g of water at 10. C. You add a 50.0 g iron ball at 90. C to the water. (s H2O = 4.18 J/ C g and s Fe = 0.45 J/ C g) The final temperature of the water is: a) Between 50 C and 90 C b) 50 C c) Between 10 C and 50 C Calculate the final temperature of the water. 18 C Copyright Cengage Learning. All rights reserved 37

Section 6.2 Enthalpy and Calorimetry Interactive Example 6.5 -Constant-Pressure Calorimetry When 1.00 L of 1.00 M Ba(NO 3 ) 2 solution at 25.0 C is mixed with 1.00 L of 1.00 M Na 2 SO 4 solution at 25.0 C in a calorimeter, the white solid BaSO 4 forms, and the temperature of the mixture increases to 28.1 C

Section 6.2 Enthalpy and Calorimetry Interactive Example 6.5 -Constant-Pressure Calorimetry (Continued) Assume that: The calorimeter absorbs only a negligible quantity of heat The specific heat capacity of the solution is 4.18 J/ C g The density of the final solution is 1.0 g/ml Calculate the enthalpy change per mole of BaSO 4 formed

Section 6.2 Enthalpy and Calorimetry Interactive Example 6.5 - Solution Where are we going? To calculate ΔHper mole of BaSO 4 formed What do we know? 1.00 L of 1.00 M Ba(NO 3 ) 2 1.00 L of 1.00 M Na 2 SO 4 T initial = 25.0 C and T final = 28.1 C Heat capacity of solution = 4.18 J/ C g Density of final solution = 1.0 g/ml

Section 6.2 Enthalpy and Calorimetry Interactive Example 6.5 - Solution (Continued 1) What do we need? Net ionic equation for the reaction The ions present before any reaction occurs are Ba 2+, NO 3, Na +, and SO 4 2 The Na + and NO 3 ions are spectator ions, since NaNO 3 is very soluble in water and will not precipitate under these conditions The net ionic equation for the reaction is: ( aq) ( aq) ( s) Ba + SO BaSO 2+ 2 4 4

Section 6.2 Enthalpy and Calorimetry Interactive Example 6.5 - Solution (Continued 2) How do we get there? What is ΔH? Since the temperature increases, formation of solid BaSO 4 must be exothermic ΔHis negative Heat evolved by the reaction = heat absorbed by the solution = specific heat capacity mass of solution increase in temperature

Section 6.2 Enthalpy and Calorimetry Interactive Example 6.5 - Solution (Continued 3) What is the mass of the final solution? Mass of solution= 2.00 L 1000 ml What is the temperature increase? How much heat is evolved by the reaction? 1 L 1.0 g ml 3 = 2.0 10 g T = Tfinal Tinitial = 28.1C 25.0C = 3.1C ( )( )( ) Heat evolved = 4.18 J/ C g 2.0 10g 3 3.1 C = 2.6 10 4 J

Section 6.2 Enthalpy and Calorimetry Interactive Example 6.5 - Solution (Continued 4) Thus, 4 q= q = H = 2.6 10 J P What is ΔHper mole of BaSO 4 formed? Since 1.0 L of 1.0 M Ba(NO 3 ) 2 contains 1 mole of Ba 2+ ions and 1.0 L of 1.0 MNa 2 SO 4 contains 1.0 mole of SO 4 2 ions, 1.0 mole of solid BaSO 4 is formed in this experiment Thus the enthalpy change per mole of BaSO 4 formed is H = 4 2.6 10 J/mol = 26 kj/mol

Section 6.2 Enthalpy and Calorimetry Constant-Volume Calorimetry Used in conditions when experiments are to be performed under constant volume No work is done sincev must change forpv work to be performed Bomb calorimeter Weighed reactants are placed within a rigid steel container and ignited Change in energy is determined by the increase in temperature of the water and other parts

Section 6.2 Enthalpy and Calorimetry Figure 6.6 - A Bomb Calorimeter

Section 6.2 Enthalpy and Calorimetry Constant-Volume Calorimetry (Continued) For a constant-volume process, ΔV= 0 Therefore, w= PΔV= 0 E= q+ w= q= q V ( constant volume) Energy released by the reaction = temperature increase energy required to change the temperature by 1 C = ΔT heat capacity of the calorimeter

Section 6.2 Enthalpy and Calorimetry Example 6.6 - Constant-Volume Calorimetry It has been suggested that hydrogen gas obtained by the decomposition of water might be a substitute for natural gas (principally methane) To compare the energies of combustion of these fuels, the following experiment was carried out using a bomb calorimeter with a heat capacity of 11.3 kj/ C

Section 6.2 Enthalpy and Calorimetry Example 6.6 - Constant-Volume Calorimetry (Continued) When a 1.50-g sample of methane gas was burned with excess oxygen in the calorimeter, the temperature increased by 7.3 C When a 1.15-g sample of hydrogen gas was burned with excess oxygen, the temperature increase was 14.3 C Compare the energies of combustion (per gram) for hydrogen and methane

Section 6.2 Enthalpy and Calorimetry Example 6.6 - Solution Where are we going? To calculate ΔH of combustion per gram for H 2 and CH 4 What do we know? 1.50 g CH 4 ΔT= 7.3 C 1.15 g H 2 ΔT = 14.3 C Heat capacity of calorimeter = 11.3 kj/ C

Section 6.2 Enthalpy and Calorimetry Example 6.6 - Solution (Continued 1) What do we need? ΔE = ΔT heat capacity of calorimeter How do we get there? What is the energy released for each combustion? For CH 4, we calculate the energy of combustion for methane using the heat capacity of the calorimeter (11.3 kj/ C) and the observed temperature increase of 7.3 C 4 ( )( C) Energy released in the combustion of 1.5 g CH = 11.3 kj/ C 7.3 = 83 kj

Section 6.2 Enthalpy and Calorimetry Example 6.6 - Solution (Continued 2) Energy released in the combustion of 1 g CH For H 2, 2 4 83 kj = = 55 kj/g 1.5 g ( )( C) Energy released in the combustion of 1.15 g H = 11.3 kj/ C 14.3 Energy released in the combustion of 1 g H 2 = 162 kj 162 kj = = 141 kj/g 1.15 g

Section 6.2 Enthalpy and Calorimetry Example 6.6 - Solution (Continued 3) How do the energies of combustion compare? The energy released in the combustion of 1 g hydrogen is approximately 2.5 times that for 1 g methane, indicating that hydrogen gas is a potentially useful fuel

Section 6.2 Enthalpy and Calorimetry Hess s Law In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps Copyright Cengage Learning. All rights reserved 54

Section 6.2 Enthalpy and Calorimetry Characteristics of Enthalpy Changes If a reaction is reversed, the sign of ΔHis also reversed Magnitude of ΔHis directly proportional to the quantities of reactants and products in a reaction If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔHis multiplied by the same integer Copyright Cengage Learning. All rights reserved 55

Section 6.2 Enthalpy and Calorimetry Problem-Solving Strategy - Hess s Law Work backward from the required reaction Use the reactants and products to decide how to manipulate the other given reactions at your disposal Reverse any reactions as needed to give the required reactants and products Multiply reactions to give the correct numbers of reactants and products Copyright Cengage Learning. All rights reserved 56

Section 6.2 Enthalpy and Calorimetry Interactive Example 6.8 - Hess s Law II Diborane (B 2 H 6 ) is a highly reactive boron hydride that was once considered as a possible rocket fuel for the U.S. space program Calculate ΔH for the synthesis of diborane from its elements, according to the following equation: ( s) ( g) BH ( g) 2B + 3H 2 2 6

Section 6.2 Enthalpy and Calorimetry Interactive Example 6.8 - Hess s Law II (Continued) Use the following data: a. Reaction ΔH 3 2B ( s) + O2( g) BO 2 3( s ) 1273 kj 2 b. ( g) ( g) BO ( s) + 3HO( g) BH + 3O 2 6 2 2 3 2 2035 kj c. d. 1 H 2( g) + O2( g) HO 2 ( l ) 286 kj 2 HO ( l) HO( g) 2 2 44 kj

Section 6.2 Enthalpy and Calorimetry Interactive Example 6.8 - Solution To obtain ΔH for the required reaction, we must somehow combine equations (a), (b), (c), and (d) to produce that reaction and add the corresponding ΔH values This can best be done by focusing on the reactants and products of the required reaction The reactants are B(s) and H 2 (g), and the product is B 2 H 6 (g)

Section 6.2 Enthalpy and Calorimetry Interactive Example 6.8 - Solution (Continued 1) How can we obtain the correct equation? Reaction (a) has B(s) as a reactant, as needed in the required equation Reaction (a) will be used as it is Reaction (b) has B 2 H 6 (g) as a reactant, but this substance is needed as a product Reaction (b) must be reversed, and the sign of ΔH must be changed accordingly

Section 6.2 Enthalpy and Calorimetry Interactive Example 6.8 - Solution (Continued 2) Up to this point we have: 3 (a) 2B ( s) + O2( g) BO 2 3( s) H = 1273 kj 2 (b) BO + 3HO ( s) ( g) BH ( g) + 3O ( g) H = ( 2035 kj) 2 3 2 2 6 2 3 Sum: BO 2 3( s) + 2B ( s) + O 2( g) + 3HO 2 ( g) 2 BO 2 3( s) + BH 2 6( g) + 3O2( g) ΔH =762 kj Deleting the species that occur on both sides gives: 3 2B ( s) + 3HO 2 ( g) BH 2 6( g) + O 2( g) H = 762 kj 2

Section 6.2 Enthalpy and Calorimetry Interactive Example 6.8 - Solution (Continued 3) We are closer to the required reaction, but we still need to remove H 2 O(g) and O 2 (g) and introduce H 2 (g) as a reactant We can do this using reactions (c) and (d) Multiply reaction (c) and its ΔHvalue by 3 and add the result to the preceding equation

Section 6.2 Enthalpy and Calorimetry Interactive Example 6.8 - Solution (Continued 4) 3 2B ( s) + 3HO 2 ( g) BH 2 6( g) + O 2( g) H = 762 kj 2 1 3 (c) 3 H 2 + O2 HO 2 = 3 286 kj g 2 g l H ( ) ( ) ( ) ( ) 3 Sum: 2B ( s) + 3H 2( g) + O 2( g) + 3HO 2 ( g) 2 3 BH 2 6( g) + O 2( g) + 3HO 2 ( l) ΔH = 96 kj 2

Section 6.2 Enthalpy and Calorimetry Interactive Example 6.8 - Solution (Continued 5) We can cancel the 3/2 O 2 (g) on both sides, but we cannot cancel the H 2 O because it is gaseous on one side and liquid on the other This can be solved by adding reaction (d), multiplied by 3: ( ) ( ) ( ) ( ) ( ) 2B s + 3H g + 3HO g BH g + 3HO l H = 96 kj 2 2 2 6 2 ( l) ( g) H ( ) 3 (d) 3 HO 2 HO 2 = 3 44 kj ( s) 2( g) 2 ( g) 2 ( l) BH ( ) + 3HO ( ) + 3HO( ) 2B + 3H + 3HO + 3HO g l g ΔH = +36 kj 2 6 2 2

Section 6.2 Enthalpy and Calorimetry Interactive Example 6.8 - Solution (Continued 6) This gives the reaction required by the problem Conclusion ( ) ( ) ( ) 2B s + 3H g BH g H = +36 kj 2 2 6 ΔH for the synthesis of 1 mole of diborane from the elements is +36 kj

Section 6.4 Standard Enthalpies of Formation Standard Enthalpy of Formation (ΔH f ) Change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states. Copyright Cengage Learning. All rights reserved 66

Section 6.4 Standard Enthalpies of Formation Conventional Definitions of Standard States For a Compound For a gas, pressure is exactly 1 atm. For a solution, concentration is exactly 1 M. Pure substance (liquid or solid) For an Element The form [N 2 (g), K(s)] in which it exists at 1 atm and 25 C. Copyright Cengage Learning. All rights reserved 67

Section 6.4 Standard Enthalpies of Formation A Schematic Diagram of the Energy Changes for the Reaction CH 4 (g)+ 2O 2 (g) CO 2 (g) + 2H 2 O(l) ΔH reaction = ( 75 kj) + 0 + ( 394 kj) + ( 572 kj) = 891 kj Copyright Cengage Learning. All rights reserved 68

Section 6.4 Standard Enthalpies of Formation Problem-Solving Strategy: Enthalpy Calculations 1. When a reaction is reversed, the magnitude of ΔH remains the same, but its sign changes. 2. When the balanced equation for a reaction is multiplied by an integer, the value of ΔHfor that reaction must be multiplied by the same integer. Copyright Cengage Learning. All rights reserved 69

Section 6.4 Standard Enthalpies of Formation Problem-Solving Strategy: Enthalpy Calculations 3. The change in enthalpy for a given reaction can be calculated from the enthalpies of formation of the reactants and products: ΔH rxn = Σn p ΔH f (products) - Σn r ΔH f (reactants) 4. Elements in their standard states are not included in the ΔH reaction calculations because ΔH f for an element in its standard state is zero. Copyright Cengage Learning. All rights reserved 70

Section 6.4 Standard Enthalpies of Formation EXERCISE! Calculate ΔH for the following reaction: 2Na(s) + 2H 2 O(l) 2NaOH(aq) + H 2 (g) Given the following information: ΔH f (kj/mol) Na(s) 0 H 2 O(l) 286 NaOH(aq) 470 H 2 (g) 0 ΔH = 368 kj Copyright Cengage Learning. All rights reserved 71

Section 6.4 Standard Enthalpies of Formation Interactive Example 6.10 -Enthalpies from Standard Enthalpies of Formation II Using enthalpies of formation, calculate the standard change in enthalpy for the thermite reaction: ( s) ( s) AlO ( s) + 2Fe( s) 2Al + FeO 2 3 2 3 This reaction occurs when a mixture of powdered aluminum and iron(iii) oxide is ignited with a magnesium fuse

Section 6.4 Standard Enthalpies of Formation Interactive Example 6.10 - Solution Where are we going? To calculate ΔH for the reaction What do we know? ΔH f for Fe 2 O 3 (s) = 826 kj/mol ΔH f for Al 2 O 3 (s) = 1676 kj/mol ΔH f for Al(s) = ΔH f for Fe(s) = 0

Section 6.4 Standard Enthalpies of Formation Interactive Example 6.10 - Solution (Continued 1) What do we need? We use the following equation: H = np H f( products) nr H f( reactants) How do we get there? ( ) ( ) H = H for AlO s H for FeO s reaction f 2 3 f 2 3 = 1676 kj ( 826 kj ) = 850 kj

Section 6.4 Standard Enthalpies of Formation Interactive Example 6.10 - Solution (Continued 2) This reaction is so highly exothermic that the iron produced is initially molten Used as a lecture demonstration Used in welding massive steel objects such as ships propellers

Section 6.4 Standard Enthalpies of Formation Interactive Example 6.13 -Comparing Enthalpies of Combustion Assuming that the combustion of hydrogen gas provides three times as much energy per gram as gasoline, calculate the volume of liquid H 2 (density = 0.0710 g/ml) required to furnish the energy contained in 80.0 L (about 20 gal) of gasoline (density = 0.740 g/ml) Calculate also the volume that this hydrogen would occupy as a gas at 1.00 atm and 25 C

Section 6.4 Standard Enthalpies of Formation Interactive Example 6.13 - Solution Where are we going? To calculate the volume of H 2 (l) required and its volume as a gas at the given conditions What do we know? Density for H 2 (l) = 0.0710 g/ml 80.0 L gasoline Density for gasoline = 0.740 g/ml H 2 (g) P =1.00 atm, T =25 C = 298 K

Section 6.4 Standard Enthalpies of Formation Interactive Example 6.13 - Solution (Continued 1) How do we get there? What is the mass of gasoline? 80.0 L 1000 ml 1 L How much H 2 (l) is needed? 0.740 g = 59,200 g ml Since H 2 furnishes three times as much energy per gram as gasoline, only a third as much liquid hydrogen is needed to furnish the same energy

Section 6.4 Standard Enthalpies of Formation Interactive Example 6.13 - Solution (Continued 2) 59,200 g Mass of H 2( l ) needed = = 19,700 g 3 Since density = mass/volume, then volume = mass/density, and the volume of H 2 (l) needed is 19,700 g 5 V = 2.77 10 ml = 277 L 0.0710 g /ml = Thus, 277 L of liquid H 2 is needed to furnish the same energy of combustion as 80.0 L of gasoline

Section 6.4 Standard Enthalpies of Formation Interactive Example 6.13 - Solution (Continued 3) What is the volume of the H 2 (g)? To calculate the volume that this hydrogen would occupy as a gas at 1.00 atm and 25 C, we use the ideal gas law: In this case: P =1.00 atm T =273 + 25 C = 298 K PV = nrt R =0.08206 L atm/k mol

Section 6.4 Standard Enthalpies of Formation Interactive Example 6.13 - Solution (Continued 4) What are the moles of H 2 (g)? n 2 Thus, V =19,700 g H 2 nrt = = P 1 mol H 2.016 g H At 1 atm and 25 C, the hydrogen gas needed to replace 20 gal of gasoline occupies a volume of 239,000 L 2 3 = 9.77 10 mol H 2 ( 3 9.77 10 mol) 0.08206 L atm ( /K mol) 298 K 1.00 atm 5 = 2.39 10 L = 239,000 L ( )