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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 Initial Position In the initial position, ball A has a higher potential energy than ball B. Copyright © Cengage Learning. All rights reserved 4 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! Is the freezing of water an endothermic or exothermic process? Explain. Copyright © Cengage Learning. All rights reserved 10 Section 6.1 The Nature of Energy CONCEPT CHECK! Classify each process as exothermic or endothermic. Explain. The system is underlined in each example. Exo a) Endo Endo b) c) Exo Endo d) e) Your hand gets cold when you touch ice. The ice gets warmer when you touch it. Water boils in a kettle being heated on a stove. Water vapor condenses on a cold pipe. 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 system and its surroundings and give the direction of 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 lower in 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 E of 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: q is positive Exothermic Process: q is 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: w is negative Surroundings do work on the system: w is positive Copyright © Cengage Learning. All rights reserved 18 Section 6.1 The Nature of Energy Work Work = P × A × Δh = PΔV P is pressure. A is area. Δh is the piston moving a distance. ΔV is the change in volume. 19 Section 6.1 The Nature of Energy Work For an expanding gas, ΔV is a positive quantity because the volume is increasing. Thus ΔV and 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 4.00×106 L to 4.50×106 L by the addition of 1.3×108 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? V1 = 4.00×106 L q = +1.3×108 J P = 1.0 atm 1 L · atm = 101.3 J V2 = 4.50×106 L Section 6.1 The Nature of Energy Interactive Example 6.3 - Solution (Continued 1) What do we need? E q w How do we get there? What is the work done on the gas? w PV What is ΔV? V V2 V1 4.50 106 L 4.00 106 L 5.0 105 L Section 6.1 The Nature of Energy Interactive Example 6.3 - Solution (Continued 2) What is the work? w PV 1.0 atm 5.0 105 L 5.0 105 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) 101.3 J w 5.0 10 L atm 5.1 107 J L atm 5 Then, E q w +1.3 108 J 5.1 107 J 8 107 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 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 ΔE for each of the following with the listed conditions: a) An endothermic process that performs work. |work| > |heat| Δ E = negative |work| < |heat| Δ E = positive b) Work is done on a gas and the process is exothermic. |work| > |heat| Δ E = positive |work| < |heat| Δ E = negative Copyright © Cengage Learning. All rights reserved 28 Section 6.2 Enthalpy and Calorimetry Change in Enthalpy State function ΔH = q at constant pressure ΔH = Hproducts – Hreactants Copyright © Cengage Learning. All rights reserved 29 Section 6.2 Enthalpy and Calorimetry EXERCISE! Consider the combustion of propane: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔH = –2221 kJ Assume that all of the heat comes from the combustion of propane. Calculate ΔH in 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 A Coffee–Cup Calorimeter Made of Two Styrofoam Cups Copyright © Cengage Learning. All rights reserved 33 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. (sH2O = 4.18 J/°C·g and sFe = 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(NO3)2 solution at 25.0°C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25.0°C in a calorimeter, the white solid BaSO4 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 BaSO4 formed Section 6.2 Enthalpy and Calorimetry Interactive Example 6.5 - Solution Where are we going? To calculate ΔH per mole of BaSO4 formed What do we know? 1.00 L of 1.00 M Ba(NO3)2 1.00 L of 1.00 M Na2SO4 Tinitial = 25.0°C and Tfinal = 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 Ba2+, NO3–, Na+, and SO42– The Na+ and NO3– ions are spectator ions, since NaNO3 is very soluble in water and will not precipitate under these conditions The net ionic equation for the reaction is: Ba 2 aq + SO24 aq BaSO4 s 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 BaSO4 must be exothermic ΔH is 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? 1000 mL 1.0 g Mass of solution 2.00 L 2.0 103 g 1 L mL What is the temperature increase? T Tfinal Tinitial 28.1C 25.0C = 3.1C How much heat is evolved by the reaction? Heat evolved = 4.18 J/ C g 2.0 103 g 3.1 C 2.6 10 4 J Section 6.2 Enthalpy and Calorimetry Interactive Example 6.5 - Solution (Continued 4) Thus, q qP H 2.6 104 J What is ΔH per mole of BaSO4 formed? Since 1.0 L of 1.0 M Ba(NO3)2 contains 1 mole of Ba2+ ions and 1.0 L of 1.0 M Na2SO4 contains 1.0 mole of SO42– ions, 1.0 mole of solid BaSO4 is formed in this experiment Thus the enthalpy change per mole of BaSO4 formed is H 2.6 104 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 since V must change for PV 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 = qV 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 H2 and CH4 What do we know? 1.50 g CH4 → ΔT = 7.3°C 1.15 g H2 → 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 CH4, 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 Energy released in the combustion of 1.5 g CH 4 11.3 kJ/ C C = 83 kJ Section 6.2 Enthalpy and Calorimetry Example 6.6 - Solution (Continued 2) 83 kJ Energy released in the combustion of 1 g CH 4 55 kJ/g 1.5 g For H2, Energy released in the combustion of 1.15 g H 2 11.3 kJ/ C C = 162 kJ 162 kJ Energy released in the combustion of 1 g H 2 141 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 ΔH is also reversed Magnitude of ΔH is 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 ΔH is 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 (B2H6) 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: 2B s + 3H2 g B2H6 g Section 6.2 Enthalpy and Calorimetry Interactive Example 6.8 - Hess’s Law II (Continued) Use the following data: Reaction a. 2B s + 3 O 2 g B2 O 3 s 2 ΔH –1273 kJ b. B2H6 g + 3O2 g B2O3 s + 3H2O g –2035 kJ 1 O2 g H 2O l 2 –286 kJ c. H 2 g + d. H2O l H2O g 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 H2(g), and the product is B2H6(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 B2H6(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 2B s + O 2 g B2 O3 s 2 (b) B2 O3 s + 3H 2 O g B2 H 6 g + 3O 2 g (a) H = 1273 kJ H = 2035 kJ 3 Sum: B2 O3 s + 2B s + O 2 g + 3H 2 O g 2 B2 O3 s + B2 H 6 g + 3O 2 g ΔH = 762 kJ Deleting the species that occur on both sides gives: 2B s + 3H 2O g B2 H 6 g + 3 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 H2O(g) and O2(g) and introduce H2(g) as a reactant We can do this using reactions (c) and (d) Multiply reaction (c) and its ΔH value by 3 and add the result to the preceding equation Section 6.2 Enthalpy and Calorimetry Interactive Example 6.8 - Solution (Continued 4) 2B s + 3H 2O g B2 H 6 g + 1 3 (c) 3 H 2 g + O 2 g H 2O l 2 3 O2 g 2 H = 762 kJ H = 3 286 kJ 3 Sum: 2B s + 3H 2 g + O 2 g + 3H 2 O g 2 3 B2 H 6 g + O 2 g + 3H 2O l ΔH = –96 kJ 2 Section 6.2 Enthalpy and Calorimetry Interactive Example 6.8 - Solution (Continued 5) We can cancel the 3/2 O2(g) on both sides, but we cannot cancel the H2O 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 + 3H2 g + 3H2O g B2H6 g + 3H2O l H = 96 kJ 3 (d) 3 H 2 O l H 2O g H = 3 44 kJ 2B s + 3H 2 g + 3H 2 O g + 3H 2 O l B2 H 6 g + 3H 2O l + 3H 2 O g ΔH = +36 kJ Section 6.2 Enthalpy and Calorimetry Interactive Example 6.8 - Solution (Continued 6) This gives the reaction required by the problem 2B s + 3H2 g B2 H6 g H = +36 kJ Conclusion 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 (ΔHf°) 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 [N2(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 CH4(g) + 2O2(g) CO2(g) + 2H2O(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 ΔH for 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: H°rxn = npHf°(products) - nrHf°(reactants) 4. Elements in their standard states are not included in the ΔHreaction calculations because ΔHf° 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 H° for the following reaction: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) Given the following information: Hf° (kJ/mol) Na(s) 0 H2O(l) –286 NaOH(aq) –470 H2(g) 0 H° = –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: 2Al s + Fe2O3 s Al2O3 s + 2Fe s 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? ΔHf°for Fe2O3(s) = – 826 kJ/mol ΔHf°for Al2O3(s) = – 1676 kJ/mol ΔHf°for Al(s) = ΔHf°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 reaction H f for Al2O3 s H f for Fe2O3 s = 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 Figure 6.11 - Energy Sources Used in the United States Copyright © Cengage Learning. All rights reserved 76 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 H2 (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 H2(l) required and its volume as a gas at the given conditions What do we know? Density for H2(l) = 0.0710 g/mL 80.0 L gasoline Density for gasoline = 0.740 g/mL H2(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? 1000 mL 0.740 g 80.0 L 59,200 g 1 L mL How much H2(l) is needed? Since H2 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) Mass of H 2 l needed = 59,200 g 19,700 g 3 Since density = mass/volume, then volume = mass/density, and the volume of H2(l) needed is V 19,700 g 0.0710 g /mL 2.77 105 mL = 277 L Thus, 277 L of liquid H2 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 H2(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: PV nRT In this case: P = 1.00 atm T = 273 + 25°C = 298 K 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 H2(g)? 1 mol H 2 n 19,700 g H 2 9.77 103 mol H 2 2.016 g H 2 Thus, nRT V P 9.77 103 mol 0.08206 L atm / K mol 298 K 1.00 atm = 2.39 105 L = 239,000 L At 1 atm and 25°C, the hydrogen gas needed to replace 20 gal of gasoline occupies a volume of 239,000 L