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PHALL not integrated into presentation Chapter 6.3-6.4 Thermochemistry THERMODYNAMICS Courtesy of lab-initio.com Chapter 6 Table of Contents • • • • • • 6.1 6.2 6.3 6.4 6.5 6.6 The Nature of Energy Enthalpy and Calorimetry Hess’s Law Standard Enthalpies of Formation Present Sources of Energy New Energy Sources Copyright © Cengage Learning. All rights reserved 3 Section 6.3 Hess’s Law Tuesday - November 12, 2013 • CW: Notes 6.3 together Hess’s Law - with new handout • CW/HW: Practice with Hess’s law w/sheet • CW: CW: Notes 6.1-6.2 should be finished in class or homework. • Calorimetry Lab - Discussion Questions t/g Return to TOC Copyright © Cengage Learning. All rights reserved 4 Section 6.3 Hess’s Law Calorimetry Lab - Discussion Questions • 1. Calorimetry is the science of measuring heat. A calorimeter is an instrument used to measure the enthalpy of a chemical or physical change. • 2.a. Calculate the heat of solution for 10.0 g of sodium carbonate. qsol = (m x c x ΔT)water + (C x ΔT)calorimeter • q = -(75.0 g x 4.18 J/(°C*g) x 7.4 C + (10 J/°C x 7.4 °C) = -2.4 kJ • 2b. Calculate the number of moles of sodium carbonate. • 10.0 g / 106 g/mol = 0.0943 moles • c. Calculate the enthalpy of solution per mole of sodium carbonate • ΔH = -2.4 kJ/0.0943 mol = -25 kJ/mol Return to TOC Copyright © Cengage Learning. All rights reserved 5 Section 6.3 Hess’s Law Calorimetry Lab Discussion Questions #3 • Heat of neutralization • qneutralization = (m x c x ΔT)water + (C x ΔT)calorimeter • q = - (110.0 g x 4.18 J/(C*g) x 6.1 C + 10 J/C x 6.1 C) • q = -2.9 kJ evolved from 60.0 mL of 0.5 M sulfuric acid and 50.0 mL of sodium hydroxide are mixed. • 3b. Calculate the number of moles of water • Sodium hydroxide is limiting reactant, so the number of moles of NaOH = # moles of water formed from eqn.. • 0.0500 L x 1.0 M NaOH = 0.050 moles of water formed. • 3c. ethalpy of neutralization per mole of water • ΔH = -2.9 kJ/0.0500 mol = -58 kJ/mol Copyright © Cengage Learning. All rights reserved 6 Return to TOC Section 6.3 Hess’s Law Hess’s Law Return to TOC Copyright © Cengage Learning. All rights reserved 7 Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. H = H (products) – H (reactants) H = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants H < 0 Hproducts > Hreactants H > 0 6.3 Table 6.2 Standard Enthalpies of Formation for Several Compounds at 25°C Copyright © Houghton Mifflin Company. All rights reserved. 6– 6.5 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 a series of steps.” np & nr - is the moles of product and reactant respectively. The other symbol (sigma) means to take the sum of the terms as done in math class. Hess’s Law Problem Example 1 Calculate H for the combustion of methane, CH4: CH4 + 2O2 CO2 + 2H2O Reaction C + 2H2 CH4 Ho -74.80 kJ C + O2 CO2 -393.50 kJ H2 + ½ O2 H2O -285.83 kJ CH4 C + 2H2 +74.80 kJ Step #1: CH4 must appear on the reactant side, so we reverse reaction #1 and change the sign on H. Hess’s Law Problem Example 1 Calculate H for the combustion of methane, CH4: CH4 + 2O2 CO2 + 2H2O Reaction Ho C + 2H2 CH4 -74.80 kJ C + O2 CO2 -393.50 kJ H2 + ½ O2 H2O -285.83 kJ CH4 C + 2H2 C + O2 CO2 +74.80 kJ -393.50 kJ Step #2: Keep reaction #2 unchanged, because CO2 belongs on the product side Hess’s Law Problem Example 1 Calculate H for the combustion of methane, CH4: CH4 + 2O2 CO2 + 2H2O Reaction C + 2H2 CH4 Ho -74.80 kJ C + O2 CO2 -393.50 kJ H2 + ½ O2 H2O -285.83 kJ CH4 C + 2H2 C + O2 CO2 2H2 + O2 2 H2O +74.80 kJ -393.50 kJ -571.66 kJ Step #3: Multiply reaction #3 by 2 Hess’s Law Problem Example 1 Calculate H for the combustion of methane, CH4: CH4 + 2O2 CO2 + 2H2O Reaction C + 2H2 CH4 C + O2 CO2 H2 + ½ O2 H2O CH4 C + O2 2H2 + O2 CH4 + 2O2 C + 2H2 CO2 2 H2O CO2 + 2H2O Ho -74.80 kJ -393.50 kJ -285.83 kJ +74.80 kJ -393.50 kJ -571.66 kJ -890.36 kJ Step #4: Sum up reaction and H Hess’s Law Problem Example 2 EXAMPLE - Calculate ΔH for the reaction from 2 steps N2(g) + 2O2(g) → 2NO2(g) ΔH1 = 68 kJ N2(g) + O2(g) → 2NO(g) ΔH2 = 180 kJ 2NO(g) + O2(g) → 2NO2(g) ΔH3 = – 112 kJ Section 6.3 Hess’s Law N2(g) + 2O2(g) → 2NO2(g) ΔH1 = 68 kJ • This reaction also can be carried out in two distinct steps, with enthalpy changes designated by ΔH2 and ΔH3. • N2(g) + O2(g) → 2NO(g) kJ 2NO(g) + O2(g) → 2NO2(g) N2(g) + 2O2(g) → 2NO2(g) • • • ΔH2 = 180 ΔH3 = – 112 kJ ΔH2 + ΔH3 = 68 kJ ΔH1 = ΔH2 + ΔH3 = 68 kJ Return to TOC Copyright © Cengage Learning. All rights reserved 17 Hess’s Law Problem Example 2 EXAMPLE - Calculate ΔH for the reaction from 2 steps N2(g) + 2O2(g) → 2NO2(g) ΔH1 = 68 kJ N2(g) + O2(g) → 2NO(g) ΔH2 = 180 kJ 2NO(g) + O2(g) → 2NO2(g) ΔH3 = – 112 kJ N2(g) + 2O2(g) → 2NO2(g) ΔH2 = 180 kJ + ΔH3 = – 112 kJ 180 kJ + (-112) kJ = 68 kJ showing that either path has the same overall net change in enthalpy. Calculation of Heat of Reaction EX. 3 Calculate H for the combustion of methane, CH4: CH4 + 2O2 CO2 + 2H2O Substance Hf CH4 -74.80 kJ O2 0 kJ CO2 -393.50 kJ H2O -285.83 kJ Calculation of Heat of Reaction EX. 3 Calculate H for the combustion of methane, CH4: CH4 + 2O2 CO2 + 2H2O Substance Hf CH4 -74.80 kJ O2 0 kJ CO2 -393.50 kJ H2O -285.83 kJ Hrxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ] Hrxn = -890.36 kJ Section 6.3 Hess’s Law The Principle of Hess’s Law Return to TOC Copyright © Cengage Learning. All rights reserved 21 Hess’s Law Click here to watch visualization. Copyright © Houghton Mifflin Company. All rights reserved. 6– Section 6.3 Hess’s Law Characteristics of Enthalpy Changes • • If a reaction is reversed, the sign of ΔH is also reversed. The 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. Return to TOC Copyright © Cengage Learning. All rights reserved 23 Section 6.3 Hess’s Law Example 4 • Consider the following data: • Calculate ΔH for the reaction Return to TOC Copyright © Cengage Learning. All rights reserved 24 Section 6.3 Hess’s Law Problem-Solving Strategy • • • Work backward from the required reaction, using 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. Return to TOC Copyright © Cengage Learning. All rights reserved 25 Section 6.3 Hess’s Law Example 4 • Reverse the two reactions: • Desired reaction: Return to TOC Copyright © Cengage Learning. All rights reserved 26 Section 6.3 Hess’s Law Example 4 • Multiply reactions to give the correct numbers of reactants and products: 4( ) • 3( ) • • ) 4( ) 3( Desired reaction: Return to TOC Copyright © Cengage Learning. All rights reserved 27 Section 6.3 Hess’s Law Example 4 • Final reactions: • Desired reaction: • ΔH = +1268 kJ Return to TOC Copyright © Cengage Learning. All rights reserved 28 Section 6.3 Hess’s Law Stop here for Hess’s Law practice • Calorimetry Lab report due on Thursday. • Hess’s Law problems due Thursday. • Finish Notes 6.1-6.3 by Thursday. • Hess’s Law Lab on Thursday (?) Return to TOC Copyright © Cengage Learning. All rights reserved 29 Need to add 6.4 to student notes outline next time for heat of formation Monday - Nov. 18, 2013 • CW: Notes 6.4 (brief) (6.3 done last Tuesday & 6.4 intro.) • QUIZ - 2 questions - Hess’s Law • CW: Finish Notes 6.1-6.2 from handout sheet • CW: Hess’s Law problems finish and turn in • HW: Quia.com Quiz (9 questions) • HW: Hess’s Law Informal Lab report due Wed. • (Be sure Calorimetry Lab has been turned in today to not lose points for being late.) • TEST ch. 6 - Tuesday - Nov. 26, 2013 before Thanksgiving Break (will not include ch. 16 for test now) Section 6.4 KEY - VOCABULARY DEFINITION Standard Enthalpies of Formation Standard Enthalpy of Formation (ΔHf°) is the • Change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states. • (SEE APPENDIX A19-A22 for Heat of formation, entropy, and Gibbs Free Energy values for common compounds and elements.) Return to TOC Copyright © Cengage Learning. All rights reserved 32 Section 6.4 Standard Enthalpies of Formation Conventional Definitions of Standard States • For a Compound If compound is 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. Heat of formation is zero at standard state. Return to TOC Copyright © Cengage Learning. All rights reserved 33 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 Return to TOC Copyright © Cengage Learning. All rights reserved 34 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. Return to TOC Copyright © Cengage Learning. All rights reserved 35 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 Copyright © Cengage Learning. All rights reserved Return to TOC 36 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° = (-470 kJ/mol)(2 mol) + 0 - [(-286 kJ/mol)(2mol) + 0] •–368 kJ Copyright © Cengage Learning. All rights reserved 37 Return to TOC Monday - Nov. 18, 2013 • CW: Notes 6.4 (brief) (6.3 done last Tuesday & 6.4 intro.) • QUIZ - 2 questions - Hess’s Law • CW: Finish Notes 6.1-6.2 from handout sheet • CW: Hess’s Law problems finish and turn in • HW: Quia.com Quiz (9 questions) • HW: Hess’s Law Informal Lab report due Wed. • (Be sure Calorimetry Lab has been turned in today to not lose points for being late.) • TEST ch. 6 - Tuesday - Nov. 26, 2013 before Thanksgiving Break (will not include ch. 16 for test now) Wednesday - Nov. 20, 2013 • TURN IN informal Hess’s Law Lab Report in box. • TURN IN Calorimetry Lab (formal lab report) in box if not already done • TURN IN Hess’s Law problems worksheet with work shown or attached. • CW: Thermodynamics Team Activity Game with Partners (selected by drawing cards) • HW: Be sure all ch. 6 notes are caught up 6.1-6.4 • TEST is TUESDAY Nov. 26, 2013 over chapter 6, calorimetry labs & thermodynamics problems & notes. • HW/CW: Quia.com Quiz (9 questions) • HW/CW: Sciencegeek.net Quiz ch. 6 - 15 questions • Over Thanksgiving Break: Read chapter 16 :) Friday - Nov. 22 2013 • • • • • • • • • Follow up late items to turn in: Hess’s Law Informal Lab Report in Calorimetry Lab (formal lab report) Hess’s Law problems worksheet with work shown Hess’s Law - 2 question quiz To be turned in with TEST on TUESDAY: 1) Thermodynamics Team Activity Game with work shown 2) Interactive ch. 6 notes 3) By Monday Quia.com Quiz (9 questions) at latest - not accepted late for points. • TODAY: Show me HW/CW: Sciencegeek.net Quiz AP ch. 6 - 15 question results and work shown if applicable. • TODAY: Thermochemistry Link Reviews • TEST is TUESDAY Nov. 26, 2013 over chapter 6, calorimetry labs & thermodynamics problems & notes. • Over Thanksgiving Break: Read chapter 16 :) Section 6.4 Standard Enthalpies of Formation QUIZ with Hess’s Law • 1. Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: N2(g) + 2O2(g) → 2NO 2(g) • N2(g) + 3H2(g) → 2NH3(g) ΔH = -115 kJ 2NH3(g) + 4H2O(l) → 2NO2(g) + 7H2(g) ΔH = -142.5 kJ H2O(l) → H2(g) + 1/2O 2(g) ΔH = -43.7 kJ Return to TOC Copyright © Cengage Learning. All rights reserved 41 Section 6.4 Standard Enthalpies of Formation QUIZ with Hess’s Law • 2. Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: CO2(g) → C(s) + O2(g) • H2O(l) → H2(g) + 1/2O2(g) ΔH = 643 kJ C2H6(g) → 2C(s) + 3H 2(g) ΔH = 190.6 kJ 2CO2(g) + 3H2O(l) → C 2H6(g) + 7/2O2(g) ΔH = 3511.1 kJ Return to TOC Copyright © Cengage Learning. All rights reserved 42 Chapter 6 Questions Thermochemistry QUESTION #1 The combustion of a fuel is an exothermic process. This means… 1. the surroundings have lost exactly the amount of energy gained by the system. 2. the potential energy of the chemical bonds in the products should be less than the potential energy of the chemical bonds in the reactants. 3. q must be positive; w must be negative 4. E would have a + overall value because the surroundings have gained energy. Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– ANSWER Choice 2 correctly summarizes the connection between reactants and products for an exothermic process. The exothermic nature of the process means that the products must be less energetic than the reactants. The energy descriptions and comparisons are from the view point of the system. Section 6.1: The Nature of Energy Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– QUESTION #2 The change in enthalpy, or heat of reaction, for chemical changes is equal to the heat flow when… 1. 2. 3. 4. the pressure is constant and only PV work is allowed. H = –q (constant pressure) H = E – PV the pressure is constant and enthalpy = internal energy (E) Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– ANSWER Choice 1 properly relates enthalpy to q. At constant pressure, any work is change in volume work. So, H = E + PV and since q (at constant pressure) = E + PV it follows that H = q (at constant P where most chemical reactions take place). Section 6.2: Enthalpy and Calorimetry Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– QUESTION #3 Acetylene torches used for welding work so effectively because of the large release of energy (H) when acetylene, C2H2, reacts with oxygen. For the combustion of acetylene, H is approximately –1,300 kJ/mol. What is the value for H when 1.0 gram of acetylene reacts in this way? 1. 2. 3. 4. Approximately +50. kJ Approximately –50. kJ Approximately +34,000 kJ Approximately –34,000 kJ Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– ANSWER Choice 2 indicates the proper sign, value, and label. The reported –1,300 kJ is for the combustion of one mole. After 1.0 gram is properly converted to moles, (1.0/26.04) multiplying by –1,300 kJ/mol will yield the answer. The negative sign indicates that heat is released to the surroundings. Section 6.2: Enthalpy and Calorimetry Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– QUESTION #4 The instant cold packs associated with athletic injuries makes use of the heat change when NH4NO3 dissolves. If 8.0 grams of NH4NO3 (molar mass = 80.1) were placed in 100.0 mL of water such as shown in the constant pressure calorimeter, and the temperature change was – 6.0°C, what would you calculate as the H for dissolving one mole of NH4NO3? 1. 2. 3. 4. –2,500 kJ/mol +2,500 kJ/mol –25 kJ/mol +25 kJ/mol Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– ANSWER Choice 4 shows both the proper sign for this endothermic process and the proper value. The constant 4.184 J/°C g was used to determine the heat absorbed from the 100.0 mL of water as its temperature dropped by 6.0°C, during dissolving. Once this was known for 8.0 grams, a ratio to 80.1 grams was calculated. Section 6.2: Enthalpy and Calorimetry Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– QUESTION #5 Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– QUESTION (continued) The heat capacity of a calorimeter of the type shown on the previous slide must be determined before other investigations, using the instrument, can be performed. When 0.5000 grams of pure carbon are combusted in the “bomb” the temperature rose by 1.842°C. (Note: use –393.5 kJ/mol as the heat of combustion for carbon.) What is the calorimeter’s heat capacity? 1. 2. 3. 4. 30.20 kJ/°C 106.8 kJ/K 8.894 kJ/°C 35.60 kJ/°C Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– ANSWER Choice 3 is correct. The heat released per mole must be converted to heat released for 0.5000 gram (0.5000/12.01) –393.5. This should be divided by the number of degrees the calorimeter registered and the sign changed to be positive. Section 6.2: Enthalpy and Calorimetry Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– QUESTION #6 A certain small piece of candy is made of 1.525 grams of sucrose (C12H22O11; molar mass = 342.30 g/mol). When placed in a bomb calorimeter with a specific heat of 5.024 kJ/°C, how much should the temperature rise when the candy undergoes combustion? (Note: use –5,640 kJ/mol as the heat released for the combustion of one mole of sucrose.) 1. 2. 3. 4. 5.00°C 126°C 25.1°C Couldn’t I just eat the candy, I am not sure what to do with the calculations. Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– ANSWER Choice 1 (although choice 4 might be tastier) is correct. Since it is given that one mole of sucrose releases 5,640 kJ, the determination of the number of moles combusted would enable a ratio to find the kJ released from 1.525 g (1.525/342.30) 5,640). Then the heat released by the combustion would be divided by the kJ/°C for the calorimeter; the result would be the change in temperature. Section 6.2: Enthalpy and Calorimetry Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– QUESTION #7 The following reaction shows the conversion of ozone to oxygen and provides the standard enthalpy change for the process: 2 O3 3 O2; H = –427 kJ What is the H value for the conversion of one mole of O2 to O3? 1. 2. 3. 4. +427 kJ +641 kJ +142 kJ +285 kJ Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– ANSWER Choice 3 provides both the proper sign and value for the new reaction. The original reaction must be reversed in order to show oxygen forming ozone. This means the given sign for H must also be reversed. Since the reversed reaction shows the process for 3 moles of oxygen, the H value must be reduced by 1/3 to determine the value for only one mole. Section 6.3: Hess’s Law Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– QUESTION #8 The following reaction depicts one way in which nitric acid could be formed: N2O5 + H2O 2 HNO3 H = ? Use the following information, and apply Hess’s law, to determine the missing value. H2 + 1/2 O2 H2O; H = –285.8 kJ ½ N2 + 3/2 O2 + ½ H2 HNO3; H = –174.1 kJ 2N2 + 5 O2 2 N2O5; H = + 28.4 kJ Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– QUESTION (continued) 1. 2. 3. 4. –362.4 kJ –69.5 kJ –76.6 kJ I am having difficulty rearranging the given equations to produce the “target” equation. Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– ANSWER Choice 3 is the answer when the three given equations are properly arranged. The arrangement that, when added, yields the desired equation is: reverse the first given equation, double the next one, and finally take ½ and reverse the third equation. Each manipulation should be done to the respective H values before tallying those to yield the final H value. Section 6.3: Hess’s Law Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– QUESTION #9 Using information from Appendix Four, determine the standard enthalpy change at 25°C for the combustion of one mole of heptane (C7H16) sometimes used in gasoline fuel mixtures. The balanced equation is shown here: C7H16 (g) + 11 O2 (g) 7 CO2 (g) + 8 H2O (l) (Note: the standard molar enthalpy change of formation of C7H16 (g) is –264 kJ/mol) 1. 2. 3. 4. –4430 kJ –4780 kJ –5310 kJ This can’t be solved without the heat of formation of oxygen. Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– ANSWER Choice 2 provides the correct sign and value. The moles of each reactant must be multiplied by the heat of formation of each component (noting their physical state), then the sum of the enthalpy for reactants is subtracted from the sum of the enthalpies of the products (noting that oxygen has a zero for enthalpy of formation). Section 6.4: Standard Enthalpies of Formation Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– QUESTION #10 Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– QUESTION (continued) After examining the graph shown here, which energy source would you like to see increase and which would you like to see decrease? Be prepared to explain your choice. 1. 2. 3. 4. Wood increase; nuclear decrease because… Hydro and nuclear increase, petroleum decrease because… Coal increase, hydro and nuclear decrease because… I have another choice besides these, it is… Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– ANSWER While there are a variety of opinions, the decrease or increase of energy sources will depend on a complex combination of politics, life style choices, economics and technology developments. Section 6.5: Present Sources of Energy Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– QUESTION #11 One source of energy being considered as a replacement for petroleum fuels is the combustion of hydrogen. There are several technical considerations to bring this about, but the heat of combustion reaction is part of the determination. How much heat, under standard conditions and 25°C, could be obtained from the combustion of 2.00 kg of H2? H2 (g) + ½ O2 (g) H2O (l) 1. 2. 3. 4. –572,000 kJ –484,000 kJ –284,000 kJ –242,000 kJ Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6– ANSWER Choice 3 has taken into account the kg to g to mole conversion for 2.00 kg of H2 and the stoichiometry. Since combustion of one mole of H2 in the equation yields 286 kJ, the number of moles of H2 multiplied by the H value for the equation (using H2O (l)) will yield the correct value. Section 6.6: New Energy Sources Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6–