chapter 6.3-6.4 thermochemistry phall not integrated into presentation

Chapter 6.3-6.4

Thermochemistry

PHALL not integrated into presentation

THERMODYNAMICSTHERMODYNAMICS

Courtesy of lab-initio.com

3

Chapter 6

Table of Contents

Copyright © Cengage Learning. All rights reserved

6.1 The Nature of Energy

6.2 Enthalpy and Calorimetry

6.3 Hess’s Law

6.4 Standard Enthalpies of Formation

6.5 Present Sources of Energy

6.6New Energy Sources

4

Section 6.3

Hess’s Law


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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

5

Section 6.3

Hess’s Law


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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

6

Section 6.3

Hess’s Law


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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

7

Section 6.3

Hess’s Law


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Hess’s LawHess’s Law

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.

ΔH = H (products) – H (reactants)

ΔH = heat given off or absorbed during a reaction at constant pressure

Hproducts < Hreactants

ΔH < 0Hproducts > Hreactants

ΔH > 0 6.3

Copyright © Houghton Mifflin Company. All rights reserved. 6–

Table 6.2 Standard Enthalpies of Formation for Several Compounds at 25°C

Hess’s LawHess’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 Hess’s Law Problem Example 1Example 1

Calculate ΔH for the combustion of methane, CH4: CH4 + 2O2 CO2 + 2H2O

Reaction ΔHo

C + 2H2 CH4 -74.80 kJ

C + O2 CO2 -393.50 kJ

H2 + ½ O2 H2O -285.83 kJ

Step #1: CH4 must appear on the reactant side, so we reverse reaction #1 and change the sign on ΔH.

CH4 C + 2H2 +74.80 kJ



Reaction ΔHo

C + 2H2 CH4 -74.80 kJ

C + O2 CO2 -393.50 kJ

H2 + ½ O2 H2O -285.83 kJ

CH4 C + 2H2 +74.80 kJ

Step #2: Keep reaction #2 unchanged, because CO2 belongs on the product side

C + O2 CO2 -393.50 kJ



Reaction ΔHo C + 2H2 CH4 -74.80 kJ

C + O2 CO2 -393.50 kJ

H2 + ½ O2 H2O -285.83 kJ

CH4 C + 2H2 +74.80 kJ C + O2 CO2 -393.50 kJ

Step #3: Multiply reaction #3 by 2

2H2 + O2 2 H2O -571.66 kJ



Reaction ΔHo C + 2H2 CH4 -74.80 kJ

C + O2 CO2 -393.50 kJ

H2 + ½ O2 H2O -285.83 kJ

CH4 C + 2H2 +74.80 kJ

C + O2 CO2 -393.50 kJ 2H2 + O2 2 H2O -571.66 kJ

Step #4: Sum up reaction and ΔH

CH4 + 2O2 CO2 + 2H2O -890.36 kJ

Hess’s Law Problem Example 2Hess’s Law Problem Example 2

EXAMPLE - Calculate ΔH for the reaction from 2 stepsN2(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

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Section 6.3

Hess’s Law


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• 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) ΔH2 = 180 kJ

2NO(g) + O2(g) → 2NO2(g) ΔH3 = – 112 kJ

N2(g) + 2O2(g) → 2NO2(g) ΔH2 + ΔH3 = 68 kJ

ΔH1 = ΔH2 + ΔH3 = 68 kJ

N2(g) + 2O2(g) → 2NO2(g) ΔH1 = 68 kJ

Hess’s Law Problem Example 2Hess’s Law Problem Example 2

EXAMPLE - Calculate ΔH for the reaction from 2 stepsN2(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. Calculation of Heat of Reaction EX. 33Calculate ΔH for the combustion of methane, CH4:

CH4 + 2O2 CO2 + 2H2O

Substance ΔHf

CH4 -74.80 kJ

O2 0 kJ

CO2 -393.50 kJ

H2O -285.83 kJ

Calculation of Heat of Reaction EX. Calculation of Heat of Reaction EX. 33Calculate ΔH for the combustion of methane, CH4:

CH4 + 2O2 CO2 + 2H2O Substance ΔHf

CH4 -74.80 kJ

O2 0 kJ

CO2 -393.50 kJ

H2O -285.83 kJ

ΔHrxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ]

ΔHrxn = -890.36 kJ

21

Section 6.3

Hess’s Law


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The Principle of Hess’s Law

Copyright © Houghton Mifflin Company. All rights reserved. 6–

Hess’s Law

Click here to watch visualization.

http://college.cengage.com/chemistry/zumdahl/chemistry/7e/assets/students/protected/fae/index.html?layer=act&src=qtiworkflowflash_6_9.xml&w=750&h=434

23

Section 6.3

Hess’s Law


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• 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.

Characteristics of Enthalpy Changes

24

Section 6.3

Hess’s Law


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• Consider the following data:

• Calculate ΔH for the reaction

Example 4

25

Section 6.3

Hess’s Law


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• 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.

Problem-Solving Strategy

26

Section 6.3

Hess’s Law


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• Reverse the two reactions:

• Desired reaction:

Example 4

27

Section 6.3

Hess’s Law


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• Multiply reactions to give the correct numbers of reactants and products:

4( ) 4( )

3( ) 3( )


Example 4

28

Section 6.3

Hess’s Law


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• Final reactions:


Example 4

ΔH = +1268 kJ

29

Section 6.3

Hess’s Law


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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 (?)

Need to add 6.4 to student notes Need to add 6.4 to student notes outline next time for heat of outline next time for heat of

formationformation

Monday - Nov. 18, 2013Monday - Nov. 18, 2013• CW: Notes 6.4 (brief) (6.3 done last Tuesday & 6.4 CW: Notes 6.4 (brief) (6.3 done last Tuesday & 6.4

intro.)intro.)• QUIZ - 2 questions - Hess’s LawQUIZ - 2 questions - Hess’s Law• CW: Finish Notes 6.1-6.2 from handout sheetCW: Finish Notes 6.1-6.2 from handout sheet• CW: Hess’s Law problems finish and turn inCW: Hess’s Law problems finish and turn in• HW: Quia.com Quiz (9 questions)HW: Quia.com Quiz (9 questions)• HW: Hess’s Law Informal Lab report due Wed.HW: Hess’s Law Informal Lab report due Wed.• (Be sure Calorimetry Lab has been turned in today to not (Be sure Calorimetry Lab has been turned in today to not

lose points for being late.)lose points for being late.)• TEST ch. 6 - Tuesday - Nov. 26, 2013 before Thanksgiving TEST ch. 6 - Tuesday - Nov. 26, 2013 before Thanksgiving

Break (will not include ch. 16 for test now)Break (will not include ch. 16 for test now)

32

Section 6.4

Standard Enthalpies of Formation


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• 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.)

Standard Enthalpy of Formation (ΔHf°) is the

KEY - VOCABULARY DEFINITION

33

Section 6.4



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• 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.

Conventional Definitions of Standard States

34

Section 6.4



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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

35

Section 6.4



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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.

Problem-Solving Strategy: Enthalpy Calculations

36

Section 6.4



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3. The change in enthalpy for a given reaction can be calculated from the enthalpies of formation of the reactants and products:

ΔH°rxn = ΣnpΔHf(products) - ΣnrΔHf(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.

Problem-Solving Strategy: Enthalpy Calculations

37

Section 6.4



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Exercise

Calculate ΔH° for the following reaction:

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

Given the following information:

ΔHf° (kJ/mol)

Na(s) 0

H2O(l) –286

NaOH(aq) –470

H2(g) 0

ΔH° = (-470 kJ/mol)(2 mol) + 0 - [(-286 kJ/mol)(2mol) + 0]

–368 kJ

Monday - Nov. 18, 2013Monday - Nov. 18, 2013• CW: Notes 6.4 (brief) (6.3 done last Tuesday & 6.4 CW: Notes 6.4 (brief) (6.3 done last Tuesday & 6.4

intro.)intro.)• QUIZ - 2 questions - Hess’s LawQUIZ - 2 questions - Hess’s Law• CW: Finish Notes 6.1-6.2 from handout sheetCW: Finish Notes 6.1-6.2 from handout sheet• CW: Hess’s Law problems finish and turn inCW: Hess’s Law problems finish and turn in• HW: Quia.com Quiz (9 questions)HW: Quia.com Quiz (9 questions)• HW: Hess’s Law Informal Lab report due Wed.HW: Hess’s Law Informal Lab report due Wed.• (Be sure Calorimetry Lab has been turned in today to not (Be sure Calorimetry Lab has been turned in today to not

lose points for being late.)lose points for being late.)• TEST ch. 6 - Tuesday - Nov. 26, 2013 before Thanksgiving TEST ch. 6 - Tuesday - Nov. 26, 2013 before Thanksgiving

Break (will not include ch. 16 for test now)Break (will not include ch. 16 for test now)

Wednesday - Nov. 20, 2013Wednesday - Nov. 20, 2013• TURN IN informal Hess’s Law Lab Report in box.TURN IN informal Hess’s Law Lab Report in box.• TURN IN Calorimetry Lab (formal lab report) in box if not TURN IN Calorimetry Lab (formal lab report) in box if not

already donealready done• TURN IN Hess’s Law problems worksheet with work TURN IN Hess’s Law problems worksheet with work

shown or attached.shown or attached.• CW: Thermodynamics Team Activity Game with Partners CW: Thermodynamics Team Activity Game with Partners

(selected by drawing cards) (selected by drawing cards) • HW: Be sure all ch. 6 notes are caught up 6.1-6.4HW: Be sure all ch. 6 notes are caught up 6.1-6.4• TEST is TUESDAY Nov. 26, 2013 over chapter 6, TEST is TUESDAY Nov. 26, 2013 over chapter 6,

calorimetry labs & thermodynamics problems & notes.calorimetry labs & thermodynamics problems & notes.• HW/CW: Quia.com Quiz (9 questions)HW/CW: Quia.com Quiz (9 questions)• HW/CW: Sciencegeek.net Quiz ch. 6 - 15 questionsHW/CW: Sciencegeek.net Quiz ch. 6 - 15 questions

• Over Thanksgiving Break: Read chapter 16 :)Over Thanksgiving Break: Read chapter 16 :)

Friday - Nov. 22 2013Friday - Nov. 22 2013• Follow up Follow up latelate items to turn in: items to turn in: • Hess’s Law Informal Lab Report in Hess’s Law Informal Lab Report in • Calorimetry Lab (formal lab report)Calorimetry Lab (formal lab report)• Hess’s Law problems worksheet with work shown Hess’s Law problems worksheet with work shown • Hess’s Law - 2 question quizHess’s Law - 2 question quiz• To be turned in with TEST on TUESDAY:To be turned in with TEST on TUESDAY:• 1) Thermodynamics Team Activity Game with work 1) Thermodynamics Team Activity Game with work

shownshown• 2) Interactive ch. 6 notes2) Interactive ch. 6 notes• 3) By Monday Quia.com Quiz (9 questions) at latest - not 3) By Monday Quia.com Quiz (9 questions) at latest - not

accepted late for points.accepted late for points.• TODAYTODAY: Show me HW/CW: : Show me HW/CW: Sciencegeek.net Quiz AP ch. 6Sciencegeek.net Quiz AP ch. 6

- 15 question results and work shown if applicable.- 15 question results and work shown if applicable.• TODAYTODAY: Thermochemistry Link Reviews: Thermochemistry Link Reviews• TEST is TUESDAY Nov. 26TEST is TUESDAY Nov. 26, 2013 over chapter 6, , 2013 over chapter 6,

calorimetry labs & thermodynamics problems & notes.calorimetry labs & thermodynamics problems & notes.

• Over Thanksgiving Break: Read chapter 16 :)Over Thanksgiving Break: Read chapter 16 :)

41

Section 6.4



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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

42

Section 6.4



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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

Chapter 6Questions

ThermochemistryThermochemistry

Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 6–

QUESTION #1The 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 negativeΔE would have a + overall value because the surroundings have

gained energy.


ANSWERChoice 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


QUESTION #2The change in enthalpy, or heat of reaction, for chemical changes is equal to the heat flow when…

1. the pressure is constant and only PV work is allowed.ΔH = –q (constant pressure)ΔH = ΔE – PΔV4. the pressure is constant and enthalpy = internal energy (E)


ANSWERChoice 1 properly relates enthalpy to q. At constant pressure, any work is change in volume work. So, ΔH = ΔE + PΔV and since q (at constant pressure) = ΔE + PΔV it follows that ΔH = q (at constant P where most chemical reactions take place).

Section 6.2: Enthalpy and Calorimetry


QUESTION #3Acetylene torches used for welding work so effectively because of the large release of energy (ΔH) when acetylene, C2H2, reacts with oxygen. For the combustion of acetylene, ΔH is approximately –1,300 kJ/mol. What is the value for ΔH when 1.0 gram of acetylene reacts in this way?

1. Approximately +50. kJ2. Approximately –50. kJ3. Approximately +34,000 kJ4. Approximately –34,000 kJ


ANSWERChoice 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.



QUESTION #4The 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 ΔH for dissolving one mole of NH4NO3?

1. –2,500 kJ/mol2. +2,500 kJ/mol3. –25 kJ/mol4. +25 kJ/mol


ANSWERChoice 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.



QUESTION #5


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. 30.20 kJ/°C2. 106.8 kJ/K3. 8.894 kJ/°C4. 35.60 kJ/°C


ANSWERChoice 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.



QUESTION #6A 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. 5.00°C2. 126°C3. 25.1°C4. Couldn’t I just eat the candy, I am not sure what to do with the

calculations.


ANSWERChoice 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.



QUESTION #7The following reaction shows the conversion of ozone to oxygen and provides the standard enthalpy change for the process:

2 O3 → 3 O2; ΔH = –427 kJ

What is the ΔH value for the conversion of one mole of O2 to O3?

1. +427 kJ2. +641 kJ3. +142 kJ4. +285 kJ


ANSWERChoice 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 ΔH must also be reversed. Since the reversed reaction shows the process for 3 moles of oxygen, the ΔH value must be reduced by 1/3 to determine the value for only one mole.

Section 6.3: Hess’s Law


QUESTION #8The following reaction depicts one way in which nitric acid could be formed:

N2O5 + H2O → 2 HNO3 ΔH = ?

Use the following information, and apply Hess’s law, to determine the missing value.

H2 + 1/2 O2 → H2O; ΔH = –285.8 kJ½ N2 + 3/2 O2 + ½ H2 → HNO3; ΔH = –174.1 kJ2N2 + 5 O2 → 2 N2O5; ΔH = + 28.4 kJ


QUESTION (continued)1. –362.4 kJ2. –69.5 kJ3. –76.6 kJ4. I am having difficulty rearranging the given equations to produce

the “target” equation.


ANSWERChoice 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 ΔH values before tallying those to yield the final ΔH value.

Section 6.3: Hess’s Law


QUESTION #9Using 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. –4430 kJ2. –4780 kJ3. –5310 kJ4. This can’t be solved without the heat of formation of oxygen.


ANSWERChoice 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


QUESTION #10


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. Wood increase; nuclear decrease because…2. Hydro and nuclear increase, petroleum decrease because…3. Coal increase, hydro and nuclear decrease because…4. I have another choice besides these, it is…


ANSWERWhile 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


QUESTION #11One 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. –572,000 kJ2. –484,000 kJ3. –284,000 kJ4. –242,000 kJ


ANSWERChoice 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 ΔH value for the equation (using H2O (l)) will yield the correct value.

Section 6.6: New Energy Sources

chapter 6.3-6.4 thermochemistry phall not integrated into presentation

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