Intersection 6: Thermochemistry

10/10/05

6.4 p228-232; 6.7 p240-242; 6.10-6.11 p 248-256

11.3-11.4 p494-507

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A Review of Thermochemistry (from Studio)

• Thermochemical EquationsC(graphite) + O2(g)→ CO2(g)  ΔH = -393.509 kJ/mol

• Endothermic and exothermic

• Heats of reactions are additive (Hess’s Law)

• Calorimetry – How heats of reaction are measured– Heat capacity q = mCT

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Problem 1

Suppose you burn 0.300 g C(graphite) in an excess of O2(g) to give CO2(g). The temperature of the calorimeter which contains 775 g of water increases from 25.00oC to 27.38oC.

What is the quantity of thermal energy evolved per mole of graphite?

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Question 1

100 mL of water at 25°C and 100 mL of alcohol at 25°C are both heated at the same rate under identical conditions. After 3 minutes the temperature of the alcohol is 50°C. Two minutes later the temperature of the water is 50°C.

Which liquid received more heat as it warmed to 50°C?

a.The water. b.The alcohol. c.Both received the same amount of heat. d.It is impossible to tell from the information given.

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Do all substances have the same heat capacity?

Explain why at the beginning of the summer, the air is really warm, but the water at the beach is not.

It takes 145 J to raise the temperature of 1 mole of air by 5oC.

Is the heat capacity of air higher, lower, or the same as water?

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Thermochemistry (in IS)

• Heat and bonds– Breaking bonds takes energy

– Making bonds releases energy

– Trends in bond enthalpy

• Calculate Heats of Reactions– Heats of reaction can be calculated using bond enthalpy

– Heat of reactions can be calculated using heats of formation

• Change of state and heat energy

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BondsWhy do bonds form?

Does it take energy or give off energy to break a bond?

Picture from: www.nios.ac.in/ sc10/ch5sc10.htm

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The energy given off to make H-H bonds is shown below.

H(g) + H(g) → H2(g)     Δ H = -435 kJ/mol        The opposite process, breaking H-H bonds requires the

input of the same amount of energy that was given off in making the bond  

H2(g) → H(g) + H(g)      Δ H = +435 kJ/mol   

Which process is endothermic? Exothermic?  

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Bond Enthalpy (kJ/mol) (average energy needed to break bonds)

H-F 566 C=C 598 C-O 336

H-Cl 431 C=N 418 C=O 695

H-Br 366 C=O 695-803 C≡O 1073

H-I 299

C-C 356

C-Si 301 C-C 356 C=C 598

C-P 264 C-N 285 C≡C 813

C-S 272 C-O 336

C-Cl 327 C-F 486

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Energy terms

Heat energy: enthalpy, H, q

Enthalpy (H) is the quantity of thermal energy transferred into a system at constant pressure. 

 Δ E = Δ H + wexpansion

"ΔH accounts for all the energy transferred except the quantity that does the work of pushing back the atmosphere, which is usually relatively small.  Whenever heat transfer is measured at constant pressure, it is ΔH that is determined." (Moore, p232)  

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Holy Contradiction!

It takes energy to break bonds….and yet we get energy from fats (and

carbohydrates and alcohol) by breaking

them down ???

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How do we get energy from food?

Food is broken down by adding O2 and converting all of the carbon and hydrogen in the food molecules to CO2 and H2O.

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Energy from Ethanol CH3CH2OH  +  O2 →   CO2 + H2O

H

C

H

H C

H

H

O O O O C OH

O

HH 3 2 3+ +

moles Bonds broken

Enthalpies kJ/mol

moles Bonds

made

Enthalpies kJ/mol

C-H 416 C=O 803

C-C 356 O-H 467

C-O 336

O-H 467

O=O 498

Total

3 2 3

5

1

1

1

3

4

6

4733 kJ (-) 6014 kJ

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A Picture of Breaking Down EthanolA

Total H

H

C

H

H C

H

H

O O O O C OH

O

HH 3 2 3+ +

4733 kJ into reactants 6014 kJ out of products

Δ Hreaction = Σ Hreactants - Σ Hproducts

= -1281 kJ

Watch your signs!!!!!!!!!!….!!!!!

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Breaking Down Glucose

C6H12O6 + ___O2 ____CO2 + ___H2O

moles bonds kJ/mol moles bonds kJ/mol

C-H 416 C=O 803

C-C 356 O-H 467

C-O 336

O-H 467

O=O 498

Total

6 6 6

7

5

7

5

6

12

12

12217 kJ 15240 kJ

Δ Hreaction = Σ Hreactants - Σ Hproducts = -3023 kJ

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HOO

C

CC

H

H

HO

H

OH

OHH

CH2

HC

C

OH

Inflaming Nitrocellulose

OO

C

CC

H

H

O

H

OO

H

CH2

HC

C

O

N

O

O

N

O

O

N

OO

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Cellulose Oxidation

C6O5H10 + 6 O2 6 CO2 + 5 H2O

Nitrocelluose Oxidation

4 C6O11N3H7 + 9 O2 24 CO2 + 14 H2O + 6 N2

OO

C

CC

H

H

HO

H

OOH

H

CH2

HC

C

OH

n

HNO3/H2SO4

Hreaction Comparison

Cellulose Combustion

Reactants Products

Moles

Bonds

kJ/mol Moles Bonds kJ/mol

7 C-H 416 12 C=O 803

5 C-C 356 10 O-H 467

7 C-O 336

3 O-H 467

6 O=O 498

Total 11,433

14306

Δ Hreaction = -2870 kJ/mol

Nitrocellulose Combustion

Reactants Products

Moles

Bonds

kJ/mol Moles Bonds kJ/mol

28 C-H 416 48 C=O 803

20 C-C 356 28 O-H 467

28 C-O 336 6 N≡N 946

- O-H 467

9 O=O 498

24 N-O 201

12 N=O 607

Total 44,776

57,296

Δ Hreaction = -12,520 kJ/mol

Conversion to food Calories• 1 Calorie = 1 kilocalorie = 1000 calories

• 1 calorie = 4.184 J

• Convert the enthalpies for glucose and ethanol to Cal and Cal/g:

3023 kJ 1000J 1 cal 1Cal

mol 1 kJ 4.184 J 1000 cal = 722.5 Cal/mol

722.5 Cal 1 mol glucose

mol glucose 180 g

For glucose

= 4.0 Cal/g

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Comparing Energy Sources

C18H36O2 + 26 O2 18 CO2 + 18 H2O H = -10622 kJ

Stearic acid

Bond Energy

Nutritional Guidelines*

Ethanol 6.7 Cal/g 7 Cal/g

Glucose 4.0 Cal/g 4 Cal/g

Stearic Acid 8.9 Cal/g 9 Cal/g

*http://www.unlimited-health.com/nutrition/

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Evolution due to fat storage….."Animals, though they store small amounts of glycogen in the

muscles and liver, and rely on glucose in the bloodstream for immediate energy needs, bank most of their energy reserves in fat.  Economy in weight is important for an organism that must work not only to move, but simply to stand up against the force of gravity.  The body of a typical woman is about 25% fat by weight (a man's body is closer to 15%).  This means that if her fat supply were converted to its energy equivalent in carbohydrates, a 120-pound woman would weigh 150 pounds.  In a world where swiftness and agility increase an animal's chances for survival, fat is clearly the preferable means of energy storage. (Harold McGee, On Food and Cooking. Scribners, New York:  1984, p. 597.)

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Who wants to count up bonds every time?

Δ Hof standard enthalpy of formation is the energy given

off or needed to form a substance from its elements in their standard states [ie, O2(g), H2(g), N2(g), Br2(l), Na(s)..etc].  

The o stands for standard conditions of 1 atm and 298K.   

The formation reaction of gaseous water would be written:

H2(g)  +  1/2 O2(g)  →  H2O(g)     

  Δ Hrxn (1 atm, 298K) = Δ Hof  H2O(g) = -241.83 kJ/mol

Δ Hof found in Appendix J of Moore or back of coursepack

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Table 6-2, p.250

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Standard Enthalpy of Formation

H2(g)  +  1/2 O2(g)  →  H2O(g)     Δ Hof H2O(g) = -241.83

kJ/mol

Do you get a different answer from bond enthalpies?

 

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moles bonds

kJ/mol moles bonds kJ/mol

H-H 436 O-H 467

O=O 498

Total

Hof : Signs and Multiplicity

C(graphite) + O2(g)→ CO2(g)   ΔH = -393.509 kJ/mol

CO2(g) → C(graphite) + O2(g)   ΔH = +393.509 kJ/mol

2C(graphite) +2 O2(g)→ 2CO2(g)   ΔH = -787.018 kJ/mol

4CO2(g) → 4C(graphite) + 4O2(g)   ΔH = +1574.04 kJ/mol

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Calculate the Hrxn using heats of formationCH3CH2CH3 (g) +   5 O2 (g)   →   3 CO2 (g) + 4 H2O (g)

3C(s,graphite)+ 4H2(g) → CH3CH2CH3 (g) H = - 103.82 kJ/mol

C(s,graphite)+ O2 (g) → CO2 (g) H = - 393.5 kJ/mol

H2(g) + ½ O2(g) → H2O(g) H = - 241.82 kJ/mol

O2 (g) H = 0 kJ/mol

CH3CH2CH3(g) → 3C(s,graphite) + 4H2(g) H = + 103.82 kJ/mol

3C(s,graphite)+ 3O2 (g) → 3CO2 (g) 3(H = - 393.5 kJ/mol)

4H2(g) + 2 O2(g) → 4H2O(g) 4(H = - 241.82 kJ/mol)

-2043.96 kJ/mol

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 Σ nΔ Hof reactants = -103.82 kJ/mol (1 mol propane)  +  0 kJ/mol (5 mol O2)

= -103.82 kJ

 Σ nΔ Hof products = -393.5 kJ/mol (3 mol CO2)  +  -241.83 kJ/mol (4 mol H2O)

= - 2147.82 kJ

Δ Hrxn = Σ nHof products - Σ nHo

f reactants 

         = -2147.82 kJ - (-103.82 kJ)        = -2044 kJ

Calculate the Hrxn using heats of formationCH3CH2CH3 (g) +   5 O2 (g)   →   3 CO2 (g) + 4 H2O (g)

Form the products (keep sign)

Unform the reactants (change sign)

Δ Hrxn = Σ nHof products - Σ nHo

f reactants

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Iron Oxidation

4 Fe + 3 O2 2 Fe2O3

Hf for Fe2O3 = -824.2 kJ/mol

What is Hf for Fe and O2??

Δ Hrxn = Σ nHof products - Σ nHo

f reactants

Δ Hrxn = Σ nHof products - Σ nHo

f reactants 

         = 2(-824.2 kJ/mol) - 0        = -1648 kJ/mol

Problem 2

2 C8H6(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g) H = -10,992 kJ

Write a thermochemical equation for:

a) The production of 4.00 mol CO2(g)

b) The combustion of 100. mol C8H6

c) How much heat energy is produced when 50 g of C8H6 is burned?

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Problem 3

The standard molar enthalpy of formation of AgCl(s) is –127.1 kJ/mol. Write a balanced thermochemical equation for which the enthalpy of reaction is –127.2 kJ

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Problem 4

The Romans used CaO as mortar in stone structures. The CaO was mixed with water to give Ca(OH)2, which slowly reacted with CO2 in the air to give limestone.

Ca(OH)2(s) + CO2(g) CaCO3(s) + H2O(g)

Calculate the enthalpy change for this reaction

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Question 2

Heat is given off when hydrogen burns in air according to the equation 2H2 + O2  2H2O

Which of the following is responsible for the heat? a. Breaking hydrogen bonds gives off energy. b. Breaking oxygen bonds gives off energy. c. Forming hydrogen-oxygen bonds gives off energy. d. Both (a) and (b) are responsible. e. (a), (b), and (c) are responsible.

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Debate: _________is the safest and best artificial sweetener

Friday, October 13th in studio– Splenda (sucralose)– Aspartame – Saccharin – Acesulfame K

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Chips Points Time

Opening 3 9 6 min

Rebuttal 3 5 3 min

Question 1 1 2 30 sec

Response 1 1 3 1 min

Question 2 1 2 30 sec

Response 2 1 3 1 min

Question 3 1 2 30 sec

Response 3 1 3 1 min

Question 4 1 2 30 sec

Response 4 1 3 1 min

Closing 3 6 3 min

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