Lecture 8: Thermochemistry - PowerPoint Presentation

Slide1

Lecture 8: Thermochemistry

Lecture 8 Topics

Brown chapter 5

8.1

:

Kinetic

vs. potential energy

5.1

8.2

:

Transferring

energy as heat & work

Thermal energy

8.3:

System

vs. surroundings

Closed systems

8.4:

First

Law of Thermodynamics

5.2

Internal energy of chemical reactions

Energy

diagrams



E, system & surroundings

8.5:

Enthalpy

5.3

Exothermic

vs. endothermic

Guidelines thermochemical equations 5.4

Hess’s

Law

5.6

8.6:

Calorimetry

Constant

pressure

calorimetry

8.7:

Enthlapy

of formation

5.7Slide2

Enthalpy of formation

Enthalpy of formation: enthalpy to form 1 mole

Enthalpies of formation can be used to calculate enthalpy of reaction.Slide3

standard states?

see table 5.3, p. 177 & appendix C

p.

183-5

The enthalpy of formation of any substance in its

most stable form is always zero

:

Carbon as graphite; oxygen as O2 gas; copper as an elemental solid (metal).

There are many forms (or expressions) of enthalpy including

H

vaporization

, Hfusion, Hcombination & Hformation. The most useful is standard enthalpy of formation (H°f): the enthalpy change that accompanies the formation of 1 mole of a compound from elements in their standard states.

Na? Hg? N2?

Write the reactions for H°f for:CH3CH2OHFePO4

2C + 1/2O2 + 3H2

 CH3CH2OH

Fe + P + 2O2  FePO4

Enthalpy of formation (ΔHf)

The

form of the element at

1 atm, 298 K/25°C gases - 1 atm solutions - 1 M

s

olid

l

iquid

g

asSlide4

p. 183-5

a. 2Na(s) + 1/2O2

(g) 

Na

2

O(s)

2K(l) + Cl

2

(g)

 2KCl(s)

C6H12O6(s)  6C(diamond) + 6H2(g) + 3O2(g)Yes - all in standard states & 1 mole product is formedNo - K should be (s) & 2 moles product are formedNo - this is a decomposition, not a formation; reverse itExample: Δ

Hf

For which reaction (at 25°C) would enthalpy change represent standard enthalpy

of formation?If not, how could you change the reaction conditions?Slide5

p. 185-7

If we know the H

f of all reactants & products, we can calculate H

rxn

:

Propane (C

3

H

8

) is combusted to form CO

2

and H2O under standard cond.C3H8 + 5O2  3CO2 + 4H2O(gas) (gas) (gas) (liquid)C3H8 --> 3C + 4H23C + 3O2 --> 3CO2

4H2 + 2O

2 --> 4H2O H1

= -Hf [C3

H8(g)]

H2 = 3Hf [CO2(g)]

H3 = 4Hf [H2O(l)]

C3H8

+ 5O2 --> 3CO2 + 4H

2OH°rxn = H1 + H2 + H3

H°rxn

= -

Hf

[C

3

H

8

(g)] + 3Hf [CO

2

(g)] + 4Hf [H

2

O(l)]

= -(-103.85 kJ) + 3(-393.5 kJ) + 4(-285.8 kJ)

= -2220 kJ

reversed

state !

stoichiometry

(

moles)(kJ

/mole)

H°rxn

=

nH°f(products

) -

nH°f(reactants

)

So this concept combines Hess’s Law with

H

f

.

Expressed via Hess’s Law:

Δ

Hfs

can be summed to calculate

ΔHrxn

Slide6

p. 185-7

Calculate the H for combustion of 1 mole benzene (C6

H6).

C

6

H

6

+ 15/2O

2



6CO2 + 3H2O(liquid) (gas) (gas) (liquid)Hrxn = [6Hf(CO2) + 3Hf(H2O)] - [Hf(C6H6) + 15/2 Hf(O2)]

= [6(-393.5 kJ) + 3(-285.8 kJ)] - [(49.0 kJ) + 15/2(0 kJ)]= (-2361 - 857.4 - 49.0) kJ

= - 3267 kJ

 products

 reactants

Use enthalpies of formation to calculate the Hrxn for: CaCO3(s) --> CaO(s

) + CO2(g)Hrxn = [Hf(CaO) + Hf(CO2)] - [Hf(CaCO3)]= [(-635.5 kJ) + (-393.5 kJ)] - [(-1207.1 kJ)] = 178.1 kJ

 products



reactantsExamples: ΔHfSlide7

p. 188-9

Carbohydrates? Glucose = C6H12O6

What about industrial fuels?

fuel value

(kJ/

g

)

Coal

31 - 32 Oil 45 Natural gas 49 Gasoline 48 H2 142Fats? Triacylglycerol (body fat) = C57H110O6C6H12O6 + 6O2  6CO2 + 6H2O Hrxn = - 2803 kJ

Average fuel value for carbohydrates (& proteins) is ~17 kJ/g.

2C57H110O6 + 163O2 --> 114CO2 + 110H2O Hrxn = - 75,520 kJ

Average

fuel value for carbohydrates (& proteins) is ~38 kJ/g.

Enthalpies of combustion of fuels