READING for Thursday: 7.3 – 7.4 HOMEWORK – DUE TUESDAY 10/20/15

HW-BW 7.1 (Bookwork) CH 7 #’s 5, 7-12 all, 14, 15, 20, 21, 24, 28-31 all, 34 HW-WS 12 (Worksheet) (from course website)

HOMEWORK – DUE THURSDAY 10/22/15 HW-BW 7.2 (Bookwork) CH 7 #’s 39, 42, 48-52 all, 55-60 all, 64, 69, 71, 72,

78, 90 HW-WS 13 (Worksheet) (from course website)

Lab Wednesday/Thursday – finish EXP 8 Next Monday/Tuesday – EXP 9

More Enthalpy DiagramDraw and enthalpy diagram and calculate the DHrxn for:

N2O(g) + 2 O2(g) N2O5(s)

Ent

halp

y

elementsN2(g) + 5/2 O2(g)

reactantsN2O(g) + 2 O2(g)

1) formation of reactants

productsN2O5(g)

DH = 1 mol (+82.05 kJ/mol) = +82.05 kJ

4) heat release in formation of products-43.1 kJ

heat of formation of reactants+82.05 kJ

DH = -43.1 kJ – (+82.05 kJ) = -125.2 kJ

2) formation of N2O5(g)DH = 1 mol (-43.1 kJ/mol) =

-43.1 kJ

Heat of Reaction for Heats of FormationJust like in the enthalpy diagrams, but with less diagram

formation, productsrxnH H C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)

2 ( ),241.826

gf H OH

kJ

mol2( ),393.5

gf COH

kJ

mol

3 8( ),105

gf C HH

kJ

mol

formation, products formation, reactantsrxnH H H

2( ),0

gf OH kJ

mol

formation, products393.

35

1H

22 CO

CO m

kJ

molol

formation, reactants105

11

H

3 83 8

C H C H

kJ mol

mol

( )2147.8 ( ) 204004 31 5rxnH kJ kJ kJ

formation, products393.5 241.826

3 4 2147.8041 1

H

2 22 2

CO H O CO H O

kJ kJ mol mol kJ

mol molformation, products393.5 241.826

3 41 1

H

2 22 2

CO H O CO H O

kJ kJ mol mol

mol molformation, products393.5

3 41

H

2 22

CO H O CO

kJ mol mol

molformation, products393.5

31

H

22

CO CO

kJ mol

molformation, products393.5

31

H

22

CO CO

kJ mol

mol

formation, reactants105

11H

3

3 88 C

HH

C m

kJ

mololformation, reactants

1051

1H

3 8

3 8

C H C H

kJ mol

molformation, reactants105

1 51

H

3 8 23 8

C H O C H

kJ mol mol

molformation, reactants105 0

1 51 1

H

3 8 23 8 2

C H O C H O

kJ kJ mol mol

mol molformation, reactants105 0

1 5 1051 1

H

3 8 23 8 2

C H O C H O

kJ kJ mol mol kJ

mol mol

Heat of Reaction for Heats of FormationJust like in the enthalpy diagrams, but with less diagram

N2O(g) + 2 O2(g) N2O5(s)

2 5( ),43.1

sf N OH

kJ

mol2 ( ),82.05

gf N OH

kJ

mol

formation, products43.1

1 43.11

H

2 52 5

N O N O

kJ mol kJ

mol

formation, reactants82.05 0

1 2 82.051 1

H

2 22 2

N O O N O O

kJ kJ mol mol kJ

mol mol

843.1( ) ( ) 12. 5 22 05 .rxnH kJ kJ kJ

2( ),0

gf OH kJ

mol

formation, products formation, reactantsrxnH H H

Heat CapacityWhen a system absorbs heat, its temperature

increasesThe increase in temperature is directly proportional

to the amount of heat absorbedThe proportionality constant is called the heat

capacity, Cunits of C are J/°C or J/K

q = C x DT

The specific heat capacity is the amount of heat energy required to raise the temperature of one gram of a substance 1°C units are J/(g∙°C)

The molar heat capacity is the amount of heat energy required to raise the temperature of one mole of a substance 1°C

Specific Heat Capacity

q Cm T q Cm T q m C T

oCcal

g oCJ

g

q m C T heat (J or cal)

mass (g)

specific heat ( or )

change in temp. (oC or K)

DT = Tfinal - Tinitial

oCcal

g oCJ

g

q m C T

Specific Heat Capacity

The amount of energy (J or cal) that is required to raise 1 g of a substance by 1 degree Celsius.

Specific heat for water = or1.00

oC cal

g

4.184oC

J

g

Specific heat for silver = or0.0566

oC cal

g

0.237oC

J

g

Specific Heat Capacity

A SMALLER specific heat means a LARGER temperature change given the same amount of energy.

Specific Heat

Temperature change when 51000 calories are added to 1000.0 grams of each:

~900 oC

oCg

cal

11.00

oCg

al c

1 0.0566

o cal

cal

g C 1 510001.00 1

o cal

cal

g C

1 510000.0566 1

o cal

cal g

g C

1 51000 11.00 1 1000.0

o cal

c

g C

gal

1 51000 10.0566 1 1000.0

g1 o

c

C

al1.00

a c l51000

g

11 1000.0

g1 o

a

C

l c0.0566

a c l51000

g

11 1000.0

g1 o

c

C

al1.00

a c l51000

g

11 1000.0

oC51

g1 o

a

C

l c0.0566

a c l51000

g

11 1000.0

oC29.0 10

75.0 mL of ethanol at 25.0 oC is poured into a perfectly insulating coffee cup calorimeter containing 50.0 g of water at 33.4 oC. What is the final temperature of the ethanol? C(water) = 4.184 J/goC, C(ethanol) = 2.42 J/goC, d(ethanol) = 0.789g/mL

c ld oo h tq q

2

oH O fo

g Jq T C

g C

50.0 4.18433.4

1 1

2EtOH H Oq q

oEtOH fo

mL g Jq T C

mL g C

75.0 0.789 2.4225.0

1 1 1

2 fo

H O o

T

g Jq C

g C

50.0 4.18433.4

1 1

ofo o f

o mL g J g JT C T

m CC

g CL g

75.0 0.789 2.4225

50.0 4.1.0

1 1 1

8433.4

1 1

o f

oofo

JT C

JT C

C C

143.203525

209.233.4.0

oof fT C T C0.68453 25. 33.40

of f

oT C T C0.68453 17.11 5 33.432

foT C1 50.513.6845 253 o o

fT C C29.9866 30.0

Enthalpy (DH)C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) DHrxn = –2202.0 kJ

If it takes 7.20x103 kJ to perfectly cook a steak, how many grams of propane are necessary (assuming perfect heat transfer)?

3 1 44.117.20 10 144

2202.0 1 3 8 3 8

3 83 8

C H C H C H

C H

mol gkJ g

kJ mol

3 3 44.017.20 10 432

2202.0 1 2 2

22

CO CO CO

CO

mol gkJ g

kJ mol

What mass of carbon dioxide is produced when enough propane is burned to release 7.20x103 kJ?

Enthalpy (DH)C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) DHrxn = –2202.0 kJ

What mass of propane would need to be burned to raise 2500 grams of water from 25.0 oC to 100.0 oC? (MMpropane = 44.11 g/mol)

2500 2 H Og 4.184

25001 1

22

H O H O

g

J g

C4.184 75.0

250011 1

22

H O H O

J

C

g

Cg

3

4.184 75.0 12500

11 1 1 10

2

2

H O H O

J C kJg

g C

J

3

4.184 75.0 1 12500

1 11 1 1 10

2

2

H O H O

J C kJg

kJg C

kJ

J

3

14.184 75.0 1 12500

1 1 2202.01 1 1 10

2

3 82

C H O

H O

HJ C k

Jg

kJg C

m

JJ

olkJ

k

3

1 44.114.184 75.0 1 12500

1 1 2202.0 11 1 1 10

3 8 3 8

32

82

H O H

C H C H

HO C

J C k

Jg

kJg C

mol gkJ

kJ mol J

3

1 44.114.184 75.0 1 12500 16

1 1 2202.0 11 1 1 10

2

3 8 3 83 8

3 82

C H C H C H O H

H CH O

mol gkJJ C kJ

g

kJ molg

kJg C J

1 kJ absorbed by the water = 1 kJ released by burning propane

+1 kJwater = –1 kJpropane

4.184

1 12 H O

J

g

C

Enthalpy Calculations1.00 grams of solid KClO3 (122.55 g/mol) is dropped into 25.00 grams of water at 25.0 oC. If the heat of solution for KClO3 is +41.38 kJ/mol, what is the final temperature of the water?

What is the system? Dissolving of KClO3

What is the surrounding? Water!!

What would we measure? Temperature of WATER!!

What would we observe? Temperature of water decreasing!

Did the water gain or lose energy? Water LOST energy!!

Where did it go? Into the process of dissolving the KClO3(s)

1 41.381.00 0.3377

122.55 1

3

33 3

KClO KClO

KClO KClO

mol kJ g kJ

g mol

31

1 104.184

0.3377 25.00 25.01 1

final finalT T

o2

2

H O 21.8 C

H O

kJoJ

o

J kJ g C

g C

Enthalpy Calculations1.00 grams of solid KClO3 (122.55 g/mol) is dropped into 25.00 grams of water at 25.0 oC. If the heat of solution for KClO3 is +41.38 kJ/mol, what is the final temperature of the water?

What is the system? Dissolving of KClO3

What is the surrounding? Water!!

What would we measure? Temperature of WATER!!

What would we observe? Temperature of water decreasing!

Did the water gain or lose energy? Water LOST energy!!

Where did it go? Into the process of dissolving the KClO3(s)

3 1 1.001 41.381 1 1 1 103.228

4.184 25.00 1 1 122.1 55 1T

3 3

3 3

2

2

H O KCl

O KClO

KClO KH CO lO

oo mol g kJ kJ

m

g C JC

J g kJ kJ ol g

3.228 25.0final finalT T T o21.8 Co oC C

Enthalpy CalculationsWhen 2.10 grams of solid KOH (56.11 g/mol) is dropped into 55.00 grams of water at 22.5 oC, the water raises to 31.9 oC. What is the heat of solution for KOH in kJ/mol?

4.18455.00 9.4 2163

1 1

2

2

H O J H O

oo

J g C

g C

31

1 101

56.11

2163 57.8

12.10

KOH KOH

KOH KOH

kJJ

mol g

J kJ

mol g

3

1 1 56.11 1 57.8

1 2.10 11 10

55.004.184 9.4

1 1 11 1

2

2

H O

H

KOH

KOH KOH KOHO

o

o

g J C

J

J kJ g kJ

mol g olg mJC

What is the system? Dissolving of KOH

What is the surrounding? Water!!

What would we measure? Temperature of WATER!!

What would we observe? Temperature of water increasing!

Did the water gain or lose energy? Water GAINED energy!!

Where did it go? From the process of dissolving the KOH(s)