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# Unit 3: Thermochemistry - PowerPoint PPT Presentation

Unit 3: Thermochemistry. Chemistry 3202. Unit Outline. Temperature and Kinetic Energy Heat/Enthalpy Calculation Temperature changes (q = mc∆T) Phase changes (q = n∆H) Heating and Cooling Curves Calorimetry (q = C∆T & above formulas). Unit Outline. Chemical Reactions

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Unit 3: Thermochemistry

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## Unit 3: Thermochemistry

Chemistry 3202

### Unit Outline

• Temperature and Kinetic Energy

• Heat/Enthalpy Calculation

• Temperature changes (q = mc∆T)

• Phase changes (q = n∆H)

• Heating and Cooling Curves

• Calorimetry (q = C∆T & above formulas)

### Unit Outline

• Chemical Reactions

• PE Diagrams

• Thermochemical Equations

• Hess’s Law

• Bond Energy

• STSE: What Fuels You?

### Temperature and Kinetic Energy

Thermochemistryis the study of energy changes in chemical and physical changes

eg. dissolving

burning

phase changes

Temperature - a measure of the average kinetic energy of particles in a substance

- a change in temperature means particles are moving at different speeds

- measured in either Celsius degrees or degrees Kelvin

Kelvin = Celsius + 273.15

The Celsius scale is based on the freezing and boiling point of water

The Kelvin scale is based on absolute zero- the temperature at which particles in a substance have zero kinetic energy.

p. 628

300 K

# of particles

500 K

Kinetic Energy

### Heat/Enthalpy Calculations

system - the part of the universe being studied and observed

surroundings - everything else in the universe

open system - a system that can exchange matter and energy with the surroundings

eg. an open beaker of water

a candle burning

closed system - allows energy transfer but is closed to the flow of matter.

isolated system – a system completely closed to the flow of matter and energy

heat - refers to the transfer of kinetic energy from a system of higher temperature to a system of lower temperature.

- the symbol for heat is q

WorkSheet: Thermochemistry #1

### Heat/Enthalpy Calculations

specific heat capacity – the energy , in Joules (J), needed to change the temperature of one gram (g) of a substance by one degree Celsius (°C).

• The symbol for specific heat capacity is a lowercase c

• A substance with a large value of c can absorb or release more energy than a substance with a small value of c.

ie. For two substances, the substance with the larger c will undergo a smaller temperature change with the same loss or gain of heat.

q = mc∆T

q = heat (J)

m = mass (g)

c = specific heat capacity

∆T = temperature change

= T2 – T1

= Tf – Ti

### FORMULA

eg. How much heat is needed to raise the temperature of 500.0 g of water from 20.0 °C to 45.0 °C?

Solve q = m c ∆T

for c, m, ∆T, T2 & T1

• p. 634 #’s 1 – 4

• p. 636 #’s 5 – 8

WorkSheet: Thermochemistry #2

heat capacity- the quantity of energy , in Joules (J), needed to change the temperature of a substance by one degree Celsius (°C)

• The symbol for heat capacity is uppercase C

• The unit is J/ °C or kJ/ °C

C = mc

q = C ∆T

C = heat capacity

c = specific heat capacity

m = mass

∆T = T2 – T1

### FORMULA

WorkSheet: Thermochemistry #3

### Enthalpy Changes

enthalpy change - the difference between the potential energy of the reactants and the products during a physical or chemical change

AKA: Heat of Reaction or ∆H

Products

∆H

Reactants

PE

Endothermic Reaction

Reaction Progress

Products

∆H

Reactants

PE

Endothermic Reaction

Enthalpy

∆H

Reaction Progress

Products

Reactants

Enthalpy

∆H is +

Endothermic

products

∆H is -

reactants

Enthalpy

Exothermic

### Enthalpy Changes in Reactions

• All chemical reactions require bond breaking in reactants followed by bond making to form products

• Bond breaking requires energy (endothermic) while bond formation releases energy (exothermic)

see p. 639

### Enthalpy Changes in Reactions

endothermic reaction - the energy required to break bonds is greater than the energy released when bonds form.

ie. energy is absorbed

exothermic reaction - the energy required to break bonds is less than the energy released when bonds form.

ie. energy is produced

### Enthalpy Changes in Reactions

∆H can represent the enthalpy change for a number of processes

• Chemical reactions

∆Hrxn – enthalpy of reaction

∆Hcomb – enthalpy of combustion

(see p. 643)

• Formation of compounds from elements

∆Hof– standard enthalpy of formation

The standard molar enthalpy of formation is the energy released or absorbed when one mole of a compound is formed directly from the elements in their standard states. (see p. 642)

eg.

C(s) + ½ O2(g) → CO(g) ΔHfo = -110.5 kJ/mol

### Use the equations below to determine the ΔHfo for CH3OH(l) and CaCO3(s)

2 C(s) + 4 H2(g) + O2(g) → 2 CH3OH(l) + 477.2 kJ

2 CaCO3(s) + 2413.8kJ → 2 Ca(s) + 2 C(s) + 3 O2(g)

• Phase Changes (p.647)

∆Hvap – enthalpy of vaporization

∆Hfus – enthalpy of melting

∆Hcond – enthalpy of condensation

∆Hfre – enthalpy of freezing

eg. H2O(l)  H2O(g) ΔHvap =

Hg(l)  Hg(s) ΔHfre =

+40.7 kJ/mol

-23.4 kJ/mol

• Solution Formation(p.647, 648)

∆Hsoln– enthalpy of solution

• eg.

• ΔHsoln, of ammonium nitrate is +25.7 kJ/mol.

• NH4NO3(s) + 25.7 kJ → NH4NO3(aq)

• ΔHsoln, of calcium chloride is −82.8 kJ/mol.

• CaCl2(s) → CaCl2(aq) + 82.8 kJ

Three ways to represent an enthalpy change:

1. thermochemical equation - the energy term written into the equation.

2.enthalpy term is written as a separate expression beside the equation.

3.enthalpy diagram.

eg. the formation of water from the elements produces 285.8 kJ of energy.

1. H2(g) + ½ O2(g) → H2O(l) + 285.8 kJ

2. H2(g) + ½ O2(g) → H2O(l) ∆Hf = -285.8 kJ/mol

thermochemical equation

H2(g) + ½ O2(g)

∆Hf = -285.8 kJ/mol

H2O(l)

enthalpy diagram

3.

Enthalpy

(H)

examples:pp. 641-643

questionsp. 643 #’s 15-18

WorkSheet: Thermochemistry #4

FORMULA:

q = n∆H

q = heat (kJ)

n = # of moles

∆H = molar enthalpy

(kJ/mol)

### Calculating Enthalpy Changes

eg. How much heat is released when 50.0 g of CH4forms from C and H ?

(p. 642)

q = nΔH

= (3.115 mol)(-74.6 kJ/mol)

= -232 kJ

eg. How much heat is released when 50.00 g of CH4 undergoes complete combustion?

(p. 643)

q = nΔH

= (3.115 mol)(-965.1 kJ/mol)

= -3006 kJ

eg. How much energy is needed to change 20.0 g of H2O(l) at 100 °C to steam at 100 °C ?

Mwater = 18.02 g/molΔHvap = +40.7 kJ/mol

q = nΔH

= (1.110 mol)(+40.7 kJ/mol)

= +45.2 kJ

∆Hfre and ∆Hcond have the opposite sign of the above values.

eg. The molar enthalpy of solution for ammonium nitrate is +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?

q = nΔH

= (0.4996 mol)(+25.7 kJ/mol)

= +12.8 kJ

What mass of ethane, C2H6, must be burned to produce 405 kJ of heat?

ΔH = -1250.9 kJ

q = - 405 kJ

q = nΔH

n = 0.3238 mol

m = n x M

= (0.3238 mol)(30.08 g/mol)

= 9.74 g

Complete:p. 643 #’s 15 - 18

p. 645; #’s 19 – 23

pp. 648 – 649; #’s 24 – 29

19. (a) -8.468 kJ (b) -7.165 kJ

20. -1.37 x103 kJ

21. (a) -2.896 x 103 kJ

21. (b) -6.81 x104 kJ

21. (c) -1.186 x 106 kJ

22. -0.230 kJ

23. 3.14 x103 g

24. 2.74 kJ

25.(a) 33.4 kJ (b) 33.4 kJ

26.(a) absorbed (b) 0.096 kJ

27.(a) NaCl(s) + 3.9 kJ/mol → NaCl(aq)

(b) 1.69 kJ

(c) cool; heat absorbed from water

28. 819.2 g

29. 3.10 x 104 kJ

p. 638 #’ 4 – 8

pp. 649, 650 #’s 3 – 8

p. 657, 658 #’s 9 - 18

WorkSheet: Thermochemistry #5

### Heating and Cooling Curves

Demo: Cooling of p-dichlorobenzene

KE

PE

KE

80

Temp.

(°C )

liquid

50

freezing

solid

20

Time

KE

KE

PE

### Heating curve for p-dichlorobenzene

80

Temp.

(°C )

50

20

Time

What did we learn from this demo??

• During a phase change temperature remains constant and PE changes

• Changes in temperature during heating or cooling means the KE of particles is changing

p. 651

p. 652

p. 656

### Heating Curve for H20(s) to H2O(g)

A 40.0 g sample of ice at -40 °C is heated until it changes to steam and is heated to 140 °C.

• Sketch the heating curve for this change.

• Calculate the total energy required for this transition.

q = mc∆T

140

100

q = n∆H

Temp.

(°C )

q = mc∆T

q = n∆H

0

q = mc∆T

-40

Time

Data:

cice = 2.01 J/g.°C

cwater = 4.184 J/g.°C

csteam = 2.01 J/g.°C

ΔHfus = +6.02 kJ/mol

ΔHvap= +40.7 kJ/mol

warming ice:

q = mc∆T

= (40.0)(2.01)(0 - -40)

= 3216 J

warming water:

q = mc∆T

= (40.0)(4.184)(100 – 0)

= 16736 J

warming steam:

q = mc∆T

= (40.0)(2.01)(140 -100)

= 3216 J

melting ice:

q = n∆H

= (2.22 mol)(6.02 kJ/mol)

= 13.364 kJ

boiling water:

q = n∆H

= (2.22 mol)(40.7 kJ/mol)

= 90.354 kJ

n = 40.0 g

18.02 g/mol

= 2.22 mol

moles of water:

90.354 kJ

13.364 kJ

3216 J

3216 J

16736 J

127 kJ

### Practice

WorkSheet:

Thermochemistry #6

p. 655: #’s 30 – 34

pp. 656: #’s 1 - 9

p. 657 #’s 2, 9

p. 658 #’s 10, 16 – 20

30.(b) 3.73 x103 kJ

31.(b) 279 kJ

32.(b) -1.84 x10-3 kJ

33.(b) -19.7 kJ -48.77 kJ

34. -606 kJ

### Law of Conservation of Energy (p. 627)

The total energy of the universe is constant

∆Euniverse = 0

Universe = system + surroundings

∆Euniverse = ∆Esystem + ∆Esurroundings

∆Euniverse = ∆Esystem + ∆Esurroundings = 0

OR ∆Esystem = -∆Esurroundings

OR qsystem = -qsurroundings

First Law of Thermodynamics

### Calorimetry (p. 661)

calorimetry - the measurement of heat changes during chemical or physical processes

calorimeter - a device used to measure changes in energy

2 types of calorimeters

1. constant pressure or simple calorimeter (coffee-cup calorimeter)

2. constant volume or bomb calorimeter.

### SimpleCalorimeter

p.661

• a simple calorimeter consists of an insulated container, a thermometer, and a known amount of water

• simple calorimeters are used to measure heat changes associated with heating, cooling, phase changes, solution formation, and chemical reactions that occur in aqueous solution

• to calculate heat lost or gained by a chemical or physical change we apply the first law of thermodynamics:

qsystem = -qcalorimeter

Assumptions:

• the system is isolated

• c (specific heat capacity) for water is not affected by solutes

• heat exchange with calorimeter can be ignored

eg.

A simple calorimeter contains 150.0 g of water. A 5.20 g piece of aluminum alloy at 525 °C is dropped into the calorimeter causing the temperature of the calorimeter water to increase from 19.30°C to 22.68°C.

Calculate the specific heat capacity of the alloy.

eg.

The temperature in a simple calorimeter with a heat capacity of 1.05 kJ/°C changes from 25.0 °C to 23.94 °C when a very cold 12.8 g piece of copper was added to it.

Calculate the initial temperature of the copper. (c for Cu = 0.385 J/g.°C)

### Homework

• p. 664, 665 #’s 1b), 2b), 3 & 4

• p. 667, #’s 5 - 7

p. 665 # 4.b)

(60.4)(0.444)(T2 – 98.0) = -(125.2)(4.184)(T2 – 22.3)

26.818(T2 – 98.0) = -523.84(T2 – 22.3)

26.818T2 - 2628.2 = -523.84T2 + 11681

550.66T2 = 14309.2

T2 = 26.0 °C

Calorimeter

v = 100 ml

so m = 100 g

c = 4.184

T2 = 40.7

T1 = 20.4

6. System (Mg)

m = 0.50 g

= 0.02057 mol

Find ΔH

qMg = -qcal

nΔH = -mcΔT

7. System

ΔH = -53.4 kJ/mol

n = CV

= (0.0550L)(1.30 mol/L)

= 0.0715 mol

Calorimeter

v = 110 ml

so m = 110 g

c = 4.184

T1 = 21.4

Find T2

Bomb Calorimeter

### Bomb Calorimeter

• used to accurately measure enthalpy changes in combustion reactions

• the inner metal chamber or bomb contains the sample and pure oxygen

• an electric coil ignites the sample

• temperature changes in the water surrounding the inner “bomb” are used to calculate ΔH

• to accurately measure ΔH you need to know the heat capacity (kJ/°C) of the calorimeter.

• must account for all parts of the calorimeter that absorb heat

Ctotal = Cwater +Cthermom.+ Cstirrer + Ccontainer

NOTE: C is provided for all bomb

calorimetry calculations

eg.A technician burned 11.0 g of octane in a steel bomb calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.

What is the enthalpy of combustion for octane?

eg.

1.26 g of benzoic acid, C6H5COOH(s), is burned in a bomb calorimeter. The temperature of the calorimeter and contents increases from 23.62 °C to 27.14 °C. Calculate the heat capacity of the calorimeter. (∆Hcomb = -3225 kJ/mol)

Homework

p. 675 #’s 8 – 10

WorkSheet: Thermochemistry #7

### Hess’s Law of Heat Summation

• the enthalpy change (∆H) of a physical or chemical process depends only on the beginning conditions (reactants) and the end conditions (products)

• ∆H is independent of the pathway and/or the number of steps in the process

• ∆H is the sum of the enthalpy changes of all the steps in the process

### eg. production of carbon dioxide

Pathway #1: 2-step mechanism

C(s) + ½ O2(g) → CO(g)∆H = -110.5 kJ

CO(g) + ½ O2(g) → CO2(g)∆H = -283.0 kJ

C(s) + O2(g) → CO2(g) ∆H = -393.5 kJ

### eg. production of carbon dioxide

Pathway #2: formation from the elements

C(s) + O2(g) → CO2(g) ∆H = -393.5 kJ

### Using Hess’s Law

• We can manipulate equations with known ΔH to determine the enthalpy change for other reactions.

NOTE:

• Reversing an equation changes the sign of ΔH.

• If we multiply the coefficients we must also multiply the ΔH value.

eg.

Determine the ΔH value for:

H2O(g) + C(s) → CO(g) + H2(g)

using the equations below.

C(s) + ½ O2(g) → CO(g) ΔH = -110.5 kJ

H2(g) + ½ O2(g) → H2O(g) ΔH = -241.8 kJ

Switch

eg.

Determine the ΔH value for:

4 C(s) + 5 H2(g) → C4H10(g)

using the equations below.

ΔH (kJ)

C4H10(g) + 6½ O2(g) → 4 CO2(g) + 5 H2O(g) -110.5

H2(g) + ½ O2(g) → H2O(g)-241.8

C(s) + O2(g) → CO2(g)-393.5

Multiply by 5

Multiply by 4

4 CO2(g) + 5 H2O(g) →C4H10(g) + 6½ O2(g)+110.5

5(H2(g) + ½ O2(g) → H2O(g)-241.8)

4(C(s) + O2(g) → CO2(g)-393.5)

4 CO2(g) + 5 H2O(g) →C4H10(g) + 6½ O2(g)+110.5

5 H2(g) + 2½ O2(g) → 5 H2O(g)-1209.0

4C(s) + 4 O2(g) → 4 CO2(g)-1574.0

Ans: -2672.5 kJ

### Practice

WorkSheet:

Thermochemistry #8

pg. 681 #’s 11-14

### Review

∆Hof(p. 642, 684, & 848)

The standard molar enthalpy of formation is the energy released or absorbed when one mole of a substance is formed directly from the elements in their standard states.

∆Hof = 0 kJ/mol

for elements in the standard state

The more negative the ∆Hof, the more stable the compound

### Using Hess’s Law and ΔHf

Use the formation equations below to determine the ΔH value for:

C4H10(g) + 6½ O2(g) → 4 CO2(g) + 5 H2O(g)

ΔHf (kJ/mol)

4 C(s) + 5 H2(g) → C4H10(g)-2672.5

H2(g) + ½ O2(g) → H2O(g)-241.8

C(s) + O2(g) → CO2(g)-393.5

### Using Hess’s Law and ΔHf

ΔHrxn = ∑ΔHf(products) - ∑ΔHf(reactants)

eg. Use ΔHf , to calculate the enthalpy of reaction for the combustion of glucose.

C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g)

Use the molar enthalpy’s of formation to calculate ΔH for the reaction below

Fe2O3(s) + 3 CO(g)→ 3 CO2(g) + 2 Fe(s)

p. 688 #’s 21 & 22

Eg.

The combustion of phenol is represented by the equation below:

C6H5OH(s) + 7 O2(g) →6 CO2(g) + 3 H2O(g)

If ΔHcomb = -3059 kJ/mol, calculate the heat of formation for phenol.

### Bond Energy Calculations (p. 688)

• The energy required to break a bond is known as the bond energy.

• Each type of bond has a specific bond energy (BE).

(table p. 847)

• Bond Energies may be used to estimate the enthalpy of a reaction.

### Bond Energy Calculations (p. 688)

ΔHrxn = ∑BE(reactants) - ∑BE (products)

eg. Estimate the enthalpy of reaction for the combustion of ethane using BE.

2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g)

Hint: Drawing the structural formulas for all reactants and products will be useful here.

→ 4 O=C=O + 6 H-O-H

2

+ 7 O = O

C

C

[2(347) + 2(6)(338) + 7(498)]

- [4(2)(745) + 6(2)(460)]

8236 - 11480

= -3244 kJ

p. 690 #’s 23,24,& 26

p. 691 #’s 3, 4, 5, & 7

### Energy Comparisons

• Phase changes involve the least amount of energy with vaporization usually requiring more energy than melting.

• Chemical changes involve more energy than phase changes but much less than nuclear changes.

• Nuclear reactions produce the largest ΔH

• eg. nuclear power, reactions in the sun

### STSE

• What fuels you? (Handout)

aluminum alloywater

m = 5.20 gm = 150.0 g

T1 = 525 ºCT1 = 19.30 ºC

T2 = ºCT2 = 22.68 ºC

FIND c for Alc = 4.184 J/g.ºC

qsys = - qcal

mcT = - mc T

(5.20)(c)(22.68 - 525 ) = -(150.0)(4.184)(22.68 – 19.30)

-2612 c = -2121

c = 0.812 J/g.°C

BACK

calorimeter

C = 1.05 kJ/°C

T1 = 25.00 ºC

T2 = 23.94 ºC

copper

m = 12.8 g

T2 = ºC

c = 0.385 J/g.°C

FIND T1 for Cu

qsys = - qcal

mcT = - CT

(12.8)(0.385)(23.94 – T1) = -(1050)(23.94 – 25.0)

4.928 (23.94 – T1) = 1113

23.94 – T1= 1113/4.928

23.94 – T1= 225.9

T1= -202 ºC

BACK