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 CurvesCalorimetry (q = C∆T & above

formulas)

Unit Outline

Chemical ReactionsPE DiagramsThermochemical EquationsHess’s LawBond Energy

STSE: What Fuels You?

Temperature and Kinetic Energy

Thermochemistry is the study of energy changes in chemical and physical changes

eg. dissolving a solid

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

p. 628

K 50.15 450.15

°C 48 -200

# of

par

ticle

s

500 K

300 K

Kinetic Energy

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.

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

Part A: Thought Lab (p. 631)

Part B: Thought Lab

specific heat capacity - the amount of 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

Heat/Enthalpy Calculations

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 amount of heat applied

FORMULA

q = mc∆T

q = heat (J)

m = mass (g)

c = specific heat capacity

∆T = temperature change

= T2 – T1

= Tf – Ti

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

FORMULA

C = mc

q = C ∆T

C = heat capacity

c = specific heat capacity

m = mass

∆T = T2 – T1

Your Turn p.637 #’s 11-14

WorkSheet: Thermochemistry #3

Enthalpy Changes

The difference between the potential energy of the reactants and the products during a physical or chemical change is the Enthalpy change or ∆H.

AKA: Heat of Reaction

Reaction Progress

PE

Reactants

Products

∆H

Endothermic Reaction

Reaction Progress

PE

Reactants

Products

∆H ∆HEnthalpy

Endothermic Reaction

∆H is +Enthalpy

Reactants

Products

Endothermic

Enthalpy

products∆H is -

Exothermic

reactants

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.

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

Enthalpy Changes in Reactions

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

1. Chemical reactions

∆Hrxn – enthalpy of reaction

∆Hcomb – enthalpy of combustion

(see p. 643)

2. 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)

3. Phase Changes (p.647)

∆Hvap – enthalpy of vaporization

∆Hfus – enthalpy of melting

∆Hcond – enthalpy of condensation

∆Hfre – enthalpy of freezing

4. Solution Formation

∆Hsoln – enthalpy of solution

There are three ways to represent any 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

3.

H2O(l)

H2(g) + ½ O2(g)

∆Hf = -285.8 kJ/molEnthalpy (H)

enthalpy diagram

examples: pp. 641-643questions p. 643 #’s 15-18

WorkSheet: Thermochemistry #4

Calculating Enthalpy Changes

FORMULA:

q = n∆H

q = heat (kJ)

n = # of moles

∆H = molar enthalpy

(kJ/mol)

n m

M

eg. How much heat is released when 50.0 g of CH4 forms from C and H ?

p. 642 n 50.0 g

16.05 g / m ol

3 .115 m ol

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?

n 50.0 g

16.05 g / m ol

3 .115 m ol

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/mol ΔHvap = +40.7 kJ/mol

n 20.0 g

18.02 g / m ol

1 .110 m ol

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?

n 40.0 g

80.06 g / m ol

0 .4996 m ol

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

H

q n

q = nΔH

kJ 1250.9

kJ 405- n

n = 0.3238 mol

m = n x M = (0.3238 mol)(30.08 g/mol) = 9.74 g

Complete: 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 Time (s) Temperature

(°C)Time (s) Temperature (°C)

Cooling curve for p-dichlorobenzene

Temp. (°C )

50

80

KE

PE

KE

Time

solidfreezing

liquid

20

Heating curve for p-dichlorobenzene

Temp. (°C )

50

20

80

KE

KE

PE

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.

1. Sketch the heating curve for this change.

2. Calculate the total energy required for this transition.

Time

Temp. (°C )

-40

0

100

140

q = mc∆T

q = n∆H

q = mc∆T

q = mc∆T

q = n∆H

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:

Total Energy

90.354 kJ

13.364 kJ

3216 J

3216 J

16736 J

127 kJ

Practicep. 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

WorkSheet: Thermochemistry #6

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

(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

6. System (Mg) m = 0.50 g or 0.02057 mol Find ΔH

Calorimeter 100 ml or m = 100 g c = 4.184 T2 = 40.7 T1 = 20.4

7. System ΔH = -53.4 kJ/mol n = CV = (0.0550L)(1.30 mol/L) = 0.0715 mol

Calorimeter 110 ml or m = 110 g c = 4.184 T1 = 21.4 Find T2

6.

qMg = -qcal

nΔH = -mcΔT

n = 0.50 g

24.31 g / m ol 0 .02057 m ol

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)

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

Homework

p. 675 #’s 8 – 10

WorkSheet: Thermochemistry #6

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

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

Switch

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)

Ans: -2672.5 kJ

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

Practice

pg. 681 #’s 11-14

WorkSheet: Thermochemistry #7

Review∆Ho

f (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 stateThe more negative the ∆Ho

f, the more stable the compound

Determine the ΔH value for:

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

using the formation equations below.

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

Using Hess’s Law and ΔHf

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

eg. Calculate the enthalpy of reaction for the complete 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.

+ 7 O = OCC

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

→ 4 O=C=O + 6 H-O-H

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

p. 690 #’s 23,24,& 26p. 691 #’s 3, 4, 5, & 7

= -3244 kJ

2

8236 - 11480

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 ΔHeg. nuclear power, reactions in the sun

STSE

What fuels you? (Handout)

#’s 1 – 4, 7, 9, 10, 13, 14, & 16

(List only TWO advantages for #2)

aluminum alloy water

m = 5.20 g m = 150.0 g

T1 = 525 ºC T1 = 19.30 ºC

T2 = ºC T2 = 22.68 ºC

FIND c for Al c = 4.184 J/g.ºC

qsys = - qcal

mcT = - 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

copper

m = 12.8 g

T2 = ºC

c = 0.385 J/g.°C

FIND T1 for Cu

qsys = - qcal

mcT = - CT(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

calorimeter

C = 1.05 kJ/°C

T1 = 25.00 ºC

T2 = 23.94 ºC

q heat J or kJ

c Specific heat capacity

J/g.ºC

C Heat capacity kJ/ ºC or J/ ºC

ΔH Molar heat or molar enthalpy

kJ/mol

Group 2 Total Mass = 2.05 g

Group 5 Total Mass = 1.86 g