Unit Outline

Temperature and Kinetic Energy

Heat/Enthalpy Calculation

Temperature changes (q = mcΔT)

Phase changes (q = nΔH)

Heating and Cooling Curves

Calorimetry (q = CΔT & above formulas)

Chemical Reactions PE Diagrams

Thermochemical Equations

Hess’s Law

Bond Energy

STSE: What Fuels You?

Temperature and Kinetic Energy

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

eg. dissolving

burning

phase changes

Temperature, T, measures the average kinetic energy of particles.

A change in temperature means particles are moving at different speeds

Temperature is 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.

K 321.15 50.15 73.15 450.15

°C 48 -223 -200 177

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

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

the specific heat capacity, c, of a substance reflects how well a substance can store energy

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 heat loss or gain.

FORMULA

q = mcΔT q = heat (J)

m = mass (g)

c = specific heat capacity

ΔT = 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? (cH2O = 4.184 J/g.oC) 17

What is the temperature change for a 15 g piece of iron that absorbs 26.5 J of heat?

(cFe = 0.444 J/g.oC)

A) 0.25 oC

B) 0.78 oC

C) 1.3 oC

D) 4.0 oC

If 23.9 g of an unknown metal requires 343 J of energy to change its temperature from

24.5 oC to 85.0 oC, what is the specific heat capacity of the metal?

A) 0.237 J/g.oC

B) 0.568 J/g.oC

C) 4.22 J/g.oC

D) 868 J/g.oC

Make sure you are comfortable in taking

q = m c ΔT and solving for c, m, ΔT, T2 & T1

p. 634 #’s 1 – 3 p. 632-->c values

p. 636 #’s 5 – 8 p. 659--> answers

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

Your Turn p.637 #’s 11-14

31. Which is the amount of energy required to raise the temperature of 1.0 g of a substance by 1.0 oC?

(A) heat capacity

(B) molar enthalpy

(C) molar heat

(D) specific heat capacity (June ’09)

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 ΔH

Enthalpy of Reaction

● Endothermic Reaction

PEEnthalpy

Reactants

ΔH

Products

ΔH

Reaction Progress

Enthalpy Diagram

Reactants

Enthalpy

Products

ΔH is +

Endothermic

Enthalpy Diagram

Enthalpy

Reactants

Products

ΔH is - 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)

Enthalpy Changes in Reactions

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

ie. energy is absorbed

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

ie. energy is produced

Enthalpy Changes in Reactions

ΔH can represent the enthalpy change for a number of processes

1.Chemical reactions

ΔHrxn – enthalpy of reaction

ΔHcomb – enthalpy of combustion

(see p. 643)

2.Formation of compounds from elements

ΔHo

f – 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) ΔHo

f = -110.5 kJ/mol

Use the equation below to determine the ΔHo

f for

CH3OH(l)

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

1 C(s) + 2 H2(g) + ½ O2(g) → 1 CH3OH(l) + 238.6 kJ

ΔHo

f= -238.6 kJ/mol

Use the equation below to determine the ΔHo

f for

CaCO3(s)

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

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

1 Ca(s) + 1 C(s) + 1.5 O2(g) → 1 CaCO3(s) + 1206.9 kJ

ΔHo

f = -1206.9 kJ/mol

Use the equation below to determine the ΔHo

f for

PH3(g)

4 PH3(g) → P4(s) + 6 H2(g) + 21.6 kJ

a) +21.6 kJ/mol

b) -21.6 kJ/mol

c) +5.4 kJ/mol

d) -5.4 kJ/mol

3.Phase Changes (p.647)

ΔHvap – enthalpy of vaporization (l → g)

ΔHfus – enthalpy of melting (fusion: s → l)

ΔHcond – enthalpy of condensation (g → l)

ΔHfre – enthalpy of freezing (l → s)

eg.

H2O(l) H2O(g) ΔHvap = +40.7 kJ/mol

Hg(l) Hg(s) ΔHfre = -23.4 kJ/mol

4.Solution Formation (p.647, 648)

ΔHsoln – 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. A thermochemical equation has the energy term written into the equation.

2. The enthalpy term is written separate beside the equation.

3. An 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 (Thermochemical equation)

2. H2(g) + ½ O2(g) → H2O(l) ΔHo

f = -285.8 kJ/mol

3. H2(g) + ½ O2(g)

ΔHo

f = -285.8 kJ/mol

H2O(l)

Enthalpy (H)

examples: pp. 641-643 questions p. 643 #’s 15-18 WorkSheet: Thermochemistry #4

Enthalpy Diagram

Calculating Enthalpy Changes

FORMULA:

q = heat (kJ)

q = nΔH n = # of moles = mass (n = m ) Molar mass M

ΔH = molar enthalpy (kJ/mol)

Compound ΔHo

f (kJ/mol)

CO(g) -110.5CO

2(g) -393.5

CH4(g) -74.6

eg. How much heat is released when 50.0 g of CH4 forms from C and H ?(Pg. 642)

eg. How much heat is released when 50.00 g of CH4

undergoes complete combustion?

(Pg.643)

eg. How much energy is needed to change 20.0 g of H2O(l)

at 100 °C to steam at 100 °C ?

ΔHfreez

and ΔHcond

have the opposite sign of the

above values. ΔH

vap = -ΔH

cond & ΔH

melt = -ΔH

freez

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 (NH

4NO

3)

dissolves?

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

FOR YOU TO COMPLETE

p. 645; #’s 19 – 23

pp. 648 – 649; #’s 24 – 29

Worksheet: Thermochemistry #5

ANSWERS: Pg. 659

Heating and Cooling Curves

Cooling of p-dichlorobenzene (aka moth balls)

https://www.youtube.com/watch?v=JKCJG-Jco_8

Cooling curve for p-dichlorobenzene

● Check your notes for the graph!

Heating curve for p-dichlorobenzene

● Check your notes for the graph!

What we learned from those graphs

During a phase change temperature remains constant and PE changes● PE increases in a heating curve and decreases

in a cooling curve

Changes in temperature during heating or cooling means the KE of particles is changing● KE increases during a heating curve and

decreases during a cooling curve

June 2009 # 38 http://www.ed.gov.nl.ca/edu/k12/evaluation/chem3202/chem3202_jun09.pdf

Pg. 651

Pg. 652

q = mcΔT

q = nΔH

Pg. 656

q = mcΔT

q = nΔH

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.

*Check your notes for the heating curve

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: (from -40 ºC to 0 ºC)

q = mcΔT

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

= 3216 J

warming water: (from 0 ºC to 100 ºC)

q = mcΔT

= (40.0)(4.184)(100 – 0)

= 16736 J

warming steam: (from 100 ºC to 140 ºC)

q = mcΔT

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

= 3216 J

moles of water: n = 40.0g

18.02g/mol

= 2.22 mol

melting ice: (fusion)

q = nΔH

= (2.22 mol)(6.02 kJ/mol)

= 13.364 kJ

boiling water: (vaporization)

q = nΔH

= (2.22 mol)(40.7 kJ/mol)

= 90.354 kJ

Total Energy

90.354 kJ

13.364 kJ

3216 J

16736 J

3216 J

90.354 kJ

13.364 kJ

3.216 J

16.736 J

3.216 J

126.9kJ

PRACTICE

p. 655: #’s 30 – 34

p. 659 – Answers

WorkSheet: Thermochemistry #6

Law of Conservation of Energy (p. 627) The total energy of the universe is constant

ΔEuniverse = 0

Universe = system + surroundings

ΔEuniverse = ΔEsystem + ΔEsurroundings

ΔEsystem + ΔEsurroundings = 0 → 1st Law of Thermodynamics

OR ΔEsystem = -ΔEsurroundings

OR qsystem = -qsurroundings

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.

Simple Calorimeter (P. 661)

Bomb Calorimeter

a simple calorimeter is 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

Example

A simple calorimeter contains 150.0 g of water. A 5.20g 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.

Solutionaluminum alloy water

m = 5.20g m = 150.0 g

T1 = 525 ºC T1 = 19.30 ºC

T2 =22.68 ºC T2 = 22.68 ºC

FIND c for Al c = 4.184 J/g.ºC

qsys

= - qcal

mcΔT = - mcΔT

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

-2612 c = -2121

c = 0.812 J/g.°C

Example

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)

Solutioncopper Calorimeter

m = 12.8 g C = 1.05 kJ/°C

T2 =23.94 ºC T1 = 25.00 ºC

c = 0.385 J/g.°C T2 = 23.94 ºC

FIND T1 for Cu

qCu

= - qcal

mcΔT = - CΔT

(12.8)(0.38s5)(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

More Book Questions!!

P. 664-665 #'s: 1,4

Thermal Equilibrium

- all components in a calorimeter have the

same temperature

In the last two examples, notice that the final temperature of the water/calorimeter is the same as the final temperature of the sample

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?

(-5700 kJ/mol)

System (octane) Calorimeter

m = 11.0g C = 28.0 kJ/ºC

Find ΔHcomb T2 = 39.6 ºC

n = 11.0g T1 = 20.0 ºC

114.26g/mol

=0.09627 mol

qsys = - qcal

n ΔH = -CΔT

(0.09627) ΔH = - (28.0)(39.6 – 20.0)

ΔH = -5.70x103 kJ/mol

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. (ΔHcomb = -3225 kJ/mol)

n = 1.26 g

122.13 g/mol

= 0.01032 mol

qsys = - qcal

n ΔH = -CΔT

(0.01032)(-3225) = - (C)(27.14 – 23.62)

C = 9.45 kJ/ ºC

Your Turn!

Pg. 675 #10

WorkSheet: Thermochemistry #7

Hess’ Law of Heat Summation

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

● ΔH is independent of the pathway and/or the number of steps in the process

● ΔH 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) ΔH = -110.5 kJ

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

____________________________________________

C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ

eg. production of carbon dioxide

Pathway #2: formation from the elements

C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ

Using Hess’ Law

● We can manipulate equations with known ΔH to determine an unknown enthalpy change.

NOTE: ● Reversing an equation changes the sign

of ΔH. ● If we multiply the coefficients we must also

multiply the ΔH value.

Ex: reverse? multiply?

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

Determine the ΔH value for:

4 C(s) + 5 H2(g) → C4H10(g)

using the equations below.

ΔH (kJ)

C4H10(g) + 13/2 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) + 13/2 O2(g) +110.5kJ

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

4(C(s) + O2(g) → CO2(g) -393.5 -393.5kJ)

4 CO2(g) + 5 H2O(g) → C4H10(g) + 13/2 O2(g) +110.5kJ

5 H2(g) + 5/2 O2(g) → 5 H2O(g) -1209.0 kJ

4C(s) + 4 O2(g) → 4 CO2(g) -1574.0kJ

Ans: -2672.5 kJ

Practice for you!!

pg. 681 #’s 12 &14

WorkSheet:

Thermochemistry #8

Recall

ΔHo

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.

ΔHo

f = 0 kJ/mol

for elements in the standard state

The more negative the ΔHo

f, the more

stable the compound

June ‘08 #39

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 = ΣΔHf (products) - ΣΔHf (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)

ΔHrxn = ΣΔHf (products) - ΣΔHf (reactants)

ΔHf

CO2(g) -393.5 kJ/mol

H2O(g) -241.8 kJ/mol

C6H12O6(s) -1274.5 kJ/mol

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

ΔHrxn = [6(-393.5) + 6(-241.8)] – [1(-1274.5)+ 6(0)]

= [-2361 + -1450.8] - [-1274.5 + 0]

= - 2537 kJ/mol

Use the molar enthalpy of formation to calculate ΔH for this reaction:

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

ΔHrxn = [3(-393.5) + 2(0) ] – [3(-110.5)+ 1(-824.2)]

= [-1180.5 + 0] - [-331.5 + -824.20]

= - 24.8 kJ

−824.2 kJ/mol -110.5kJ/mol −393.5 kJ/mol

0 kJ/mol

eg.

The combustion of phenol is given below:

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

If ΔHcomb = -3059 kJ/mol, calculate the heat of

formation for phenol. ΔHo

f

H2O(g) -241.8 kJ/mol

CO2(g) -393.5 kJ/mol

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 = ΣBE(reactants) - ΣBE (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.

C-C = 347kJ/mol; C-H = 338kJ/mol; O=O = 498kJ/mol; C=O = 745kJ/mol

H-O = 460kJ/mol

[2(347) + 2(6)(338) + 7(498)] - [4(2)(745) + 6(2)(460)]

8236 – 11480

= -3244 kJ

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

#36 June ‘06

#37 Aug. ‘08