A. Energy Changes

energy can be however the

energy is the capacity to and/or

Thermochemistry

with every energy conversion, energy is always

do work,

converted to other forms total energy

lost as heat

generate electricitygenerate heat

of the system is conserved

(1st Law of Thermodynamics)

(2nd Law of Thermodynamics)

the primary sources of energy are:

1. chemical: fossil fuels, plants

2. nuclear: uranium, hydrogen

3. solar: radiant energy, wind, hydroelectric

4. geothermal: geysers, hotsprings

there are 4 types of energy changes:

1. temperature change:

2. phase change:

3. chemical change:

4. nuclear change:

10’s kJ

10’s kJ

100-1000’s kJ

millions kJ

TYPICAL DIPLOMA QUESTION

temperature or kinetic energy (EK) is the energy of

heat is the transfer of

B. Temperature Changes

ΔEK is the change in

motion of particles…

thermal energy

temperature)

increase in temperature means an

in EKincrease

kinetic energy

(results in a change in

EK can be classified as:

1. vibrational motion:

2. rotational motion:

3. translational motion:

rapid back and forth movement of bundled atoms with no change of location -- solid, liquid, gas

molecular rotation, no change in position --liquid, gas

motion from one point to another --liquid, gas

heat capacity is the heat required to change the temperature of a "unit mass" of a substance by 1C

ΔEK = q = mcΔt

where: ΔEK =

q =m =Δt =c =

change in kinetic energy in Jheat energy in Jmass in g

change in temperature in Cspecific heat capacity in J/gC

ExampleFind the heat required to change 2.50 g of water from 10.0C to 27.0C .

q = mcΔt = (2.50 g)(4.19 J/gC) (27.0C - 10.0 C) = 178.075 J = 178 J

Your Assignment: pg 1

phase change is a change of

phase changes always involve but do not involve

C. Phase Changes1. The Basics

energy from the surroundings

state

energy changes a change in temperature

are associated with and not

separates the bonded molecules (intermolecular forces)

potential energykinetic energy

thereby increasing their potential energy, or Ep

Types of Phase Changes

liquid

gassolid

melting

freezing

evaporation

condensation

sublimation

deposition

exothermic =

melting, evaporation, sublimation

freezing, condensation, deposition

endothermic =

2. Enthalpy and Molar Enthalpy

enthalpy is

unfortunately, the enthalpy of individual substances cannot be measured directly (EK can with a thermometer

but how do you measure EP?)

the total kinetic and potential energy

changes in enthalpy occur whenever heat is

change in enthalpy is measured in

released or absorbed in a physical or chemical change…

J or kJ (H)

of a chemical system under constant pressure and temperature

fortunately, this can be measured

during a phase change

endothermic enthalpy changes are

exothermic enthalpy changes are

positive values

negative values

EK remains constant

molar enthalpy can be used to calculate the enthalpy change of a phase change:

ΔH = nH

where: ΔH = n =H =

enthalpy change in J or kJ number of moles in mol molar enthalpy in J/mol or kJ/mol

molar enthalpy is the enthalpy change

molar enthalpy is measured in

per mole of a substance

J/mol or kJ/mol (H)

Example

Find the energy required to melt 2.50 g of ice.

H = nH

= m H M

= 2.50 g (6.01 kJ/mol) 18.02 g/mol

= 0.8337957825 kJ = +0.834 kJ

Your Assignment: pg 4 & 5 in workbook

D. Total Energy Calculations

a heating curve is a showing the and changes as heat is added to a

substance over time

the total energy change that a substance goes through can be determined using a and the formulas and

during temperature changes the of the molecules change so you calculate heat using

during phase changes there only a change in so you calculate heat using

graph phasetemperature

heating curveq=mcΔt ΔH=nH

EKq = mct

EPH = nH

Heating Curve

H2O(l) H2O(g)

H2O(s) H2O(l)

H2O(l)

H2O(s)

H2O(g)

100C

0C

Temperature(C)

Time (min)

BP

MP

For Water

Cooling Curve

H2O(l) H2O(g)

H2O(s) H2O(l)

H2O(l)

H2O(s)

H2O(g)

100C

0C

Temperature(C)

Time (min)

BP

MP

For Water

Steps:

1. Always

2. On the curve, put a point where and a point

3. Determine which formulas are needed and which part of the curve they apply to.

you begin

draw the heating curve first!!!!

where you end (temperatures).

each new line segment, you have a new formula diagonal lines represent a change in

temperature q=mcΔt

horizontal lines represent a phase change ΔH=nH (vap or fus)

4. Perform the calculation.

Example

Find the total energy required to change 1.0 g of ice at -20C to steam at 110C.

Heating Curve For Water

Temperature(C)

Time (min)

0C

100C

-20C

110C

= mcΔt + nHfus + mcΔt + nHvap + mcΔt

= (1.0g)(2.00J/gC)(20C) + (1.0g/18.02g/mol) (6010J/mol) + (1.0g)(4.19J/gC)(100C) + (1.0g/18.02g/mol) (40650J/mol) + (1.0g)(2.02J/gC)(10C)

ΔEtotal = + + + +

= 3068.545172 J = 3.1 103 J

Total Energy Calculation Practice Question

Find the total energy required to changed 4.4 g of methanol from a solid at -102 ˚C to a gas at 89.0 ˚C. Fusion of methanol occurs at -98.0 ˚C while vaporization occurs at 65.0 ˚C.

E. Calorimetrycalorimetry is a technological process of

the isolated system used to determine the heat involved in a phase change or in a chemical reaction is called a

Bomb Calorimeter

ENERGY

insulation water

enclosed system(bomb)

measuring energy changes using an isolated system

calorimeter

here’s how it works:

reacting substances are placed in the bomb is placed in the

bomb

calorimeter

is recorded

reaction is initiated

is

recorded

initial temp of water

final (maximum) temp of water

it is assumed that no energy is by the system except for the energy required or released by the

calculations are based on the Principle of Heat Transfer:

HEAT LOST = HEAT GAINED

gained or lost

reaction or phase change

Example 1

A chemical reaction in a bomb calorimeter causes the temperature of 500 g of water to increase in temperature from 10.0C to 52.0C. Calculate the heat released by this reaction. Give your answer in kJ.

HL (rxn) = HG (water) q = mct q = (500 g)(4.19 J/gC)(52.0C – 10.0C ) q = 87 990 J

= 87.990 kJ = 88.0 kJ

Example 2

150 g of unknown metal X is at 100C. It is placed in a calorimeter with 200 mL of water at 23.0C. If the equilibrium temperature reached is 25.0C, what is the specific heat capacity of metal X?

HL (metal) = HG (water) mct = mct

(150g)c(100C – 25.0C) = (200g)(4.19 J/gC)(25.0C – 23.0C) 11250 c = 1676

c = 0.148977777 J/gC c = 0.149 J/gC

Your Assignment: pg 6 in workbook

Example 3

When 80.0 g of NaOH is added to 850 mL of water at 23.0C, the temperature of the water rose to 28.5C after the NaOH had dissolved. Calculate the molar enthalpy of dissolving.

HL (dissolving) = HG (water) (m/M)H = mct

(80.0g/40.00g/mol)H = (850g)(4.19J/gC)(28.5C – 23.0C)

(2 mol)H = 19588.25 J H = 9794.125 J/mol H = 9.79 x 103 J/mol

exothermic

–

or – 9.79 kJ/mol

Example 4

When 52.5 g of LiNO3 is added to 150 mL of water in a

calorimeter the initial temperature of the water was 18.0C and after the LiNO3 the temperature was 16.5C.

Calculate the molar enthalpy of dissolving.

HL (water) = HG (dissolving) mct = (m/M)H

(150g)(4.19J/gC)(18.0C – 16.5C) = (52.5 g/68.95 g/mol)H 942.75 J = (0.761… mol)H

1238.145 J/mol = H 1.24 x 103 J/mol = H

or +1.24 kJ/mol

+endothermic

Your Assignment: Molar Enthalpy Worksheet

Example 5

15 g of ice at 5.0C is placed in a beaker of water at 30C. Calculate the mass of the water in the beaker if the final temperature at equilibrium is 10C.

30C

5.0C

temperature

time

10C

HL (water) = HG (ice)mct = mct + (m/M)Hfus + mct

m(4.19 J/gC)(30C – 10C ) = (15 g)(2.00 J/gC)(0C – (–5.0C)) +(15 g/18.02g/mol)(6010 J/mol) + (15 g)(4.19J/gC)(10C – 0C)

(83.8 J/g)m = 150 J + 5002.77 J… + 628.5 J

m = 68.966……g = 69 g

Example 6

If 10 g of ice at 15C is placed in a calorimeter with 200 mL of water at 25C and stirred so that an equilibrium is reached, what is the final temperature of the mixture?

25C

15C

temperature

time

tf

HL (water) = HG (ice)mct = mct + (m/M)Hfus + mct

(200 g)(4.19 J/gC)(25C – tf)= (10 g)(2.00 J/gC)(0C – (–15.0C))

+(10 g/18.02g/mol)(6010 J/mol) + (10 g)(4.19J/gC)(tf – 0C)

20 950 J – (838 J/C) tf = 300 J + 3335.18313 J + (41.9 J/C) tf

17314.81687 J = (879.9 J/C) tf

19.678… C = tf

20C = tf

 

Assignment: Enthalpy and Phase Change

G. Chemical Changea is a transformation involving

an energy change in which one substance is converted into another substance

chemical change

uses of chemical energy (exothermic):

1.

2.

steam generators from burning fossil fuels

motor vehicles where fuel is burned

4.

3. natural gas, propane, coal, wood burned for heating

batteries

5. living organisms, cellular respiration

a calorimeter can be used to quantify the amount of heat lost or gained by a chemical reaction (still sticking to the heat lost = heat gained principle!!!!)

Example 1

A 2.65 g sample of methanol (CH3OH) was burned in a

calorimeter which contained 500 mL of water at 25.0C. If the final temperature of the water is 50.0C, what is the molar heat of combustion for methanol?

heat lost (combustion)= heat gained (water) (m/M)H = mct

(2.65g/32.05g/mol)H = (500g)(4.19J/gC)(50.0C – 25.0C)

(0.0826… mol )H = 52375 J H = 633441.038 J/mol H = – 6.33 x 105 J/mol

or –633 kJ/mol

Example 2

An 8.40 g sample of N2(g) is reacted with pure oxygen in a

bomb calorimeter containing 1.00 kg of water to produce N2O. The temperature of the water dropped by 5.82C.

What is the molar heat of reaction of N2(g)?

heat lost (water) = heat gained (formation) mct = (m/M)Hf

(1000 g)(4.19 J/gC)(5.82C) = (8.40 g/28.02 g/mol)H 24385.8 J = (0.299… mol )H

H = 81344.06143 J/mol H = + 8.13 x 104 J/mol

or +81.3 kJ/mol

Your Assignment: pg 7 in workbook

H. Industrial Bomb Calorimeters industrial calorimeters are used in to measure

modern calorimeters have

eg) volume of water used, container (bomb) material, stirrer and thermometer

the heat of combustion of food, fuel, oil, crops, and explosives

research

fixed components

Etotal =

in calculating the energy of combustion, you take all components of the calorimeter into account:

mct (H2O) + mct (stirrer) + mct (bomb) +

mct (thermometer)

all of the “mc” parts are constant so they are replaced by one constant C, the heat capacity of the entire system in kJ/C

Example 1

A 1.50 g sample of methane is completely burned in a calorimeter with a heat capacity of 11.3 kJ/C. The temperature increased from 20.15C to 27.45C. Calculate the molar enthalpy of combustion for methane.

heat lost (combustion) = heat gained (calorimeter) (m/M)H = Ct

(1.50 g/16.05 g/mol)H = (11.3 kJ/C)(27.45C – 20.15C)

(0.0934 … mol )H= 82.49 kJ H = 882.6430002 kJ/mol H = – 883 kJ/mol

Example 2

When 3.00 g of butter is burned in a bomb calorimeter with a heat capacity of 9.22 kJ/C the temperature changes from 19.62C to 31.89C. Calculate the specific enthalpy of combustion in kJ/g.

***note that in this question we are asked for enthalpy of combustion in kJ/g not kJ/mol. We substitute mass in for moles in the formula Hcomb = nHcomb

heat lost (combustion) = heat gained (calorimeter) *** mH = Ct (3.00g)H = (9.22 kJ/C)(31.89C – 19.62C)

(3.00 g)H= 113.1294 kJ H = 37.7098 kJ/g H = – 37.7 kJ/g

Your Assignment: pg 8 in workbook

I. Reaction Enthalpies the enthalpy change of a refers to changes in

and is called the

the heat of reaction, , can be expressed in 4 ways (vocabulary):

1. Outside Equation

the heat of reaction can be given as a H value outside the equation

2 SO2(g) + O2(g) 2 SO3(g) H = –197.8 kJ

reactionEP heat of reaction

H

ExampleCalculate the molar enthalpy of reaction (H) for sulphur dioxide using the following information:

2 SO2(g) + O2(g) 2 SO3(g) H = –197.8 kJ

H = –197.8 kJn = 2 mol

H = nH H = H n = –197.8 kJ

2 mol = –98.9 kJ/mol

2. Inside Equation

the heat of reaction can be written in the equation

endothermic reaction…

eg)

exothermic reaction…

eg)

heat is on reactant side

heat is on product side

H2O(l) + 285.8 kJ H2(g) + ½ O2(g)

Mg(s) + ½ O2(g) MgO(s) + 601.6 kJ

ExampleCalculate the molar enthalpy (H) for oxygen in the decomposition of water using the following information:

H2O(l) + 285.8 kJ H2(g) + ½ O2(g)

H = +285.8 kJn = ½ mol

H = nH H = H n = +285.8 kJ

½ mol = +571.6 kJ/mol

3. Molar Enthalpy, H

use the formula H = (m/M)H to calculate H

ExampleFind the molar enthalpy when 5.0 g of butane produces 850 kJ of energy.

H = –850 kJm = 5.0 gM = 58.14 g/mol

H = (m/M)H H = H (m/M) = –850 kJ (5.0 g/58.14 g/mol)

= –9883.8 kJ/mol = –9.9 x 103 kJ/mol

4. Potential Energy Diagram

reactants are separated from the products shape indicates whether the reaction is endothermic or

exothermic

Endothermic Exothermic

EP

(kJ)

EP

(kJ)

Reaction Progress Reaction Progress

reactants

products

H > 0positive

reactants

products

H < 0negative

Your Assignment: 1. pg 2 in workbook 2. pg 3 in workbook

J. Bond Energy

bond energy is the energy or the energy

in both endothermic and exothermic reactions, the energy required to the atoms in the

is called the

the activation energy is always than the energy contained in the reactants and the products

required to break a chemical bond

“pull apart” activation energy

released when a bond is formed

reactants

higher

Endothermic 2 H2O + energy 2 H2 + O2

Energy(kJ)

Reaction Progress

2 H2O

H H H H O O

2 H2 + O2

net energy for reaction

activation energy needed to break bonds

energy released when bonds form

Exothermic H2 + Cl2 2 HCl + energy

Energy(kJ)

Reaction Progress

H2 + Cl2

H H Cl Cl

2 HCl

net energy for reaction

activation energy needed to break bonds energy released when

bonds form

if a is used, it acts to for the reaction (***typical diploma question)

catalyst lower the activation energy

Endothermic 2 H2O + energy 2 H2 + O2

Energy(kJ)

Reaction Progress

2 H2O

H H H H O O

2 H2 + O2

net energy for reaction

activation energy needed to break bonds

energy released when bonds form

= catalyzed reaction

Exothermic H2 + Cl2 2 HCl + energy

Energy(kJ)

Reaction Progress

H2 + Cl2

H H Cl Cl

2 HCl

net energy for reaction

activation energy needed to break bonds energy released when

bonds form

= catalyzed reaction

K. Predicting Enthalpy (Hr) Changes

1. Using Hess’s Law

because of the law of conservation of energy, the heat of reaction is the whether the reactants are converted to the products in a or in a

G.H. Hess (1840) suggested that if two or more

are to give a final equation then the can be added to give the

samesingle reaction

series of reactions

thermochemical equationsadded

enthalpies

enthalpy for the final equation

sometimes the heat of reaction for a chemical change is not easily measured due to time of reaction, cost, rarity of reactants etc. so we use Hess’s Law to calculate Hr

Steps:1. Write the , if it is not given.

2. the given equations so they will to yield the if you multiply or divide an equation, multiply or

divide the H by theif you flip an equation, the sign on H

3. (you should end up with your net equation!)

the reactants and products where possible to

4. the component enthalpy changes to get the

net reaction

Manipulate addnet equation.

same factor

flip

Cancelsimplify

Addnet enthalpy change.

Example 1Find the heat of reaction for C(s, di) C(s, gr) using the

following reactions:C(s, gr) + O2(g) CO2(g) H = –393.5 kJ

C(s, di) + O2(g) CO2(g) H = –395.4 kJ

C(s, di) + O2(g) CO2(g)

CO2(g) C(s, gr) + O2(g) H =+393.5 kJ

flip

C(s, di) C(s, gr) H = –1.9 kJ

H = –395.4 kJ

Example 2Find the heat of reaction for H2O2(l) H2O(l) + ½ O2(g)

using the following reactions: H2(g) + O2(g) H2O2(l) H = –187.8 kJ

H2(g) + ½ O2(g) H2O(l) H = –285.8 kJ

H2(g) + ½ O2(g) H2O(l)

H2O2(l) H2(g) + O2(g) H =+187.8 kJ

flip

H2O2(l) H2O(l) H = –98.0 kJ

H = –285.8 kJ

½

+ ½ O2(g)

Example 3Find the heat of reaction for C(s) + 2 H2(g) CH4(g) using

the following reactions: C(s) + O2(g) CO2(g) H = –393.5 kJ

H2(g) + ½ O2(g) H2O(l) H = –285.8 kJ

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) H = –890.5 kJ

C(s) + O2(g) CO2(g)

CO2(g) + 2 H2O(l) CH4(s) + 2 O2(g)

H =–571.6 kJ

flip

2 H2(g) CH4(g) H = –74.6 kJ

H = –393.5 kJ

H =+890.5 kJ

2

2 H2(g) + 1 O2(g) 2 H2O(l)

C(s) +

Your Assignment: pgs 9-10 in workbook pg 11 in workbook

2. Using Standard Heats of Formation Hf sometimes it is not easy to measure the heat change for

a reaction (too slow/expensive)

in this case, H can be determined using

heats of formation (Hf ) are the changes in EP that

occur when

heats of formation

compounds are formed from their elements

Hf for elements cannot be directly measured therefore

they are designated as …all other Hf values are

in reference to this

Hf for common compounds are listed on pages 6-7

in data booklet

the Hf is an indirect measure of the stability

zero

of a compound

the more , the more

eg) List the following compounds in order from most stable to least stable.

exothermic the formation more stable the compound

(this means you have to add that energy to decompose it)

H2O(l) Hf =

C2H4(g) Hf =N2O4(g) Hf =PCl3(l) Hf =Al2O3(s) Hf =

–285.8 kJ/mol+52.4 kJ/mol+11.1 kJ/mol–319.7 kJ /mol–1675.7 kJ /mol1

4

3

2

5

Hess’s Law formula states that the is the difference between the standard heats of formation of the and the

Hr = nHf(products) nHf(reactants)

Hr

reactants products

Example 1Calculate the standard heat of combustion for 2 CO(g) + O2(g) 2 CO2(g) and draw the EP diagram for this

reaction.

2 CO(g) + O2(g) 2 CO2(g)

(2 mol)(-110.5 kJ/mol) + 0 kJ (2 mol)(-393.5 kJ/mol)

-221.0 kJ + 0 kJ -787.0 kJ

Hc = nHf(products) nHf(reactants)

= -787.0 kJ – (-221.0 kJ)= -566.0 kJ

EP

(kJ)

EP Diagram for 2 CO(g) + O2(g) 2 CO2(g)

2 CO(g) + O2(g)

2 CO2(g)

-221.0

-787.0

H = -566.0 kJ

Reaction Progress

Example 2Find the heat of combustion of ethane. The products of combustion are gases.

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

(2mol)(-84.0kJ/mol) + 0 kJ (4 mol)(-393.5 kJ/mol)

-168.0 kJ + 0 kJ -1574.0 kJ

Hc = nHf(products) nHf(reactants)

= (-1574.0 kJ + (-1450.8 kJ)) – (-168.0 kJ)= -2856.8 kJ

+ (6 mol)(-241.8 kJ/mol)

+ -1450.8 kJ

Your Assignment: pgs 12-13 in workbook

L. Energy Systems in Biological Processes

Photosynthesis:

Cellular Respiration:

energy + CO2(g) + H2O(l) C6H12O6(s) + O2(g)

C6H12O6(s) + O2(g) CO2(g) + H2O(l) + energy

**Water vapor condenses into liquid

M. Nuclear Change

enthalpy changes in nuclear reactions are the result of

there are two types of nuclear reactions:

EP changes as rearrangements among the

subatomic particles (protons and neutrons) occur ie) intranuclear forces

1. Fusion (Joining)

fusion of hydrogen to helium occurs on the sun and other stars

these types of reaction produce

require a great deal of

the greatest amount of energy and are necessary for life on Earth

heat and pressure

H + H He + n + 1.70 x 109 kJ01 11

2432

Top number =

Bottom number =

mass number

number of protons

2. Fission (Splitting)

basis for nuclear power plants

uranium atoms can be

was discovered in the late 1930's when uranium was bombarded by neutrons causing it to split

the neutrons produced by fission allow

split into two smaller nuclei which produces large quantities of energy

U + n Kr + Ba + 3 n + 1.9 x 1010 kJ123592 0 0

1 9256

14136

a chain reaction to occur to keep the reaction self sustaining

N. Society and Technological Connections

we must assess the risks and benefits of relying on fossil fuels and nuclear energy as energy sources

we are limited by our scientific knowledge and by the technology that has been developed to date

are the most common source of energy

many aspects of our society are based on the price of fuels like gasoline

fossils fuels

Advantages vs. Disadvantages of Fossil Fuels

Advantages Disadvantages

relatively low cost

readily available (market)

plant set-up, vehicle design, expertise affordable

used all over the world

deposits are large

release of gases that contribute to the greenhouse effect and acid rain when burned

mining is detrimental to the environment

non-renewable

Advantages vs. Disadvantages of Nuclear Power

Advantages Disadvantages

low cost of fuel

lots of energy from small amount of fuel

doesn’t add to greenhouse effect, acid rain

Canada has lots of uranium ore

preserves existing fossil fuels

high cost of plant set-up, expertise, decommission

cause thermal pollution

difficult to dispose of nuclear fuel wastes

possibility of catastrophic accidents

non-renewable