AP Notes Chapter 5 Principles of Reactivity: Energy and Chemical Reactions.

AP Notes Chapter 5AP Notes Chapter 5Principles of Reactivity:Principles of Reactivity:

Energy and Chemical ReactionsEnergy and Chemical Reactions

•study of energy transfer or heat flow

Energy •Kinetic Energy

Thermal - HeatMechanicalElectricalSound

•Potential Energy ChemicalGravitationalElectrostatic

Thermodynamics

Energy Energy is...is...

The ability to do work.The ability to do work. Conserved.Conserved. made of heat and work.made of heat and work.

Work is a force acting over a distance.Work is a force acting over a distance.

W=F x dW=F x d Power is work done over time.Power is work done over time.

P=P=WW

tt Heat is energy transferred between Heat is energy transferred between

objects because of temperature objects because of temperature difference.difference.

First Law of First Law of ThermodynamicsThermodynamics

Law of conservation of energyLaw of conservation of energy Total energy of the universe is constantTotal energy of the universe is constant

Temperature and heat Temperature and heat Heat is not temperatureHeat is not temperature Thermal energy = particle motionThermal energy = particle motion Total thermal energy is sum of all a materials Total thermal energy is sum of all a materials

individual energiesindividual energies

The The UniverseUniverse is divided into two halves.is divided into two halves.

the system and the surroundings.the system and the surroundings. The system is the part you are The system is the part you are

concerned with.concerned with. The surroundings are the rest.The surroundings are the rest.

q into system = -q from surroundingsq into system = -q from surroundings

System: region of space where process occursSurroundings: region of space around system

Thermodynamic Properties1. State functions

-depend on nature of process NOT method

-path independent-denoted by capital letters

E → state function

2. Path functions-depends on how process is carried out

-path dependent-denoted by lower case

Thermodynamic Properties

q & w → path functions

Work →w w = - (PV)

w = - P V - V P

Constant P wP = - P V

Constant V wV = - V P

Work →ww = - P V

assume ideal piston (i.e. no heat loss)

w = E E = - P V

in isothermicsw = -q

PV = nRTR=0.0821L atm/mol

K

In isothermic conditions

-q = w = -nRTR=8.314 J/mol K

U = q + w

In isothermics U = 0 so q=-w

U = q - PVIn constant V or constant P w=0 so U = q

SI unit → Joule

1 cal = 4.184 J1000cal = 1Kcal=4.184KJ

Jkg m

s

2

2

Calorie - Amount of heat needed to raise the temperature of 1 gram H20, 1 degree centigrade

Nutritional calorie [Calorie]= 1 Kcal

Direction of energy Direction of energy flowflow Every energy measurement has Every energy measurement has

three parts.three parts.

1.1. A unit ( Joules or calories).A unit ( Joules or calories).

2.2. A number how many.A number how many.

3.3. and a sign to tell direction.and a sign to tell direction. negative - exothermicnegative - exothermic positive- endothermicpositive- endothermic No change in energy - isothermicNo change in energy - isothermic

Factors Determining Amount of Heat

Amount of material

Temperature change

Heat capacity

Specific HeatAmount of heat needed to raise

the temperature of 1 gram of material 1 degree centigrade

oCg

JCp

Heat = qq = m Cp T

where Cp = Heat capacity

T = T2 – T1

CP (J/mol.oC)H2O(s) = 37.11H2O(l) = 75.35H2O(g) = 33.60

What amount of heat is needed to just boil away 100. mL of water in a typical lab?

System

Surroundings

Energy

E <0

Exothermic reactions release energy to the Exothermic reactions release energy to the surroundings.surroundings.

q < 0 (-)

Exothermic

CH + 2O CO + 2H O + Heat4 2 2 2

CH + 2O 4 2

CO + 2 H O 2 2

Pote

nti

al en

erg

y

Heat

System

Surroundings

Energy

E >0

Endothermic reactions absorb energy Endothermic reactions absorb energy from the surroundings.from the surroundings.

q > 0 (+)

Endothermic

N + O2 2

Pote

nti

al en

erg

y

Heat

2NO

N + O 2NO2 2 + heat

1. How much heat is absorbed when 10.0 g of H20 are heated from 200C to 250C?

2. How much heat is needed to completely vaporize 10.0 g of H2O at its boiling point?

Heat of VaporizationHvap(H2O) = 40.66 kJ/mol

Heat of FusionHfus(H2O) = 6.01 kJ/mol

Energy processes:•Within a phase

q=mHv or mHf

•Between phasesq=mCpT

5

1iitotal qq

3. How much total heat is used to raise the temperature of 100.0 g H2O from-15o to 125oC?

Some rules for heat and Some rules for heat and workwork Heat given off is negative.Heat given off is negative.

Heat absorbed is positive.Heat absorbed is positive. Work done by system on Work done by system on

surroundings is positive.surroundings is positive. Work done on system by Work done on system by

surroundings is negative.surroundings is negative. Thermodynamics- The study of Thermodynamics- The study of

energy and the changes it undergoes.energy and the changes it undergoes.

First Law of First Law of ThermodynamicsThermodynamics The energy of the universe is constant.The energy of the universe is constant.

Law of conservation of energy.Law of conservation of energy. q = heatq = heat w = workw = work E = q + wE = q + w Take the systems point of view to Take the systems point of view to

decide signs.decide signs.

What is work?What is work? Work is a force acting over a distance.Work is a force acting over a distance. w= F x w= F x dd P = F/ areaP = F/ area d = V/aread = V/area w= (P x area) x w= (P x area) x (V/area)= P (V/area)= PVV Work can be calculated by multiplying Work can be calculated by multiplying

pressure by the change in volume at pressure by the change in volume at constant pressure.constant pressure.

units of liter - atm L-atmunits of liter - atm L-atm

Work needs a signWork needs a sign If the volume of a gas increases, the If the volume of a gas increases, the

system has done work on the surroundings.system has done work on the surroundings. work is negativework is negative w = - Pw = - PVV Expanding work is negative.Expanding work is negative. Contracting, surroundings do work on the Contracting, surroundings do work on the

system w is positive.system w is positive. 1 L atm = 101.3 J1 L atm = 101.3 J

ExamplesExamples

What amount of work is done when 15 L What amount of work is done when 15 L of gas is expanded to 25 L at 2.4 atm of gas is expanded to 25 L at 2.4 atm pressure?pressure?

If 2.36 J of heat are absorbed by the gas If 2.36 J of heat are absorbed by the gas above. what is the change in energy?above. what is the change in energy?

How much heat would it take to change How much heat would it take to change the gas without changing the internal the gas without changing the internal energy of the gas? energy of the gas?

CalorimetryCalorimetry Measuring heat.Measuring heat. Use a calorimeter.Use a calorimeter. Two kindsTwo kinds Constant pressure calorimeter (called a Constant pressure calorimeter (called a

coffee cup calorimeter)coffee cup calorimeter) heat capacity for a material, C is heat capacity for a material, C is

calculated calculated C= heat absorbed/ C= heat absorbed/ T = T = H/ H/ TT specific heat capacity = C/mass specific heat capacity = C/mass

CalorimetryCalorimetry molar heat capacity = C/molesmolar heat capacity = C/moles heat = specific heat x m x heat = specific heat x m x TT heat = molar heat x moles x heat = molar heat x moles x TT Make the units work and you’ve done the Make the units work and you’ve done the

problem right.problem right. A coffee cup calorimeter measures A coffee cup calorimeter measures H.H. An insulated cup, full of water. An insulated cup, full of water. The specific heat of water is 1 cal/gºCThe specific heat of water is 1 cal/gºC Heat of reaction= Heat of reaction= H = sh x mass x H = sh x mass x TT

Bomb Calorimeter

Calorimeter Constantheat capacity that is constant over atemperature range

where

q(cal) = CC x TC

(CC = calorimeter constant)

4. Determine the calorimeter constant for a bomb calorimeter by

mixing 50.0 mL of water at 25oC with 50.0 mL of water at 60.0 oC. The final temp. attained is 40oC.

5. What is the molar heat of combustion of sulfur, if 2.56 grams of solid flowers of sulfur are burned in excess oxygen in the bomb calorimeter in #4?

6. Determine the molar heat of solution of LiOH. HINT: Find the calorimeter constant first.

7. The heat was absorbed by 815 g of water in calorimeter which had an initial temp of 21.25 C. After equilibrium was reached, the final temperature was 26.72 C.

ExamplesExamples The specific heat of graphite is 0.71 The specific heat of graphite is 0.71

J/gºC. Calculate the energy needed to J/gºC. Calculate the energy needed to raise the temperature of 75 kg of raise the temperature of 75 kg of graphite from 294 K to 348 K.graphite from 294 K to 348 K.

A 46.2 g sample of copper is heated to A 46.2 g sample of copper is heated to 95.4ºC and then placed in a calorimeter 95.4ºC and then placed in a calorimeter containing 75.0 g of water at 19.6ºC. containing 75.0 g of water at 19.6ºC. The final temperature of both the water The final temperature of both the water and the copper is 21.8ºC. What is the and the copper is 21.8ºC. What is the specific heat of copper?specific heat of copper?

CalorimetryCalorimetry Constant volume calorimeter is Constant volume calorimeter is

called a bomb calorimeter.called a bomb calorimeter. Material is put in a container with Material is put in a container with

pure oxygen. Wires are used to start pure oxygen. Wires are used to start the combustion. The container is put the combustion. The container is put into a container of water.into a container of water.

The heat capacity of the calorimeter The heat capacity of the calorimeter is known and tested.is known and tested.

Since Since V = 0, PV = 0, PV = 0, V = 0, E = q E = q

Bomb CalorimeterBomb Calorimeter

thermometerthermometer

stirrerstirrer

full of waterfull of water

ignition wireignition wire

Steel bombSteel bomb

samplesample

PropertiesProperties

intensive properties not related to intensive properties not related to the amount of substance.the amount of substance.

density, specific heat, temperature.density, specific heat, temperature. Extensive property - does depend on Extensive property - does depend on

the amount of stuff.the amount of stuff. Heat capacity, mass, heat from a Heat capacity, mass, heat from a

reaction.reaction.

EnthalpyEnthalpy abbreviated Habbreviated H H = E + PV (that’s the definition) PV = wH = E + PV (that’s the definition) PV = w at constant pressure.at constant pressure. H = H = E + PE + PVV

the heat at constant pressure qthe heat at constant pressure qpp can be can be

calculated fromcalculated from

E = qE = qpp + w = q + w = qpp - P - PVV

qqpp = = E + P E + P V = V = HH

Hess’s LawHess’s Law Enthalpy is a state function.Enthalpy is a state function. It is independent of the path.It is independent of the path. We can add equations to come up We can add equations to come up

with the desired final product, and with the desired final product, and add the add the HH

Two rulesTwo rules If the reaction is reversed the sign of If the reaction is reversed the sign of H H

is changedis changed If the reaction is multiplied, so is If the reaction is multiplied, so is HH

Enthalpy and Enthalpy and HHff

NN22 + 2O + 2O22 2NO 2NO22

NN22 + 2O + 2O22 + 68KJ/mol + 68KJ/mol 2NO 2NO22

Which side is the heat written on? Why?Which side is the heat written on? Why?

Look on your Heat Sheets! Look on your Heat Sheets!

Why is it only 34KJ/mol?Why is it only 34KJ/mol?

N2 + 2O2

68 kJ

2NO2H (

kJ)

N2 + 2O2

O2 + NO2

2NO2180 kJ

-112 kJ

H (

kJ)

N2 + 2O2

O2 + NO2

68 kJ

2NO2180 kJ

-112 kJ

H (

kJ)

Hess’ LawReactants → Products

The enthalpy change is the same whether the reaction occurs in one step or in a series of steps

Molar heat capacity

degmol

J

8. Calculate the heat needed to decomposeone mole of calciumcarbonate.

Standard EnthalpyStandard Enthalpy The enthalpy change for a reaction at The enthalpy change for a reaction at

standard conditions (25ºC, 1 atm , 1 standard conditions (25ºC, 1 atm , 1 M solutions)M solutions)

Symbol Symbol HºHº When using Hess’s Law, work by When using Hess’s Law, work by

adding the equations up to make it adding the equations up to make it look like the answer. look like the answer.

The other parts will cancel out.The other parts will cancel out.

H (g) + 1

2O (g) H (l) 2 2 2O

C(s) + O (g) CO (g) 2 2Hº= -394 kJ

Hº= -286 kJ

C H (g) + 5

2O (g) 2CO (g) + H O( ) 2 2 2 2 2 l

ExampleExample

GivenGiven calculate calculate Hº for this reactionHº for this reaction

Hº= -1300. kJ

2C(s) + H (g) C H (g) 2 2 2

H (g) + 1

2O (g) H (l) 2 2 2O

C(s) + O (g) CO (g) 2 2 394 kJ

286 kJ

C H (g) + 5

2O (g) 2CO (g) + H O( ) 2 2 2 2 2 l

ExampleExample

calculate calculate Hº for this reactionHº for this reaction

1300. kJ

2C(s) + H (g) C H (g) 2 2 2

GivenGiven

H (g) + 1

2O (g) H (l) 2 2 2O

C(s) + O (g) CO (g) 2 2 394 kJ

286 kJ

C H (g) + 5

2O (g) 2CO (g) + H O( ) 2 2 2 2 2 l

ExampleExample

calculate calculate Hº for this reactionHº for this reaction

1300. kJ

2C(s) + H (g) C H (g) 2 2 2

GivenGiven

1300.KJ + 2CO2(g) + H20(l) C2H2 (g) + 5/2 O2 (g)

H (g) + 1

2O (g) H (l) 2 2 2O

C(s) + O (g) CO (g) 2 2 394 kJ)

286 kJ

ExampleExample

calculate calculate Hº for this reactionHº for this reaction2C(s) + H (g) C H (g) 2 2 2

GivenGiven

1300.KJ + 2CO2(g) + H20(l) C2H2 (g) + 5/2 O2 (g)

22 2

H (g) + 1

2O (g) H (l) 2 2 2O

C(s) + O (g) CO (g) 2 2 kJ

286 kJ

ExampleExample

calculate calculate Hº for this reactionHº for this reaction2C(s) + H (g) C H (g) 2 2 2

GivenGiven

1300.KJ + 2CO2(g) + H20(l) C2H2 (g) + 5/2 O2 (g)

42 22

H (g) + 1

2O (g) H (l) 2 2 2O

C(s) + O (g) CO (g) 2 2 kJ

286 kJ

ExampleExample

calculate calculate Hº for this reactionHº for this reaction2C(s) + H (g) C H (g) 2 2 2

GivenGiven

1300.KJ + 2CO2(g) + H20(l) C2H2 (g) + 5/2 O2 (g)

42 22

ExampleExample

calculate calculate Hº for this reactionHº for this reaction2C(s) + H (g) C H (g) 2 2 2

GivenGiven

1300.KJ C2H2 (g)

2C(s) 788KJ

H2(g) 286KJ

1300.KJ 788kJ + 286KJ 1300.KJ 1074KJ

ExampleExample

calculate calculate Hº for this reactionHº for this reaction2C(s) + H (g) C H (g) 2 2 2

GivenGiven

1300.KJ 1074KJ

226KJ

Hf = 226KJ (+sign shows Endothermic)

226KJ +

Hf = 1300KJ – 1074KJ

ExampleExample

O (g) + H (g) 2OH(g) 2 2 O (g) 2O(g)2 H (g) 2H(g)2

O(g) + H(g) OH(g)

Given

Calculate Hº for this reaction

Hº= +77.9kJHº= +495 kJ

Hº= +435.9kJ

Now, You try it!

Standard Enthalpies of Standard Enthalpies of FormationFormation

Hess’s Law is much more useful if you know Hess’s Law is much more useful if you know lots of reactions.lots of reactions.

Made a table of standard heats of formation. Made a table of standard heats of formation. The amount of heat needed to for 1 mole of The amount of heat needed to for 1 mole of a compound from its elements in their a compound from its elements in their standard states.standard states.

Standard states are 1 atm, 1M and 25ºCStandard states are 1 atm, 1M and 25ºC For an element it is 0For an element it is 0 There is a table in Appendix 4 (pg A22)There is a table in Appendix 4 (pg A22)

Standard Enthalpies of Standard Enthalpies of FormationFormation

Need to be able to write the equations.Need to be able to write the equations. What is the equation for the formation of What is the equation for the formation of

NONO22 ? ?

½N½N2 2 (g) + O(g) + O22 (g) (g) NO NO22 (g) (g)

Have to make Have to make one mole one mole to meet the definition.to meet the definition. Write the equation for the formation of Write the equation for the formation of

methanol CHmethanol CH33OH.OH.

Since we can manipulate Since we can manipulate the equationsthe equations

We can use heats of formation to We can use heats of formation to figure out the heat of reaction.figure out the heat of reaction.

Lets do it with this equation.Lets do it with this equation. CC22HH55OH +3OOH +3O22(g) (g) 2CO 2CO22 + 3H + 3H22OO

which leads us to this rule.which leads us to this rule.

Since we can manipulate Since we can manipulate the equationsthe equations

We can use heats of formation to We can use heats of formation to figure out the heat of reaction.figure out the heat of reaction.

Lets do it with this equation.Lets do it with this equation. CC22HH55OH +3OOH +3O22(g) (g) 2CO 2CO22 + 3H + 3H22OO

which leads us to this rule.which leads us to this rule.

( H products) - ( H reactants) = Hfo

fo o

9. Calculate Hf of SO3(g)from the data given. Forms of Equations:

2SO2(g) + O2(g) 2SO3(g)

H = -197.8 kj

or

2SO2(g) + O2(g) 2SO3(g) +197.8 kj

10. If the metabolism of glucose is combustion of C6H12O6(s), how much heat is produced by the metabolism of 1.00 g of glucose? mol

kJ0f 1268H

H = Enthalpy

H = E + PV

H = E + PV

Additivity of Heats of Reaction

at constant pressure:E = qP + w

or E = qP - PV

now, at constant P:

VP = 0thus, H = E + PV & if E = qP - PV

then H = qP

H = Enthalpy

a “state” function

cannot be measured

measure H

H = Hproducts - Hreactants

Standard Enthalpyof FormationChange in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states

reference point- for ELEMENTS in standard state

0H0f

Thermodynamic DataAppendix L of text p. A31

Thermodynamic Standard StateT = 25oC and P = 1 atm

11. Combustion of Methane

HC = ?

)l(OH 2)g(CO)g(O 2)g(CH 2224

)()(O )( 3.

)()()( .2

)()(H 2)( 1.

2221

2

22

42

lOHggH

gCOgOsC

gCHgsC

Write equations and find ΔHf0 for

methane,carbon dioxide, & water

)l(OH 2)g(CO)g(O 2)g(CH 2224

321

2224

3222

222

124

2)(

)( 2)()( 2)(

H )( 2)(O )( 2

H )()()(

H- )(H 2)()(

HHHH

lOHgCOgOgCH

lOHggH

gCOgOsC

gsCgCH

C

HC = ?