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Chapter 6Chapter 6

EnergyEnergy

ThermodynamicsThermodynamics

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Energy is...Energy is... The ability to do work.The ability to do work. Conserved.Conserved. made of heat and work.made of heat and work. a state function.a state function. independent of the path, or how you independent of the path, or how you

get from point A to B.get from point A to B. Work is a force acting over a distance.Work is a force acting over a distance. Heat is energy transferred between Heat is energy transferred between

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

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The universeThe universe 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. Exothermic reactions release Exothermic reactions release

energy to the surroundings.energy to the surroundings. Endo thermic reactions absorb Endo thermic reactions absorb

energy from the surroundings.energy from the surroundings.

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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

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N + O2 2

Pote

nti

al en

erg

y

Heat

2NO

N + O 2NO2 2 + heat

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DirectionDirection 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

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System

Surroundings

Energy

E <0

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System

Surroundings

Energy

E >0

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Same rules for heat and workSame rules for heat and work Heat given off Heat given off by the system by the system is is

negative.negative. Heat absorbed Heat absorbed by the system by the system is is

positive.positive. Work done Work done by the system by the system on on

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

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First Law of ThermodynamicsFirst Law of Thermodynamics ****The energy of the universe is ****The energy of the universe is

constant. (“cannot be created or constant. (“cannot be created or destroyed”)destroyed”)

AKA, Law of conservation of energy.AKA, Law of conservation of energy. E = q + w, where E = q + w, where

q = heat w = work

Take Take the system’s the system’s point of view to point of view to decide signs.decide signs.

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What is work?What is work? From physics, work is a force acting From physics, work is a force acting

over a distance.over a distance.1.1. w= F x w= F x d d (physics)(physics)2.2. P = F/ areaP = F/ area3.3. d = V/aread = V/area4.4. w= (P x area) x w= (P x area) x (V/area)= P (V/area)= PV V

(chemistry)(chemistry) 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: liter - atm L-atmUnits: liter - atm L-atm

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

the system has done work the system has done work on the on the surroundingssurroundings..

work is work is negativenegative w = - Pw = - PVV Expanding, work is negative.Expanding, work is negative. Contracting, surroundings do work Contracting, surroundings do work

on the system on the system --- w is --- w is positivepositive.. 1 L1 L••atm = 101.3 Jatm = 101.3 J

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ExamplesExamples1.1. What amount of work is done What amount of work is done

when 15 L of gas is expanded to when 15 L of gas is expanded to 25 L at 2.4 atm pressure?25 L at 2.4 atm pressure?

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

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

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Enthalpy (H)Enthalpy (H) H = E + PV (that’s the definition)H = E + PV (that’s the definition) ***at constant pressure.***at constant pressure. H = H = E + PE + PVV

the heat at constant pressure (qthe heat at constant pressure (qpp ) )

can be calculated from:can be calculated from:

E = qp + w = qp – PV (w = -PΔV)

qp = E + P V = H, so

qp = ΔH

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CalorimetryCalorimetry

Measures heat; uses a calorimeter.Measures heat; uses a calorimeter. Two kinds:Two kinds:

1. Constant pressure calorimeter (called a coffee cup calorimeter)

2. Constant volume calorimeter (called a bomb calorimeter)

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CalorimetryCalorimetry Specific heat capacity:Specific heat capacity: the energy the energy

required to raise the temperature of one required to raise the temperature of one gram of a substance by one degree gram of a substance by one degree CelsiusCelsius

Heat capacity for a material (C) is Heat capacity for a material (C) is calculated: calculated: C = heat absorbed/ T = H/ T specific heat capacity = J/g•°C molar heat capacity = J/moles •°C

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CalorimetryCalorimetry Energy released by a reaction (heat) = Energy released by a reaction (heat) =

specific heat capacity x mass x specific heat capacity x mass x TT heat = molar heat capacity x moles x heat = molar heat capacity x moles x

TT Make the units work and you’ve done Make the units work and you’ve done

the problem right!the problem right! A coffee cup calorimeter measures A coffee cup calorimeter measures H.H.

– An insulated cup, full of water. – The specific heat of water is 1 cal/gºC or 4.18

J/gºC– Heat of reaction= H = q = c x m x T

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ExamplesExamples4.4. 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.

5.5. A 24.3 g sample of copper is heated A 24.3 g sample of copper is heated to 95.4ºC and then placed in a to 95.4ºC and then placed in a calorimeter containing 75.0 g of calorimeter containing 75.0 g of water at 19.6ºC. The final water at 19.6ºC. The final temperature of both the water and temperature of both the water and the copper is 21.8ºC. What is the the copper is 21.8ºC. What is the specific heat of copper?specific heat of copper?

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ASIM LabsASIM Labs

1.1. Mixing Warm and Cold Mixing Warm and Cold (Calorimetry)(Calorimetry)

2.2. Specific Heat of MetalsSpecific Heat of Metals

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

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

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

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Bomb CalorimeterBomb Calorimeter thermometerthermometer

stirrerstirrer

full of waterfull of water

ignition wireignition wire

Steel bombSteel bomb

samplesample

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PropertiesProperties InIntensive properties not related to tensive properties not related to

((inindependent of) the amount of dependent of) the amount of substance.substance.

EX: density, specific heat, EX: density, specific heat, temperature.temperature.

Extensive property - does depend Extensive property - does depend on the amount of stuff.on the amount of stuff.

EX: heat capacity, mass, heat from EX: heat capacity, mass, heat from a reaction.a reaction.

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Hess’s LawHess’s Law Enthalpy is a state function; it is Enthalpy is a state function; it is

independent of the path.independent of the path. We can add reaction equations to We can add reaction equations to

come up with the desired final come up with the desired final product, and add the product, and add the H’s.H’s.

Two rules:Two rules:1. If the reaction is reversed, the sign of H is

changed.

2. If the reaction is multiplied, so is H.

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N2 + O2

+O22NO

68 kJ

2NO2180 kJ

-112 kJ

H (

kJ)

N2 + 2O2 → 2NO2

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

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

Symbol: Symbol: H°H° 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.

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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

ExampleExample6.6. GivenGiven

nn

calculate calculate Hº for this reactionHº for this reaction

Hº= -1300. kJ

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

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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)

7. Given

Calculate Hº for this reaction

Hº= +77.9kJHº= +495 kJ

Hº= +435.9kJ

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Standard Enthalpies of FormationStandard Enthalpies of Formation Hess’s Law is much more useful if you Hess’s Law is much more useful if you

know lots of reactions.know lots of reactions. Made a table of standard heats of Made a table of standard heats of

formation: the amount of heat needed formation: the amount of heat needed to form 1 mole of a compound from its to form 1 mole of a compound from its elements in their standard states.elements in their standard states.– Standard states are 1 atm, 1M and 25ºC– For an element it is 0

There is a table in Appendix 4 (pg There is a table in Appendix 4 (pg A21)A21)

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Standard Enthalpies of FormationStandard Enthalpies of Formation Need to be able to write the equations, Need to be able to write the equations,

starting with the individual elements.starting with the individual elements. What is the equation for the formation What is the equation for the formation

of NOof NO22 ? ?

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 to meet the definition.definition.

Write the equation for the formation of Write the equation for the formation of methanol CHmethanol CH33OH.OH.

– Start with the elemental forms!!!

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Since we can manipulate the Since we can manipulate the equations: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.

This leads us to this rule:This leads us to this rule:

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

( H products) - ( H reactants) = Hfo

fo o