chapter 6 - thermochemistry (new) · chapter 6 thermochemistry section 6.1 ... to convert between...

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Chapter 6

Thermochemistry

Section 6.1The Nature of Energy

What drives chemical reactions? How do they occur?

The first is answered by THERMODYNAMICS and the second by KINETICS.

We have already seen a number of “driving forces” for reactions that are PRODUCT‐FAVORED.

• formation of a precipitate

• gas formation

• H2O formation (acid‐base reaction)

• electron transfer in a battery


ENERGY is the capacity to do work or produce and transfer heat.

WORK ‐ force acting over a distance.

HEAT is the form of energy that flows between 2 samples because of their difference in temperature.

Other forms of energy —

light electricalnuclear

kinetic potential


Temperature reflects random motions of particles, therefore related to kinetic energy of the systemenergy of the system.

Heat (form of energy) involves a transfer of energy between 2 objects due to a temperature difference


The First Law of Thermodynamics = energy can be converted from one form to another but can be neither created nor destroyed.y

The total energy content of the universe is constant.

(Euniverse is constant)


Potential: due to position or composition ‐ can be converted to work

PE = mgh

(m = mass, g = acceleration of gravity, and h = height)

Kinetic: due to motion of the object

KE = 1/2 mv2

(m = mass, v = velocity)

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In the initial position, ball A has a higher potential energy than ball B.

Initial Position

Copyright © Cengage Learning. All rights reserved 7


After A has rolled down the hill, the potential energy lost by A has been converted to random motions of the components of the hill (frictional heating) and to the increase in the potential energy of B.

Final Position


p gy


Potential energy —energy a motionlessenergy a motionless body has by virtue of

its position.


Kinetic Kinetic energy energy --energy energy of of gygymotion.motion.

• Translation• Translation

• Rotation• Rotation

• Vibration• Vibration


1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0

oC. (Not an SI Unit)

1000 cal = 1 kilocalorie = 1 kcal

1 kcal = 1 Calorie (a food “calorie”)

But we use the derived SI unit called the JOULE

1 cal = 4.184 joulesJames JouleJames Joule

18181818--188918892

2111

s

mkgmeternewtonJoule


Extensive & Intensive Properties

Extensive properties depends directly on the amount of substance

Intensive properties is not related to the amount of substance.

present.

•mass

•volume

•heat

•heat capacity (C)

•temperature

•concentration

•pressure

•specific heat (s)

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State Function is a property that depends only on the present state of the system ‐ not how it arrived there.

It is independent of pathway.

Energy change is independent of the pathway (and, therefore, a state function), while work and heat are dependent on the pathway.


ENERGY IS A STATE FUNCTION

A person standing at the top of Mt. Everest has the same potential energy whether they got there bywhether they got there by hiking up, or by falling down from a plane

WORK IS NOT A STATE FUNCTION

WHY NOT???


System – part of the universe on which we wish to focus attention.

Surroundings – include everything else in the

Chemical Energy


universe.

Universe = System + Surroundings


Endothermic Reaction:

Heat flow is into a system.

Absorb energy from the surroundings.

Chemical Energy


gy g

Exothermic Reaction:

Energy flows out of the system.

Energy gained by the surroundings must be equal to the energy lost by the system.


SurroundingsSurroundings

SystemSystem

heatheat

SurroundingsSurroundings

SystemSystem

heatheat

qqsystemsystem > 0> 0w > 0w > 0

qqsystemsystem < 0< 0w < 0w < 0

ENDOTHERMICENDOTHERMIC EXOTHERMICEXOTHERMICE(system) goes upE(system) goes up E(system) goes downE(system) goes down


Is the freezing of water an endothermic or exothermic process? Explain.

CONCEPT CHECK!CONCEPT CHECK!


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Classify each process as exothermic or endothermic. Explain. The system is underlined in each example.



a) Your hand gets cold when you touch ice.

b) The ice gets warmer when you touch it.

c) Water boils in a kettle being heated on a stove.

d) Water vapor condenses on a cold pipe.

e) Ice cream melts.

Exo

Endo

Endo

Exo

Endo


For each of the following, define a system and its surroundings and give the direction of energy transfer.



gy

a) Methane is burning in a Bunsen burner in a laboratory.

b) Water drops, sitting on your skin after swimming, evaporate.


Hydrogen gas and oxygen gas react violently to form water. Explain.



Which is lower in energy: a mixture of hydrogen and oxygen gases, or water?


• Internal energy (E) of a system is the sum of the kinetic and potential energies of all the “particles” in the system.

Thermodynamics and Internal Energy


To change the internal energy of a system:ΔE = q + w

q represents heat

w represents work


Work vs Energy Flow


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Thermodynamic quantities consist of two parts:

Number gives the magnitude of the change.

Internal Energy


Sign indicates the direction of the flow.

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Internal Energy

Sign reflects the system’s point of view.

Endothermic Process:

q is positive

Exothermic Process:

q is negative



Sign reflects the system’s point of view.

System does work on surroundings:

w is negative

Internal Energy


Surroundings do work on the system:

w is positive


Expansion Compression (i ) (d )

w P V

+V (increase) -V (decrease)

-w results +w results

Esystem decreasesWork has been done by the system on the surroundings

Esystem increasesWork has been done on the system by the surroundings


Work = P× A× Δh = PΔV

P is pressure.

A i

Work

28

A is area.

Δh is the piston moving a distance.

ΔV is the change in volume.


For an expanding gas, ΔV is a positive quantity because the volume is increasing. Thus ΔV and wmust have opposite signs:

Work


w = –PΔV

To convert between L∙atm and Joules, use 1 L∙atm = 101.3 J.


w & Δ V must have opposite signs, since work is being done by the system to expand the P = F/AP = F/A

F PAF PA gas.

wsystem = ‐P Δ V

1 Latm = 101.3 J

1 J = kgm2/s2

F = PAF = PAw = F w = F ∆ hhw = PA w = PA ∆ hhw = P w = P ∆ VV

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Which of the following performs more work?

a) A gas expanding against a pressure

EXERCISE!EXERCISE!


a) A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L.

b) A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L.

They perform the same amount of work.


Determine the sign of ΔE for each of the following with the listed conditions:

a) An endothermic process that performs work



a) An endothermic process that performs work.

|work| > |heat|

|work| < |heat|

b) Work is done on a gas and the process is exothermic.

|work| > |heat|

|work| < |heat|

Δ E = negative

Δ E = positive

Δ E = positive

Δ E = negative

Section 6.2Enthalpy and Calorimetry

Change in Enthalpy

State function

ΔH = q at constant pressure

ΔH = H – H ΔH = Hproducts – Hreactants



Enthalpy = H = E + PV

E = H PVH = E + PV

At constant pressure,

qP = E + PV, where qP = H at constant pressure

H = energy flow as heat (at constant pressure)


H = Hfinal ‐ Hinitial

If H > H then H is positive If Hfinal > Hinitial then H is positive Process is ENDOTHERMIC

If Hfinal < Hinitial then H is negative Process is EXOTHERMIC


Endothermic:

Reactions in

+ qsystem - qsurroundings

Reactions in which energy flows into the system as the reaction proceeds.

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Exothermic:

Reactions in which energy

- qsystem + qsurroundings

which energy flows out of the system as the reaction proceeds.


Consider the combustion of propane:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

EXERCISE!EXERCISE!

ΔH = –2221 kJ

Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which 5.00 g of propane is burned in excess oxygen at constant pressure.

–252 kJ



Calorimetry

Science of measuring heat

Specific heat capacity:

The energy required to raise the temperature of one The energy required to raise the temperature of one gram of a substance by one degree Celsius. =

J/OC x g or J/K x g

Molar heat capacity:

The energy required to raise the temperature of one mole of substance by one degree Celsius. = J/OC x mol or J/K x mol



Substance Spec. Heat (J/g•K)

H2O 4.184WaterWater

Al 0.902

glass 0.84

AluminumAluminum


Calorimetry

If two reactants at the same temperature are mixed and the resulting solution gets warmer, this means the reaction taking place is exothermic.g p

An endothermic reaction cools the solution.



A Coffee–Cup Calorimeter Made of Two Styrofoam Cups


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Calorimetry

Energy released (heat) = Cp×m× ΔT

Cp = specific heat capacity (J/g∙OC)

m = mass of solution (g)(g)

ΔT = change in temperature (OC)

or.......

Energy released (heat) = Cn× n× ΔT

Cn = molar heat capacity (J/mol∙OC)

n = number of moles of substance (mol)

ΔT = change in temperature (OC)Copyright © Cengage Learning. All rights reserved 43


If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al?

where T = Tfinal ‐ Tinitial

heat gain/lost = q = m C T


If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al?

where T = Tfinal ‐ Tinitial q = (0.902 J/g•K)(25.0 g)(37 ‐ 310)K

q = ‐ 6160 J

heat gain/lost = q = m CT


If 25.0 g of Al cool from 310 oC to 37 oC,

how many joules of heat energy are lost

by the Al?

q = ‐ 6160 J

Notice that the negative sign on q signals

heat “lost by” or transferred out of Al.


75.0 g of Zn at 310. oC is dropped into 50.0 g of water at 25.0OC. What is the final t t f th t ?temperature of the system?

Cp of Zn = 0.387 J/g x OCheat gain by water = heat lost by metal

86.9OC


06_75

StirrerThermometer

Ignitionwires

Water

Insulatingcontainer

Reactantsin samplecup

Steelbomb

Bomb CalorimeterBomb Calorimeter

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Calculate heat of combustion of octane.

C8H18 + 25/2 O2 ‐‐> 8 CO2 + 9 H2O

Measuring Heats of ReactionMeasuring Heats of ReactionCALORIMETRYCALORIMETRY


C8H18 + 25/2 O2 > 8 CO2 + 9 H2O

Burn 1.00 g of octane Temp rises from 25.00 to 33.20 oC Calorimeter contains 1200. g water Heat capacity of bomb = 837 J/K

Hcm = ‐(C Δ t + ms Δ t) where Hc is heat of combustion.


Step 1 Calc. heat transferred from reaction to water.

q = (4.184 J/g•K)(1200. g)(8.20 K) = 41,170.56 J

St 2 C l h t t f d f ti t b b



Step 2 Calc. heat transferred from reaction to bomb.

q = C Δ t = (837 J/K)(8.20 K) = 6863.4J

Step 3 Total heat evolved

41,170.56 J + 6863.4 J = 48,033.96J

Heat of combustion of 1.00 g of octane = ‐ 48.0 kJ

Heat of combustion of 1.00 mol of octane =

‐ 48.0 kJ/g x 96.26 g/mol = ‐ 4620 kJ/mol


A 100.0 g sample of water at 90°C is added to a 100.0 g sample of water at 10°C.


The final temperature of the water is:

a) Between 50°C and 90°C

b) 50°C

c) Between 10°C and 50°C



A 100.0 g sample of water at 90.°C is added to a 500.0 g sample of water at 10.°C.




b) 50°C


Calculate the final temperature of the water.

23°C



You have a Styrofoam cup with 50.0 g of water at 10.°C. You add a 50.0 g iron ball at 90.°C to the water. (sH2O = 4.18 J/°C∙g and sFe = 0.45 J/°C∙g)




b) 50°C


Calculate the final temperature of the water.

18°CCopyright © Cengage Learning. All rights reserved 53

Section 6.3Hess’s Law

In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a p pseries of steps.


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This reaction also can be carried out in two distinct steps, with enthalpy changes designated by ΔH2 and ΔH3.

N2(g) + 2O2(g) → 2NO2(g) ΔH1 = 68 kJ

N2(g) + O2(g) → 2NO(g) ΔH2 = 180 kJ

2NO(g) + O2(g) → 2NO2(g) ΔH3 = – 112 kJ

N2(g) + 2O2(g) → 2NO2(g) ΔH2 + ΔH3 = 68 kJ

ΔH1 = ΔH2 + ΔH3 = 68 kJ



The Principle of Hess’s Law




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Characteristics of Enthalpy Changes

If a reaction is reversed, the sign of ΔH is also reversed.

The magnitude of ΔH is directly proportional to the quantities of reactants and products in a reaction. If thequantities of reactants and products in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is multiplied by the same integer.



Example

Consider the following data:

3 2 21 3

2 2NH ( ) N ( ) H ( ) H = 46 kJ

( ) O ( ) O( )

g g g

Calculate ΔH for the reaction


2 2 22 H ( ) O ( ) 2 H O( ) H = 484 kJ g g g

2 2 2 32 N ( ) 6 H O( ) 3 O ( ) 4 NH ( ) g g g g


Problem‐Solving Strategy

Work backward from the required reaction, using the reactants and products to decide how to manipulate the other given reactions at your p g ydisposal.

Reverse any reactions as needed to give the required reactants and products.

Multiply reactions to give the correct numbers of reactants and products.


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Example

Reverse the two reactions:

2 2 31 3

2 2N ( ) H ( ) NH ( ) H = 46 kJ g g g

Desired reaction:


2 2 2 32 N ( ) 6 H O( ) 3 O ( ) 4 NH ( ) g g g g

2 2 22 H O( ) 2 H ( ) O ( ) H = +484 kJ g g g


Example

Multiply reactions to give the correct numbers of reactants and products:

4( ) 4( )

3( ) 3( )

Desired reaction:

2 2 2 32 N ( ) 6 H O( ) 3 O ( ) 4 NH ( ) g g g g

2 2 3

2 2 2

1 3

2 2N ( ) H ( ) NH ( ) H = 46 kJ

2 H O( ) 2 H ( ) O ( ) H = +484 kJ

g g g

g g g


Example

Final reactions:

2 2 32 N ( ) 6 H ( ) 4 NH ( ) H = 184 kJ

6 H O( ) 6 H ( ) 3 O ( ) H = +1452 kJ

g g g

g g g

Desired reaction:

ΔH = +1268 kJ


2 2 2 32 N ( ) 6 H O( ) 3 O ( ) 4 NH ( ) g g g g

2 2 26 H O( ) 6 H ( ) 3 O ( ) H = +1452 kJ g g g

Section 6.4Standard Enthalpies of Formation

Standard Enthalpy of Formation (ΔHf°)

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



Conventional Definitions of Standard States

For a Compound

For a gas, pressure is exactly 1 atm.

For a solution concentration is exactly 1M For a solution, concentration is exactly 1 M.

Pure substance (liquid or solid)

For an Element

The form [N2(g), K(s)] in which it exists at 1 atm and 25°C.



A Schematic Diagram of the Energy Changes for the Reaction CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

ΔH°reaction = –(–75 kJ) + 0 + (–394 kJ) + (–572 kJ) = –891 kJ


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Problem‐Solving Strategy: Enthalpy Calculations

1. When a reaction is reversed, the magnitude of ΔHremains the same, but its sign changes.

2. When the balanced equation for a reaction is multiplied2. When the balanced equation for a reaction is multiplied by an integer, the value of ΔH for that reaction must be multiplied by the same integer.



Problem‐Solving Strategy: Enthalpy Calculations

3. The change in enthalpy for a given reaction can be calculated from the enthalpies of formation of the reactants and products:

Hrxn° = npHf(products) nrHf(reactants)

H is an extensive property‐‐kJ/mol

4. standard states are not included in the ΔHreaction

calculations because ΔHf° for an element in its standard state is zero.



Calculate the heat of combustion of

methanol, i.e., Horxn for

CH3OH(g) + 3/2 O2(g) ‐‐> CO2(g) + 2 H2O(g)

Horxn = Ho

f (prod) ‐ Hof (react)


CH3OH(g) + 3/2 O2(g) ‐‐> CO2(g) + 2 H2O(g)

Horxn = Ho

f (prod) ‐ Hof (react)

Ho = Hof (CO2) + 2 Ho

f (H2O) ‐H rxn H f (CO2) + 2 H f (H2O)

{3/2 Hof (O2) + Ho

f (CH3OH)}

= (‐393.5 kJ) + 2 (‐241.8 kJ) ‐ {0 + (‐201.5 kJ)}

Horxn = ‐675.6 kJ per mol of methanol

Horxn is always in terms of moles of reactant.


CH4(g)

C(s)

CO2(g)

2H2(g)

(a) (c)

2H2O(l)

2O2(g) 2O2(g)

Reactants Elements Products

(b)

(d)

Pathway for the Combustion of MethanePathway for the Combustion of Methane


Calculate ΔH° for the following reaction:

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

EXERCISE!EXERCISE!

Given the following information:

Δ Hf° (kJ/mol)

Na(s) 0

H2O(l) –286

NaOH(aq) –470

H2(g) 0

Δ H° = –368 kJCopyright © Cengage Learning. All rights reserved 72

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Section 6.5Present Sources of Energy

Fossil Fuels

Petroleum, Natural Gas, and Coal

Wood Wood

Hydro

Nuclear



Energy Sources Used in the United States



The Earth’s Atmosphere

Transparent to visible light from the sun.

Visible light strikes the Earth, and part of it is changed to infrared radiation.to infrared radiation.

Infrared radiation from Earth’s surface is strongly absorbed by CO2, H2O, and other molecules present in smaller amounts in atmosphere.

Atmosphere traps some of the energy and keeps the Earth warmer than it would otherwise be.



The Earth’s Atmosphere


Section 6.6New Energy Sources

Coal Conversion

Hydrogen as a Fuel

Other Energy Alternatives Other Energy Alternatives

Oil shale

Ethanol

Methanol

Seed oil


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