Thermochemistry - Chapter 6 Rev 8 - Student Version · Chapter 6 — Thermochemistry 1 © 2009...

Chapter 6 — Thermochemistry

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© 2009 Brooks/Cole - Cengage

Thermochemistry

CHAPTER 6

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Objectives• Definitions of energy, heat, work and how they

interrelate

• 1st Law of Thermodynamics

• Heat Capacity

• Energy Transfer

• Define Universe, System, and Surroundings

• Enthalpy

• Hess’s Law

• Calorimetry

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Geothermal power —Wairakei North Island, New Zealand

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Ch. 6.1 - Energy & Chemistry

• Burning peanuts supply sufficient energy to boil a cup of water.

• Burning sugar (sugar reacts with KClO3, a strong oxidizing agent)

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Energy & Chemistry

• These reactions are PRODUCT FAVORED

• They proceed almost completely from reactants to products, perhaps with some outside assistance.

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Energy & ChemistryENERGY is the capacity to

do work or transfer heat.

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

Other forms of energy —

• light

• electrical

• kinetic and potential


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Potential & Kinetic Energy

Potential energy —energy a motionless body has by virtue of its position.

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• Positive and negative particles (ions) attract one another.

• Two atoms can bond

• As the particles attract they have a lower potential energy

Potential Energyon the Atomic Scale


NaCl — composed of Na+ and Cl- ions.

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• Positive and negative particles (ions) attract one another.

• Two atoms can bond

• As the particles attract they have a lower potential energy



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Potential & Kinetic Energy

Kinetic energy— energy of motion

• Translation

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Potential & Kinetic EnergyKinetic energy —

energy of motion.

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Kinetic and Potential Energy Summary

• Chemical Energy is due to the potential energy stored in the arrangements of the atoms in a substance.

• If you run a reaction that releases heat, you are releasing the energy stored in the bonds.

• Energy because of temperature is associated with the kinetic (movement) energy of the atoms and molecules.


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Internal Energy (E)*Internal Energy (E)*

• PE + KE = Internal energy (E)

(1st Law of Thermodynamics)

• Internal energy of a chemical system depends on

• number of particles

• type of particles

• temperature*Note: Internal energy is sometimes

symbolized by U

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Internal Energy (E)Internal Energy (E)PE + KE = Internal energy (E)

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Internal Energy (E)Internal Energy (E)

• The higher the T the higher the internal energy

• So, use changes in T (∆T) to monitor changes in energy (∆E).

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ThermodynamicsThermodynamics• Thermodynamics is the science of

energy transfer as heat.

Heat energy is associated with molecular motions.

Energy transfers as heat until thermal equilibrium is established.∆T measures energy transferred.

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System and Surroundings

• SYSTEM– The object under study

• SURROUNDINGS– Everything outside the

system

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Directionality of Energy Transfer• Energy transfer as heat is always from a

hotter object to a cooler one.

• EXOthermic: energy transfers from SYSTEM to SURROUNDINGS.

T(system) goes downT(surr) goes up


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Directionality of Energy Transfer• Energy transfer at heat is always from a

hotter object to a cooler one.

• ENDOthermic: heat transfers from SURROUNDINGS to the SYSTEM.

T(system) goes upT (surr) goes down

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Energy and Temperature Summary

1) Temperature determines the direction of thermal energy transfer.

2) The higher the temperature of a given object, the greater the thermal energy (molecular motion) of its atoms, ions, or molecules.

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Energy and Temperature Summary

3) Heating and cooling are processes by which energy is transferred as heat from an object at higher temperature to one at lower temperature.

• Heat is not a substance but a ‘process quantity.’ That is heating is a process that changes the energy of a system.

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Energy & ChemistryAll of thermodynamics depends

on the law of

CONSERVATION OF ENERGY.• The total energy is unchanged

in a chemical reaction.

• If potential energy (PE) of products is less than reactants, the difference must be released as kinetic energy (KE).

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Energy Change in Chemical ProcessesEnergy Change in

Chemical Processes

PE of system dropped. KE increased. Therefore, you often feel a T increase.

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Pot

enti

al e

ner

gy

Heat

Exothermic Reaction – reaction give off energy.

CH4 + 2O2 CO2 + 2H2O + Heat

CH4 + 2O2

CO2 + 2H2O


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Pot

enti

al e

ner

gy

Heat

Endothermic Reaction – requires external energy to proceed. Very strong bonds have low potential

energy.N2 + O2 + Heat 2NO

N2 + O2

2NO

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UNITS OF ENERGY1 calorie = heat required to

raise temp. of 1.00 g of H2O by 1.0 oC.

1000 cal = 1 kilocalorie = 1 kcal

1 kcal = 1 Calorie (a food

“calorie”)

But we use the unit called the JOULE

1 cal = exactly 4.184 joules

James Joule1818-1889

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Work vs Energy Flow

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• Work = P× A× Δh = PΔV

– P is pressure.

– A is area.

– Δh is the piston moving a distance.

– ΔV is the change in volume.

Work

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

w = –PΔV

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

Work30

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Which of the following performs morework?

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.

Practice


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FIRST LAW OF THERMODYNAMICS

FIRST LAW OF THERMODYNAMICS

∆E = q + w

heat energy transferred

energychange

work doneby the system

Energy is conserved!

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energy transfer out(exothermic), -q

energy transfer in(endothermic), +q

SYSTEMSYSTEM

∆E = q + w∆E = q + w

w transfer in(+w)

w transfer out(-w)

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Practice Problems• Calculate ∆E, and determine whether the process

is endothermic or exothermic for the following cases.

a)q = 1.62 kJ and w = -874 J

b)A system releases 113 kJ of heat to the surroundings and does 39 kJ of work on the surroundings.

c)The system absorbs 77.5 kJ of heat while doing 63.5 kJ of work on the surroundings.

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Ch. 6.2 - HEAT CAPACITYThe heat required to raise an

object’s T by 1 ˚C.

Which has the larger heat capacity?

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Specific Heat Capacity

How much energy is transferred

due to T difference?

The heat (q) “lost” or “gained” is

related to a) sample mass

b) change in T and

c) specific heat capacity (Symbolized by s or Cp)

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Specific Heat CapacitySubstance Spec. Heat (J/g•K)

H2O 4.184

Ethylene glycol 2.39

Al 0.897

glass 0.84

Aluminum


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Specific Heat Capacity ExampleIf 25.0 g of Al cool

from 310oC to 37oC, what amount of energy (J) is lost by the Al?

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Specific Heat Capacity ExampleIf 25.0 g of Al cool from 310 oC to 37 oC, what amount of

energy (J) has been transferred by the Al?

heat gain/lose = q = (mass)(Sp. Heat)(∆T)or q = m x Cp x ∆T

where ∆T = Tfinal - Tinitial

q = (25.0 g) (0.897 J/g•K)(37 - 310)K

q = - 6120 J

Notice that the negative sign on q signals heat “lost by” or transferred OUT of Al.

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Specific Heat Capacity PracticeIn an experiment, it was determined that 59.8 J was required to raise the temperature of 25.0 g of ethylene glycol by 1.00 K. Calculate the specific heat capacity of ethylene glycol from these data.

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Energy TransferEnergy Transfer• Use energy transfer as a

way to find specific heat capacity, Cp

• 55.0 g Fe at 99.8 ˚C• Drop into 225 g water at

21.0 ˚C• Water and metal come to

23.1 ˚C• What is the specific heat

capacity of the metal?

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Energy Transfer Important Points• Iron and the water are the system, the beaker

holding the water and everything else is the surroundings.

• The iron and water end up at the same temp.

• The energy transferred as heat from the iron to the water is negative (temp of Fe drops). The qwaterincrease is positive because the temp of water increases.

• The values of qwater and qiron are numerically equal but opposite in sign.

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Because of conservation of energy,

q(Fe) = –q(H2O) (energy out of Fe = energy into H2O)

or q(Fe) + q(H2O) = 0

q(Fe) = (55.0 g)(Cp)(23.1 ˚C – 99.8 ˚C)

q(Fe) = –4219 • Cp

q(H2O) = (225 g)(4.184 J/K•g)(23.1 ˚C – 21.0 ˚C)

q(H2O) = 1977 J

q(Fe) + q(H2O) = –4219 Cp + 1977 = 0

Cp = 0.469 J/K•g

Energy TransferEnergy Transfer


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ENTHALPYENTHALPYMost chemical reactions occur at constant P, so

and so ∆E = ∆H + w (and w is usually small)

∆H = energy transferred as heat at constant P ≈ ∆E

∆H = change in heat content of the system

∆H = Hfinal - Hinitial

and so ∆E = ∆H + w (and w is usually small)

∆H = energy transferred as heat at constant P ≈ ∆E

∆H = change in heat content of the system

∆H = Hfinal - Hinitial

Heat transferred at constant P = qp

qp = ∆H where H = enthalpy

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If Hfinal < Hinitial then ∆H is negativeProcess is EXOTHERMIC

If Hfinal > Hinitial then ∆H is positiveProcess is ENDOTHERMIC

ENTHALPYENTHALPY∆H = Hfinal – Hinitial

Enthalpy is an extensive property!!

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Hydrogen peroxide decomposes to water and oxygen by the following reaction:

2H2O2(l) 2H2O(l) + O2(g) ∆H = ‐196 kJ

–Calculate the value of energy transferred when 5.00 g of H2O2(l) decomposes at constant pressure.

Hints: What is enthalpy of reaction if only 1 mole of peroxide is decomposed?

Practice Using EnthalpyPractice Using Enthalpy 46


– Hydrogen peroxide decomposes to water and oxygen by the following reaction:

2H2O2(l) 2H2O(l) + O2(g) ∆H = ‐196 kJ -196 kJ/2 mol peroxide = -98 kJ/mol of peroxide

5.0 g * (1 mol/34 g H2O2) = 0.147 mol H2O2

= 0.147 mol H2O2 * -98 kJ/mol peroxide

= -14.4 kJ

Heat is a stoichiometric value which works the same way as last year.

Practice Using EnthalpyPractice Using Enthalpy

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Consider the formation of water

H2(g) + 1/2 O2(g) H2O(g) + 241.8 kJ

USING ENTHALPYUSING ENTHALPY

Exothermic reaction — energy is a “product” and∆H = – 241.8 kJ

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Making liquid H2O from H2 + O2 involves two exothermic steps.

USING ENTHALPYUSING ENTHALPY

H2 + O2 gas

Liquid H2OH2O vapor


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Calorimetry

• Definition of calorimetry: measure heat flow in a chemical reaction.

• Two kinds of calorimeters (devices that measure heat flow):–Constant pressure (coffee cup calorimeter)

–Constant volume (bomb calorimeter)

• Heat Capacity – amount of heat needed to raise an object’s temperature by 1°C or 1 K.

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Constant Pressure Calorimetry

m = mass of solutionC = heat capacity of the

calorimeter (J/g-⁰C) (or K)∆T = the change in temperature

in ⁰C or K

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Constant Pressure CalorimetryPractice 1

• When a 9.55 g sample of solid NaOH dissolves in 100.0g of water in a coffee-cup calorimeter, the temperature rises from 23.6˚C to 47.4˚C. Calculate the ∆H (in kJ/mol NaOH) for the solution process

NaOH(s) Na+(aq) + OH-(aq)

• Assume the specific heat of the solution is the same as that for water (4.184 J/g-˚C)

• Hint: what is system, what are surroundings?

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Constant Pressure CalorimetryPractice 2

• A 150.0 g sample of a metal at 75.0°C is added to 150.0 g H2O at 15.0°C. The temperature of the water rises to 18.3°C. Calculate the specific heat capacity of the metal.

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Measuring Heats of ReactionConstant Volume CALORIMETRYConstant Volume CALORIMETRY 54


Measuring Heats of ReactionCALORIMETRYCALORIMETRY

Constant Volume “Bomb” Calorimeter

• Burn combustible sample.

• Measure heat evolved in a reaction.

• Derive ∆E for reaction.


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CalorimetrySome heat from reaction warms waterqwater = (sp. ht.)(water mass)(∆T)

Some heat from reaction warms “bomb”qbomb = (heat capacity, J/K)(∆T)

Total heat evolved = qtotal = qwater + qbomb

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Calculate energy of combustion (∆E) of octane. 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

Measuring Heats of ReactionCALORIMETRY - Example

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Step 1 Calc. energy transferred from reaction to water.

q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J

Step 2 Calc. energy transferred from reaction to bomb.

q = (bomb heat capacity)(∆T)

= (837 J/K)(8.20 K) = 6860 J

Step 3 Total energy evolved

41,200 J + 6860 J = 48,060 J

Energy of combustion (∆E) of 1.00 g of octane

= - 48.1 kJ

Measuring Heats of ReactionCALORIMETRY - Example

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A 1.00 g sample of sucrose (C12H22O11) is burned in a bomb calorimeter. The temperature of 1500. g of water in the calorimeter rises from 25.00˚C to 27.32˚C. The heat capacity of the bomb is 837 J/K and the specific heat capacity of water is 4.18 J/g-K. Calculate the heat evolved

a)Per gram of sucrose

b)Per mole of sucrose (M = 343 g/mol)

‘BOMB’ CALORIMETRY PRACTICE

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Making H2O from H2 involves two steps.

H2(g) + 1/2 O2(g) H2O(g) ∆rH˚ =-242 kJ

H2O(g) H2O(liq) ∆rH˚ =-44 kJ-----------------------------------------------------------------------

H2(g) + 1/2 O2(g) H2O(liq) ∆rH˚ =-286 kJ

Example of HESS’S LAW—

If a rxn. is the sum of 2 or more others, the net ΔH is the sum of the ΔH’s of the other rxns.

Ch. 6.3 – Hess’s Law 60

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Hess’s Law


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Hess’s Law & Energy Level DiagramsForming H2O can occur

in a single step or in a two steps. ∆rHtotal is the same no matter which path is followed.

Active Figure 5.16

NOTE: ∆rH stands for the enthalpy change for a

reaction, r

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Hess’s Law & Energy Level Diagrams

Forming CO2 can occur in a single step or in a two steps. ∆rHtotal is the same no matter which path is followed.

Active Figure 5.16

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Enthalpy and Hess’s Law• If a reaction is reversed, the sign of ∆H is also

reversed.

Xe(g) + 2F2(g) XeF4(s) ∆H=-251 kJ

XeF4(s) Xe(g) + 2F2(g) ∆H = +251 kJ

• ∆H is also proportional to the quantities of the reactants and product (mol-rxn). If you multiply the coefficients by an integer, the ∆H value is multiplied by the same integer.

2Xe(g) + 4F2(g) 2XeF4(s) ∆H =-502 kJ

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Simple Hess’s Law problem

• C(graphite) + O2(g) → CO2(g)

∆H° = –393.51 kJ mol–1

• C(diamond) + O2(g) → CO2(g)

∆H° = –395.40 kJ mol–1

Calculate the ∆H for the conversion of graphite to diamond.

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Simple Hess’s Law problem

• C(graphite) + O2(g) → CO2(g)

∆H° = –393.51 kJ mol–1

• CO2(g) → C(diamond) + O2(g)

∆H° = +395.40 kJ mol–1

Add rxns together, cancel out substances that appear on both sides and get:

C(graphite) C(diamond) ∆H = 1.9 kJ

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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 disposal.

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

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

Copyright © Cengage Learning. All rights reserved 66


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Given:

C(s) + O2(g) CO2(g) ∆H= -393.5kJ/mol

S(s) + O2(g) SO2(g) ∆H= -296.8 kJ/mol

CS2(l) + 3O2(g) CO2(g) + 2SO2(g)

∆H = -1103.9 kJ/mol

Use Hess’s Law to calculate enthalpy change for the formation of CS2(l) from C(s) and S(s)

C(s) + 2S(s) CS2(l)

Harder Example of Hess’s Law68


Multiply equations as needed:

C(s) + O2(g) CO2(g) ∆H= -393.5kJ/mol

2S(s) + 2O2(g) 2SO2(g) ∆H=2*(-296.8 kJ/mol)

=-593.6 kJ

CO2(g) + 2SO2(g) CS2(l) + 3O2(g)

∆H = 1103.9 kJ/mol

C(s) + 2S(s) CS2(l) ∆H = 116.8 kJ

Reaction 1 was used as is

Reaction 2 was multiplied by 2 (SO2 cancels)

Reaction 3 was multiplied by -1 (Get CS2 on

product side)

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• This equation is valid because ∆H is a STATE FUNCTION

• These depend only on the state of the system and not how it got there.

• V, T, P, energy — and your bank account!

• Unlike V, T, and P, one cannot measure absolute H. Can only measure ∆H.

∆H along one path =∆H along another path

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Standard Enthalpy ValuesStandard Enthalpy ValuesMost ∆H values are labeled ∆Ho

Measured under standard conditions:• For a Compound

– For a gas, pressure is exactly 1 atm.

– 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 25⁰C.

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Ch. 6.4 – Standard Enthalpies of Formation

NIST (Nat’l Institute for Standards and Technology) gives values of

ΔHfo = standard molar enthalpy of

formation— the enthalpy change when 1 mol of

compound is formed from elements under standard conditions.

See Appendix 4 (A19 – A22)

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ΔHfo, standard molar

enthalpy of formation

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

∆Hfo (H2O, g) = -241.8 kJ/mol

By definition,

∆Hfo = 0 for elements in their

standard states.


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Enthalpy of Formation Practice• Show the ∆Hf˚ reaction for the

following compounds:

a) C2H2(g)

b) KCl(s)

• Why is the following not a ∆Hf˚?

4Na(s) + O2(g) 2Na2O(s)

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Hess’s Law and ΔHf° Practice Given:

C2H2(g) + 5/2 O2(g) 2CO2(g) + H2O(l)

∆Hr° = -1300. kJ/mol-rxn

C(s) + O2(g) CO2(g) ∆Hr° = -394. kJ/mol-rxn

H2(g) + ½O2(g) H2O(l) ∆Hr° = -286. kJ/mol-rxn

Determine the ∆Hf° for C2H2 (acetylene).

First, determine the overall reaction. Manipulate above reactions to obtain overall reaction.

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Hess’s Law and ΔHf° Practice Answer:

-1( )C2H2(g) + 5/2 O2(g) 2CO2(g) + H2O(l)

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Using Standard Enthalpy ValuesUsing Standard Enthalpy Values

Use ∆H˚’s to calculate enthalpy change for

H2O(g) + C(graphite) H2(g) + CO(g)

(product is called “water gas”)

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H2O(g) + C(graphite) H2(g) + CO(g)

From reference books we find

• H2(g) + 1/2 O2(g) H2O(g)

∆fH˚ of H2O vapor = - 242 kJ/mol

• C(s) + 1/2 O2(g) CO(g)

∆fH˚ of CO = - 111 kJ/mol

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H2O(g) H2(g) + 1/2 O2(g) ∆Hro (= –∆Hr˚) = +242 kJ

C(s) + 1/2 O2(g) CO(g) ∆Hro = -111 kJ

------------------------------------------------------------------------------------------------------

To convert 1 mol of water to 1 mol each of H2

and CO requires 131 kJ of energy. The “water gas” reaction is ENDOthermic.

H2O(g) + C(graphite) H2(g) + CO(g)∆Hr

onet = +131 kJ


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In general, when ALLenthalpies of formation are known,

Calculate ∆H of reaction?

∆Hro = ∆fHf

o (products) - ∆Hf

o (reactants)

Remember that ∆ always = final – initial

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Heats of Reactions(ΔH⁰(r or rxn) = generic reaction)

• ∆H⁰combustion or ∆H⁰comb is amount of heat release when 1 mol of a substance is combusted (burned).

Ex: CH4 + 2O2 CO2 + 2H2O(g) ∆H = -890 kJ/mol

• ∆H⁰neutralization is the amount of heat released or absorbed when an acid is neutralized by a base

Ex: HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

∆Hneut =-57 kJ/mol-rxn

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Heats of (Physical) Process (ΔH⁰process = generic process)

• ∆H⁰solution is the amount of heat released or absorbed when a solid compound dissolves and becomes aqueous.

Ex: NaOH(s) Na+(aq) + OH-(aq) ∆H⁰sol = -43 kJ/mol

• ∆H⁰fusion is the amount of heat needed to change 1 mole of a solid substance to a liquid, i.e. melting.

Ex: H2O(s) H2O(l) ∆H ⁰fus= 6.0 kJ/mol

• ∆H⁰vaporization is the amount of heat needed to change 1 mole of a liquid to a vapor. Can use opposite too.

Ex: H2O(l) H2O(g) ∆H ⁰vap= 40.7 kJ/mol

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Calculate the heat of combustion of methanol, i.e., ∆Ho

comb for

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

∆Hocomb = ∆Hf

o (prod) - ∆Hfo (react)

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∆Hocomb = ∆Hf

o (CO2) + 2 ∆Hfo (H2O)

- {3/2 ∆Hfo (O2) + ∆Hf

o (CH3OH)}

= (-393.5 kJ) + 2 (-241.8 kJ)

- {0 + (-201.5 kJ)}

∆Hocomb = -675.6 kJ per mol of methanol

CH3OH(g) + 3/2 O2(g) CO2(g) + 2 H2O(g)∆Ho

comb = ∆Hfo (prod) - ∆Hf

o (react)

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Using Enthalpies of Formation calculate the Enthalpy of Combustion for the reaction below.

C3H8(g) + 5O2(g)→ 3CO2(g) + 4H2O(l)

∆H° = 3∆H°(CO2) + 4 ∆H°(H2O(l))–[∆H°(C3H8) + 5∆H°f(O2)]

Hocomb = 3 (-393.5) + 4 (-285.8) – [-103.8 + 5(0)]

H˚comb= -2220 kJ

Standard Enthalpy Values Example


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∆Hrxn = ∆H1 + ∆H2 + ∆H3

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Using Enthalpy of Reaction

Think of reaction like this:

3C(s) + 4H2(g) 5O2 3C + 3O2 4H2+2O2

C3H8 + 5O2 3CO2 + 4H2O

∆rH = -∆Hf1 – ∆Hf2 + 3∆Hf3 + 4∆Hf4

= +3∆Hf3 + 4∆Hf4 – (∆Hf1 + ∆Hf2)

-∆Hf1° -∆Hf2° 3∆Hf3° 4∆Hf4°

Formation of elements Regroup and form new substances

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Enthalpy Practice Problem

Calculate the standard enthalpy change for the reaction below using the standard enthalpies of formation.

SiCl4(l) + 2H2O(l) SiO2(s) + 4HCl(aq)

∆H°f’s:

Thermochemistry - Chapter 6 Rev 8 - Student Version · Chapter 6 — Thermochemistry 1 © 2009...

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