Chapter 6

Thermochemistry: Energy Flow and Chemical Change

If you are doing this lecture “online” then print the lecture notes available as a word document, go through this ppt lecture, and do all the example and practice assignments for discussion time.

Review of PhysicsReview of Physics

Velocity and acceleration: v = d/t, a = Velocity and acceleration: v = d/t, a = v/tv/tForce and work: force is what is required to Force and work: force is what is required to

overcome inertia or to change the velocity overcome inertia or to change the velocity of an object; Newton = kg*m/sof an object; Newton = kg*m/s22

Work = force acting through distance,Work = force acting through distance,w = f*d in w = f*d in Joules or kg*mJoules or kg*m22/s/s22

Energy is capacity to do work or produce heat, Energy is capacity to do work or produce heat, unit is Joules unit is Joules Potential (PE) and nonpotential energy (nPE)Potential (PE) and nonpotential energy (nPE)

Heat: energy transferred as result of Heat: energy transferred as result of temperature difference; transfer temperature difference; transfer alwaysalways from from warmerwarmer body to body to coldercolder body body

LEARNING OBJECTIVES:LEARNING OBJECTIVES:

1. Distinguish between heat & work; 1. Distinguish between heat & work; temperature, thermal energy & heattemperature, thermal energy & heat

2. Relate q, 2. Relate q, T, cT, cpp, thru heat transfer equations , thru heat transfer equations q = m * cq = m * cpp * * T and q = T and q = HHphasephase * amount * amount

3. Calculate 3. Calculate HHrxnrxn from calorimeter data from calorimeter data4. Understand how 4. Understand how HHff

oo values are established values are established for compoundsfor compounds

5. Calculate 5. Calculate HHrxnrxn from either heats of from either heats of formation or Hess' lawformation or Hess' law

6. Understand all the symbols:6. Understand all the symbols: HHrxnrxn

oo HHffo o HHvapvap

oo HHfusfusoo HHcombcomb

oo

Figure 6.1 from 4th ed, not in Prin

A chemical system and its surroundings.

the system

the surroundings

The system is the chemical reaction being studied.The surroundings are everything else in the universe!

DefinitionsDefinitionsThermodynamics is the study of heat and its transformations

Thermochemistry is a branch of thermodynamics that deals with the heat involved with chemical and physical changes

Fundamental premiseWhen energy is transferred from one object to another, it appears as work and/or as heat

For our work we must define a system to study; everything else then becomes the surroundings

The system is composed of particles with their own internal energies (E or U).

Therefore the system has an internal energy. When a change occurs, the

internal energy changes

MORE ON ENERGYMORE ON ENERGY

First law of thermodynamicsFirst law of thermodynamics or the Law or the Law of Conservation of Energy: of Conservation of Energy:

In an isolated system energy can't be In an isolated system energy can't be created or destroyed, so the total energy created or destroyed, so the total energy can't changecan't change

Energy is in two major forms - nonpotential Energy is in two major forms - nonpotential (often kinetic) and potential(often kinetic) and potential

Energy can be transferred between the two Energy can be transferred between the two forms, but the total, nPE + PE, has to forms, but the total, nPE + PE, has to remain the sameremain the same

ENERGY CONTINUED:ENERGY CONTINUED:

1. POTENTIAL ENERGY - relative to position (stored)1. POTENTIAL ENERGY - relative to position (stored)2. TYPES OF NONPOTENTIAL ENERGY:2. TYPES OF NONPOTENTIAL ENERGY:

KINETIC ENERGY - MOVING OBJECT = ½mvKINETIC ENERGY - MOVING OBJECT = ½mv22

HEAT - CHANGING TEMPERATURE – FROM HEAT - CHANGING TEMPERATURE – FROM ENERGY TRANSER ENERGY TRANSER Thermal energy from the motion of atoms or Thermal energy from the motion of atoms or molecules in s, l or gmolecules in s, l or gHeat involves the transfer of energy between two Heat involves the transfer of energy between two objects due to a temperature differenceobjects due to a temperature differenceOTHER FORMS OF nPE: OTHER FORMS OF nPE: Radiant energy from the sunRadiant energy from the sunElectrical energy generated by mechanical Electrical energy generated by mechanical devices or by redox reactionsdevices or by redox reactions

ENERGY CONTINUED:ENERGY CONTINUED:

Chemical reactions either release or Chemical reactions either release or absorb energy as: absorb energy as:

heat - thermal energy**heat - thermal energy**

light - radiant energylight - radiant energy

electrical current - electrical energyelectrical current - electrical energy

What does high thermal energy What does high thermal energy mean? mean?

One relative measure of thermal energy One relative measure of thermal energy is temperature, but temperature does is temperature, but temperature does not equal thermal energy not equal thermal energy

Compare a cup of coffee at 102°F to a Compare a cup of coffee at 102°F to a tub of water at 102°F tub of water at 102°F

Which contains more thermal energy? Which contains more thermal energy? WHY? WHY?

MORE THERMOCHEMISTRY:MORE THERMOCHEMISTRY:

SYSTEM = the substance being evaluated for SYSTEM = the substance being evaluated for energy content in a thermodynamic processenergy content in a thermodynamic process

SURROUNDINGS = everything outside the SURROUNDINGS = everything outside the system in the processsystem in the process

ENDOTHERMIC PROCESS = a thermodynamic ENDOTHERMIC PROCESS = a thermodynamic process in which energy transfers into a process in which energy transfers into a system from its surroundingssystem from its surroundings

EXOTHERMIC PROCESS = from system to EXOTHERMIC PROCESS = from system to surroundingssurroundings

ENTHALPY CHANGE = energy transferred at a ENTHALPY CHANGE = energy transferred at a constant pressure, constant pressure, HH

"A bit of physics again" "A bit of physics again"

There are 2 ways to transfer energy between There are 2 ways to transfer energy between system and surroundings: work or heatsystem and surroundings: work or heat

Work = action of a force that causes Work = action of a force that causes displacement within the system in a displacement within the system in a measurable dimension as “-measurable dimension as “-X” if linear or as X” if linear or as ““V” if volumeV” if volume

Work = -F*Work = -F*h h Since Area*Since Area*h = h = V and Pressure = F/AreaV and Pressure = F/AreaThen w = - (F/Area) * Area*Then w = - (F/Area) * Area*h = - Ph = - PV V for work done at constant pressurefor work done at constant pressure

Figure 6.5

Pressure-volume work.

When the volume of a system increases by an amount V against an external pressure, P, they system pushes back and does PV work on the surr (w=-PV).

Heat Transfer and the Meaning of Heat Transfer and the Meaning of Enthalpy Change, Enthalpy Change, HH

As previously discussed, if no force is involved, then As previously discussed, if no force is involved, then energy flows spontaneously as heat from regions of energy flows spontaneously as heat from regions of higher T to regions of lower T. Most rxns take place at higher T to regions of lower T. Most rxns take place at constant pressure. We also know constant pressure. We also know EErxnrxn = E = Eff - E - Eii = q + w, = q + w, and that w = -Pand that w = -PV.V.

At constant P,At constant P, EErxnrxn = q + w = q + w

EErxnrxn = q = qPP - P - PVV

At constant P, At constant P, HHrxnrxn = = EErxnrxn + P + PV V

substituting:substituting: HHrxnrxn = (q = (qpp - P - PV) + PV) + PV V

HHrxnrxn = q = qpp

Therefore, Enthalpy of Rxn = Heat transfer Therefore, Enthalpy of Rxn = Heat transfer HHrxnrxn = q = qpp

Heat Transfer and the Meaning Heat Transfer and the Meaning of Enthalpy Change, of Enthalpy Change, HH

If If H < 0 rxn is exothermic, if >0 rxn is H < 0 rxn is exothermic, if >0 rxn is endothermicendothermic

For chemical reactions: For chemical reactions:

HHff = final enthalpy of products = final enthalpy of products

HHii = initial enthalpy of reactants = initial enthalpy of reactants

HHrxnrxn = H = Hproductsproducts - H - Hreactantsreactants

E = Efinal - Einitial = Eproducts - Ereactants

Figure 6.1 Energy diagrams for the transfer of internal energy (E) between a system and its surroundings.

When internal energy of a system decreases, E is lost to the surroundings and is negative.

When internal energy of a system increases, E is gained by the surroundings and is positive.

Figure 6.2 A system transferring energy as heat only.

Hot water transfers energy as heat, q, to the surr until Tsys=Tsurr; E<0 so q is neg.

Ice water gains energy as heat, q, from surr until Tsys = Tsurr; E>0 so q is pos.

See notes in box below or in textbook.

Figure 6.3 A system losing energy as work only.

Energy, E

Zn(s) + 2H+(aq) + 2Cl-(aq)

H2(g) + Zn2+(aq) + 2Cl-(aq)

E<0work done onsurroundings

The internal energy of the system decreases as the reactants form products because the H2(g) does work, w, on the surr by pushing back the piston. The vessel is insulated, so q = 0. Here E<0 and sign of w is positive.

Table 6.1 The Sign Conventions* for q, w and E

(from perspective of the system)

q w+ = E

+

+

-

-

-

-

+

+

+

-

depends on sizes of q and w

depends on sizes of q and w

* For q: + means system gains heat; - means system loses heat.

* For w: + means work done on system; - means work done by system.

Euniverse = Esystem + Esurroundings

Units of Energy

Joule (J)

Calorie (cal)

= energy to heat exactly 1 g H2O by exactly 1.00°C at 14.5°C

(British Thermal Unit

1 cal = 4.184 J

1 kcal = 1000 cal = 1 Cal (nutritional)

1 J = 1 kg*m2/s2

1 Btu = 1055 J)

Figure 6.5 in 4th ed. only

Some interesting quantities of energy.

Sample Problem 6.1 Determining the Change in Internal Energy of a System

PROBLEM: When gasoline burns in a car engine, the heat released causes the products CO2 and H2O to expand, which pushes the pistons outward. Excess heat is removed by the car’s cooling system. If the expanding gases do 451 J of work on the pistons and the system loses 325 J to the surroundings as heat, calculate the change in energy (E) in J, kJ, and kcal.

SOLUTION:

PLAN: Define system and surroundings, assign signs to q and w and calculate E. The answer should be converted from J to kJ and then to kcal.

q = - 325 J w = - 451 J

E = q + w = -325 J + (-451 J) = -776 J

-776J103J

kJ= -0.776kJ -0.776kJ

4.184kJ

kcal= -0.185 kcal

Figure 6.4 Two different paths for the energy change of a system.

The change in internal energy when a given amount of octane burns in air is the same no matter how the energy is transferred. On left, fuel is burned in open can and energy is lost almost entirely as heat. On right, fuel is burned in car engine so a portion of energy is lost as work to move the car and less energy is lost as heat.

Figure 6.5

Pressure-volume work.

When the volume of a system increases by an amount V against an external pressure, P, they system pushes back and does PV work on the surr (w=-PV).

Figure 6.6

Enthalpy diagrams for exothermic and endothermic processes.

Ent

halp

y, H

Ent

halp

y, H

CH4 + 2O2

CO2 + 2H2O

Hinitial

HinitialHfinal

Hfinal

H2O(l)

H2O(g)

heat out heat inH < 0 H > 0

A Exothermic process B Endothermic process

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

Sample Problem 6.2 Drawing Enthalpy Diagrams and Determining the Sign of H

PROBLEM: In each of the following cases, determine the sign of H, state whether the reaction is exothermic or endothermic, and draw an enthalpy diagram. These are thermochemical equations!

SOLUTION:

PLAN: Determine whether heat is a reactant or a product. As a reactant, the products are at a higher energy and the reaction is endothermic. The opposite is true for an exothermic reaction

(a) H2(g) + 1/2O2(g) H2O(l) + 285.8 kJ (H = -285.8 kJ)

(b) 40.7 kJ + H2O(l) H2O(g) (or H = +40.7 kJ

(a) The reaction is exothermic.

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

H2O(l) (products)

EXOTHERMIC

(products)H2O(g)

(reactants)H2O(l)

H = -285.8kJ H = +40.7kJENDOTHERMIC

(b) The process is endothermic.

Phase Changes: Latent HeatPhase Changes: Latent Heat

THREE MAJOR EXPRESSIONS FOR CHANGE IN THREE MAJOR EXPRESSIONS FOR CHANGE IN STATE: STATE:

HHfusfus for change from solid to liquid (= - for change from solid to liquid (= -HHfreezefreeze from liquid to solid)from liquid to solid)

HHvapvap for change from liquid to gas (= - for change from liquid to gas (= -HHcondcond from gas to liquid)from gas to liquid)

HHsublsubl for change from solid to gas (= - for change from solid to gas (= -HHdepdep from from gas to solid)gas to solid)

There is There is no temperature changeno temperature change, so we call it , so we call it a a latent heat transferlatent heat transfer::

q = q = HHphasephase * quantity * quantity

Enthalpy of Phase ChangeEnthalpy of Phase Change

SubstanceSubstance HHfusfus HHvapvap

Benzene, CBenzene, C66HH66 126 J/g126 J/g 395 J/g395 J/g

Bromine, BrBromine, Br22 67.8 J/g67.8 J/g 187 J/g187 J/g

Ethanol, CHEthanol, CH33CHCH22OHOH 104 J/g104 J/g 854 J/g854 J/g

Mercury, HgMercury, Hg 11.6 J/g11.6 J/g 296 J/g296 J/g

Sodium ChlorideSodium Chloride 517 J/g517 J/g 3100 J/g3100 J/g

WaterWater 333 J/g333 J/g 2257 J/g2257 J/g

Practice heat transferPractice heat transfer

1. For 237 mL of ice at 0.001. For 237 mL of ice at 0.00ooC find the heat C find the heat required to melt it and then the heat required required to melt it and then the heat required to warm it to 25.0to warm it to 25.0ooC. Finally find the total of C. Finally find the total of the two heat transfers. Dthe two heat transfers. Diceice is 0.917 g/mL. is 0.917 g/mL.

qqphasephase = = HHfusfus * quantity * quantity

= 333 J/g * 237 mL * 0.917g/mL= 333 J/g * 237 mL * 0.917g/mL

= 72,371 J= 72,371 J

qqwarmwarm = m*c = m*cpp**T = 237 mL * 0.917 g/mL T = 237 mL * 0.917 g/mL *4.184J/g*4.184J/gooC*(25.0-0.00)C*(25.0-0.00)ooC = 22,730 JC = 22,730 J

qqtotaltotal = 9.51x10 = 9.51x104 4 JJ

Practice heat transfer now you Practice heat transfer now you do it!do it!

2. Find the total heat transfer energy 2. Find the total heat transfer energy required to heat 1.00 mole of ice from required to heat 1.00 mole of ice from below the freezing point at -10.0below the freezing point at -10.0ooC to C to above the boiling point as steam at above the boiling point as steam at 110.0110.0ooC. (That includes two phase C. (That includes two phase changes!) Given this data: cchanges!) Given this data: cp(ice)p(ice) = = 37.7 J/mol37.7 J/molooC; cC; cp(water)p(water) = 75.3 J/mol = 75.3 J/molooC; C; ccp(steam)p(steam) = 36.4 J/mol = 36.4 J/molooC; C; HHfusfus = 333 J/g; = 333 J/g; HHvapvap = 2257 J/g. (Total q ~ 55000 J) = 2257 J/g. (Total q ~ 55000 J)

TO BEGIN TO UNDERSTAND TO BEGIN TO UNDERSTAND CALORIMETRY:CALORIMETRY:

Heat capacity, cHeat capacity, cpp,, of a substance = quantity of heat of a substance = quantity of heat that can be transferred through the substance per that can be transferred through the substance per quantity/per change in temperaturequantity/per change in temperature

CCpp = = quantity of heat transferredquantity of heat transferredT * gram or moleT * gram or mole

Specific heat capacitySpecific heat capacity is the heat energy required is the heat energy required to warm up 1.00 g of a substance by 1.00°Cto warm up 1.00 g of a substance by 1.00°C

Ex: Water’s specific heat is 4.184 J/g-degreeEx: Water’s specific heat is 4.184 J/g-degree Glass’ specific heat is 0.80 J/g-degreeGlass’ specific heat is 0.80 J/g-degree

TO BEGIN TO UNDERSTAND TO BEGIN TO UNDERSTAND CALORIMETRY:CALORIMETRY:

Specific heat capacity CSpecific heat capacity Cpp is a physical is a physical property of a substanceproperty of a substance

CCpp can also be expressed as molar can also be expressed as molar heat capacity by multiplying by heat capacity by multiplying by molar mass molar mass

For example, for water: For example, for water: 4.184 J/g-4.184 J/g-ooC * 18.015 g/mol =75.37 J/mol-C * 18.015 g/mol =75.37 J/mol-ooCC

Thermal energyThermal energy

Thermal energy transferred is called sensible Thermal energy transferred is called sensible heat transfer, q,heat transfer, q, uses specific heat uses specific heat capacity in this heat transfer equation:capacity in this heat transfer equation:

q = m*cq = m*cpp*(T*(Tff - T - Tii))The direction of heat transfer is determined The direction of heat transfer is determined by which is warmer body: heat flows from by which is warmer body: heat flows from hot to coldhot to coldHeat has to be transferred from something to Heat has to be transferred from something to something. The warmer body loses heat; the something. The warmer body loses heat; the colder body gains heat.colder body gains heat.

qqlostlost = -q = -qgainedgained

Thermal energy exampleThermal energy example

You drink 1 cup of coffee, 250. mL, at 60.0°C & your body is You drink 1 cup of coffee, 250. mL, at 60.0°C & your body is at 37.0°C. Assume 0.997 g/mL density, and water’s cat 37.0°C. Assume 0.997 g/mL density, and water’s cpp can can be used for both coffee and your body with 60.0 kg mass. be used for both coffee and your body with 60.0 kg mass. Heat is transferred from the coffee to your body. What is Heat is transferred from the coffee to your body. What is your body’s final temperature?your body’s final temperature?

qq(coffee)(coffee)= 249.25 g * 4.184 J/g-degree (T= 249.25 g * 4.184 J/g-degree (Tff-60.0) -60.0)

qq(body)(body)= 60.0 kg (10= 60.0 kg (1033 g/kg) 4.184 J/g-degree (T g/kg) 4.184 J/g-degree (Tff – 37.0) – 37.0)

Set qSet q(coffee) (coffee) = = --qq(body)(body)

249.25 g * 4.184 J/g-degree (T249.25 g * 4.184 J/g-degree (Tff-60.0) = -60.0) =

--[60.0 kg (10[60.0 kg (1033 g/kg) 4.184 J/g-degree (T g/kg) 4.184 J/g-degree (Tff – 37.0) – 37.0)

Solve for TSolve for Tff

TTff = 37.1°C in sig figs = 37.1°C in sig figs

Table 6.2 Specific Heat Capacities of Some Elements, Compounds, and Materials

Specific Heat Capacity (J/g*K)

SubstanceSpecific Heat Capacity (J/g*K)

Substance

Compounds

water, H2O(l)

ethanol, C2H5OH(l)

ethylene glycol, (CH2OH)2(l)

carbon tetrachloride, CCl4(l)

4.184

2.46

2.42

0.864

Elements

aluminum, Al

graphite,C

iron, Fe

copper, Cu

gold, Au

0.900

0.711

0.450

0.387

0.129

wood

cement

glass

granite

steel

Materials

1.76

0.88

0.84

0.79

0.45

Figure 6.7

Coffee-cup calorimeter.

Still used in school labs to measure heat at constant pressure, qp. Surroundings actually include water, cups, thermometer, and stirrer.

Sample Problem 6.4 Determining the Heat of a Reaction

PROBLEM: You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at 25.000C and carefully add 25.0 mL of 0.500 M HCl, also at 25.000C. After stirring, the final temperature is 27.210C. Calculate qsoln (in J) and Hrxn (in kJ/mol). (Assume the total volume is the sum of the individual volumes and that the final solution has the same density and specfic heat capacity as water: d = 1.00 g/mL and c = 4.184 J/g*K)

PLAN: Find mass of solution: m = 75.0 mL * 1.00 g/mL = 75.0 g

Find T = Tf – Ti = 27.21oC – 25.00oC = 2.21oC

qsol’n = m*cp*T = 75.0 g * 4.184 J/g-deg * 2.21oC = 693 J

Find Hrxn = - qsol’n = -693 J (opposite of q from previous slide)

(-693 J/0.0125 mol H2O)(kJ/103 J) = -55.4 kJ/ mol H2O formed

SOLUTION:

For NaOH 0.500 M x 0.0500 L = 0.0250 mol OH-

For HCl 0.500 M x 0.0250 L = 0.0125 mol H+

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

H+(aq) + OH-(aq) H2O(l)

HCl is the limiting reactant.

0.0125 mol of H2O will form during the rxn.

Sample Problem 6.4 Determining the Heat of a Reaction

continued

CALORIMETRY:CALORIMETRY:

The bomb calorimeter is used for combustible substances that The bomb calorimeter is used for combustible substances that are solid or liquidare solid or liquid

SYSTEM = SUBSTANCE + OXYGEN SYSTEM = SUBSTANCE + OXYGEN HHcalorcalor = q = qbombbomb + q + qwaterwater

SURROUNDINGS = BOMB + WATER SURROUNDINGS = BOMB + WATER HHrxnrxn = - = - HHcalorcalor

There is no There is no V, therefore qV, therefore qrxnrxn = -(q = -(qbombbomb + q + qwaterwater))

Note: bomb and water have different heat capacities even if Note: bomb and water have different heat capacities even if change in temperature is the same. Must know or change in temperature is the same. Must know or determine heat capacity of calorimeterdetermine heat capacity of calorimeter

Figure 6.8

A bomb calorimeter

Combustion of octane in bomb Combustion of octane in bomb calorimeter:calorimeter:

1. Balanced eq'n*: C1. Balanced eq'n*: C88HH18(l)18(l) + 25/2 O + 25/2 O2(g)2(g) 8 CO 8 CO2(g)2(g) + 9 H + 9 H22OO(g)(g)

Given this data: 1.00 g of octane, 1.20 kg water, Given this data: 1.00 g of octane, 1.20 kg water, initial T of bomb + water is 25.00°Cinitial T of bomb + water is 25.00°Cfinal T is 33.20°Cfinal T is 33.20°CHeat capacity of water is 4.184 J/g-degHeat capacity of water is 4.184 J/g-degHeat capacity of calorimeter is determined to be 837 J/degHeat capacity of calorimeter is determined to be 837 J/deg

2. Calculations2. Calculationsqqbombbomb = C = Cbombbomb * * T = 837 J/deg * (33.20 - 25.00) = +6.86 x 10T = 837 J/deg * (33.20 - 25.00) = +6.86 x 1033 J Jqqwaterwater = C = Cww * m * * m * T = 4.184 J/g-deg * 1200 g * (33.20 - 25.00)T = 4.184 J/g-deg * 1200 g * (33.20 - 25.00)

= +41.2 x 10= +41.2 x 1033 J Jtotal q = -(qtotal q = -(qbombbomb + q + qww) = -(6.86 + 41.2) kJ) = -(6.86 + 41.2) kJ

= -48.1 kJ/gram octane= -48.1 kJ/gram octane3. Find molar 3. Find molar H: -48.1 kJ/gram * 114.2 g/mol = -5490 kJ/molH: -48.1 kJ/gram * 114.2 g/mol = -5490 kJ/mol

Note: in thermochemistry the substance of interest has a coefficient Note: in thermochemistry the substance of interest has a coefficient of one, and other coefficients can be fractions!of one, and other coefficients can be fractions!

PracticePractice

Work on problems 23, 25 and 32.Work on problems 23, 25 and 32.

23: Calc q when 0.10 g ice is cooled from 10.023: Calc q when 0.10 g ice is cooled from 10.0ooC to -C to -75.075.0ooC, given cC, given cpp of ice is 2.087 J/g-degree. of ice is 2.087 J/g-degree.

25: A 27.7 g sample of ethylene glycol loses 688 J of 25: A 27.7 g sample of ethylene glycol loses 688 J of heat. Given Theat. Given Tff is 32.5 is 32.5ooC and cC and cpp of 2.42 J/g-degree, of 2.42 J/g-degree, determine Tdetermine Tii..

32: 32: When 25.0 mL of 0.500 M HWhen 25.0 mL of 0.500 M H22SOSO44 is added to 25.0 mL of is added to 25.0 mL of 1.00M KOH in a calorimeter at 23.50°C, the temperature 1.00M KOH in a calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate rises to 30.17°C. Calculate H of this reaction. (Assume H of this reaction. (Assume that the total volume is the sum of the individual volumes that the total volume is the sum of the individual volumes and that the density and specific heat capacity of the and that the density and specific heat capacity of the solution are the same as for pure water.)solution are the same as for pure water.)

AMOUNT (mol)

of compound A

AMOUNT (mol)

of compound B

HEAT (kJ)

gained or lost

molar ratio from balanced equation

Hrxn (kJ/mol)

Figure 6.9

Summary of the relationship between amount (mol) of substance and the heat (kJ) transferred during a reaction.

H is function of quantity of reactants H is function of quantity of reactants based on mass or moles based on mass or moles

Burning 441 g propane in backyard grill, how Burning 441 g propane in backyard grill, how much heat is transferred to surroundings, much heat is transferred to surroundings, including meat, air, grill, etc.?including meat, air, grill, etc.?

1. Always have a balanced chemical equation: 1. Always have a balanced chemical equation: CC33HH8(g)8(g) + 5 O + 5 O2(g)2(g) 3 CO 3 CO2(g)2(g) + 4 H + 4 H22OO(l)(l)

2. Look up 2. Look up HHcombcomb : -2220 kJ/mol propane : -2220 kJ/mol propane3. How many moles consumed? 3. How many moles consumed?

441 g/44.09 g/mol = 10.0 mol441 g/44.09 g/mol = 10.0 mol4. How much heat?4. How much heat? -2220 kJ/mol * 10.0 mol = -22,200 kJ-2220 kJ/mol * 10.0 mol = -22,200 kJ

Writing a Thermochemical Writing a Thermochemical Equation and an exampleEquation and an example

A thermochemical equation is a A thermochemical equation is a balanced chemical equation that balanced chemical equation that includes the enthalpy of reaction. includes the enthalpy of reaction.

Intro to Hess’ LawIntro to Hess’ LawGiven the combustion reaction:Given the combustion reaction:

2 H2 H2(g)2(g) + O + O2(g)2(g) 2 H 2 H22OO(g)(g) HHrxn rxn = -484 kJ= -484 kJ

If we cut in half the reaction in half to make the If we cut in half the reaction in half to make the formation reaction of one mol of Hformation reaction of one mol of H22O:O:

HH2(g)2(g) + 1/2 O + 1/2 O2(g) 2(g) H H22OO(g) (g)

Then we have to cut the value of the first reaction Then we have to cut the value of the first reaction in half: in half: HHrxnrxn = - 242 kJ/mol H = - 242 kJ/mol H22O = O = HHff

If we do some multiple of a known reaction we If we do some multiple of a known reaction we have to apply the same multiplier to the have to apply the same multiplier to the HHrxnrxn

Intro to Hess’ LawIntro to Hess’ Law

Given the vaporization of liquid water to steam:Given the vaporization of liquid water to steam:HH22OO(l)(l) H H22OO(g)(g) HHvapvap = 44 kJ/mol = 44 kJ/mol

TThen what is hen what is HHcondcond ? ? HH22OO(g)(g) H H22OO(l)(l)

Reverse of vaporization is condensation:Reverse of vaporization is condensation:HHcondcond = -44 kJ/mol = - = -44 kJ/mol = -HHvapvap

If a reaction is reversed the sign of If a reaction is reversed the sign of H is changed:H is changed:HHrev rxnrev rxn = - = - HHfwd rxnfwd rxn

Intro to Hess’ LawIntro to Hess’ Law

We can use the formation of water and the We can use the formation of water and the condensation of steam data to determine condensation of steam data to determine for the combustion reaction if the product for the combustion reaction if the product is liquid water.is liquid water.

Rxn 1: Rxn 1: HH2(g)2(g) + 1/2 O + 1/2 O2(g) 2(g) H H22OO(g) (g) HHrxnrxn = - 242 kJ/mol = - 242 kJ/mol

Rxn 2: Rxn 2: HH22OO(g)(g) H H22OO(l)(l) HHcondcond = -44 kJ/mol = -44 kJ/mol

Add the two reactions to obtain this reaction:Add the two reactions to obtain this reaction: HH2(g)2(g) + ½ O + ½ O2(g)2(g) H H22OO(l)(l)

HHrxnrxn = -242kJ + -44 kJ = -286 kJ/mol H = -242kJ + -44 kJ = -286 kJ/mol H22OO(l)(l)

Intro to Hess’ LawIntro to Hess’ Law

HESS' LAW OF CONSTANT HEAT HESS' LAW OF CONSTANT HEAT SUMMATION:SUMMATION:If a reaction is the sum of 2 or more If a reaction is the sum of 2 or more other reactions, then other reactions, then H for overall H for overall process must be the process must be the sum sum of of H of H of constituent reactionsconstituent reactions

Hess’ Law ExampleHess’ Law Example

Find Find HHrxnrxn for 2 SO for 2 SO2(g)2(g) + O + O2(g)2(g) 2 SO2 SO3(g)3(g), , given the info for the following two rxns:given the info for the following two rxns:

SOSO2(g)2(g) S S(s)(s) + O + O2(g)2(g) HHrxnrxn = +297 kJ = +297 kJ

2 S2 S(s)(s) + 3 O + 3 O2(g)2(g) 2 SO 2 SO3(g)3(g) HHrxnrxn = -791 kJ = -791 kJ

Double rxn 1 and add to rxn 2, cancel Double rxn 1 and add to rxn 2, cancel common species. Sum of common species. Sum of H's is -197kJH's is -197kJ

For practice see problems 45-48!For practice see problems 45-48!

Enthalpy of Formation:Enthalpy of Formation:

HHoo is standard quantity: 1 atm is standard quantity: 1 atm pressure, 1.000 M, 298.15 Kpressure, 1.000 M, 298.15 K

HHffoo is standard enthalpy of formation is standard enthalpy of formation

of compound, ion, etc., at standard of compound, ion, etc., at standard conditions from the elements in their conditions from the elements in their standard states (recall diatomics, standard states (recall diatomics, etc.)etc.)

Elements by definition have Elements by definition have HHffoo = 0.0 = 0.0

Enthalpy of Formation:Enthalpy of Formation:

Another way to use Hess’ Law:Another way to use Hess’ Law:

HHoorxnrxn = = HHff

oo(products) - (products) - HHffoo(reactants)(reactants)

Example: Find Example: Find HHoorxn rxn for combustion:for combustion:

HH2(g)2(g) + ½ O + ½ O2(g)2(g) H H22OO(l)(l)

Use Reference to find Use Reference to find HHffoo for the reactants and for the reactants and

products.products.HHoo

rxnrxn = [ = [HHffoo(H(H22OO(l)(l))] - [)] - [HHff

oo(H(H2(g)2(g)) + ½ ) + ½ HHffoo(O(O2(g)2(g))])]

= [-285.8 kJ/mol * 1mol] - [0.0 kJ/mol * 1 mol = [-285.8 kJ/mol * 1mol] - [0.0 kJ/mol * 1 mol + 0.0 kJ/mol * ½ mol]+ 0.0 kJ/mol * ½ mol]

= -285.8 kJ= -285.8 kJ

Table 6.3 Selected Standard Heats of Formation at 250C(298K)

Formula H0f(kJ/mol)

calciumCa(s)CaO(s)CaCO3(s)

carbonC(graphite)C(diamond)CO(g)CO2(g)CH4(g)CH3OH(l)HCN(g)CSs(l)

chlorineCl(g)

0-635.1

-1206.9

01.9

-110.5-393.5-74.9

-238.6135

87.9

121.0

hydrogen

nitrogen

oxygen

Formula H0f(kJ/mol)

H(g)H2(g)

N2(g)NH3(g)NO(g)

O2(g)O3(g)H2O(g)

H2O(l)

Cl2(g)

HCl(g)

0

0

0

-92.30

218

-45.990.3

143-241.8

-285.8

107.8

Formula H0f(kJ/mol)

silverAg(s)AgCl(s)

sodium

Na(s)Na(g)NaCl(s)

sulfurS8(rhombic)S8(monoclinic)SO2(g)

SO3(g)

0

0

0

-127.0

-411.1

2-296.8

-396.0

ExampleExample

Find Find HHoorxnrxn for the decomposition of calcium carbonate to for the decomposition of calcium carbonate to

calcium oxide and carbon dioxide gas.calcium oxide and carbon dioxide gas.1. Write balanced chemical equation of formation and look 1. Write balanced chemical equation of formation and look

up heats of formation:up heats of formation: CaCOCaCO3(s)3(s) CaO CaO(s)(s) + CO + CO2(g)2(g)

CaCOCaCO3(s) 3(s) -1206.9 kJmol CaO-1206.9 kJmol CaO(s)(s) -635.1 kJ/mol CO -635.1 kJ/mol CO2(g)2(g) -393.5 kJ/mol -393.5 kJ/mol

2. Set up 2. Set up HHff in equation in equation

HHoorxnrxn = = HHff

oo(products) - (products) - HHffoo(reactants)(reactants)

[-635.1 kJ/mol*1mol + -393.5 kJ/mol*1mol] – (-1206.9 [-635.1 kJ/mol*1mol + -393.5 kJ/mol*1mol] – (-1206.9 kJ/mol*1mol) = +178.8 kJ/molkJ/mol*1mol) = +178.8 kJ/mol

3. Write the finished thermochemical equation:3. Write the finished thermochemical equation:

CaCOCaCO3(s)3(s) CaO CaO(s)(s) + CO + CO2(g)2(g) HHrxnrxn = + 178.8kJ/mol = + 178.8kJ/mol

For more practice, see problems 53-56.For more practice, see problems 53-56.

Sample Problem 6.8 Writing Formation Equations

PROBLEM: Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include H0

f.

SOLUTION:

PLAN:

(a) Silver chloride, AgCl, a solid at standard conditions.

(b) Calcium carbonate, CaCO3, a solid at standard conditions.

Use the table of heats of formation for values.

(c) Hydrogen cyanide, HCN, a gas at standard conditions.

H0f = -127.0 kJ(a) Ag(s) + 1/2Cl2(g) AgCl(s)

H0f = -1206.9 kJ(b) Ca(s) + C(graphite) + 3/2O2(g) CaCO3(s)

H0f = 135 kJ(c) 1/2H2(g) + C(graphite) + 1/2N2(g) HCN(g)

Figure 6.10

The general process for determining H0rxn from H0

f values.

Ent

halp

y, H

Elements

Reactants

Products

H0rxn = mH0

f(products) - nH0f(reactants)

-H0f H0

f

form

atio

nH0

rxn

Hinitial

Hfinal

See notes below or in textbook.

Sample Problem 6.9 Calculating the Heat of Reaction from Heats of Formation

SOLUTION:

PLAN:

PROBLEM: Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia:

Calculate H0rxn from H0

f values.

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

Look up the H0f values and use Hess’s Law to find Hrxn.

Hrxn = mH0f (products) - nH0

f (reactants)

Hrxn = [4(H0f NO(g) + 6(H0

f H2O(g)] - [4(H0f NH3(g) + 5(H0

f O2(g)]

= (4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol) -

[(4 mol)(-45.9 kJ/mol) + (5 mol)(0 kJ/mol)]

Hrxn = -906 kJ

PracticePractice

Work on problems 36, 45, 48, 52, and Work on problems 36, 45, 48, 52, and 54.54.

36. Given MgCO36. Given MgCO3(s)3(s) MgO MgO(s)(s) + CO + CO2(g)2(g) with with HHrxnrxn = +117.3kJ. = +117.3kJ.

a. Is heat absorbed or released?a. Is heat absorbed or released?

b. *What is b. *What is HHrxnrxn when 5.50mol of the when 5.50mol of the reactant decompose? reactant decompose?

You look up the rest!You look up the rest!

Figure 9.16 Using bond energies to calculate H0

rxn.H0

rxn = H0reactant bonds broken + H0

product bonds formed

H01 = + sum of

BEH0

2 = - sum of BE

H0rxn

Ent

halp

y, H

BOND BREAKING

BOND FORMATION

See notes below or in textbook.

Figure 9.17 Using bond energies to calculate H0rxn of

methane.

Enth

alp

y,

H

BOND BREAKING4BE(C-H)= +1652kJ

2BE(O2)= + 996kJ

H0(bond breaking) = +2648kJ BOND FORMATION

4[-BE(O-H)]= -1868kJ

H0(bond forming) = -3466kJ

H0rxn= -

818kJ

2[-BE(C O)]= -1598kJ

Treating the combustion of methane as a hypothetical two-step process (see Figure 9.16) means breaking all the bonds in the reactants and forming all the bonds in the products.

SAMPLE PROBLEM 9.3 Calculating Enthalpy Changes from Bond

Energies

SOLUTION:

PROBLEM: Use Table 9.2 (button at right) to calculate H0rxn for the

following reaction:

CH4(g) + 3Cl2(g) CHCl3(g) + 3HCl(g)

PLAN: Write the Lewis structures for all reactants and products and calculate the number of bonds broken and formed.

H

C H

H

H + Cl Cl3

Cl

C Cl

Cl

H + H Cl3

bonds broken

bonds formed

SAMPLE PROBLEM 9.3 Calculating Enthalpy Changes from Bond

Energiescontinued

bonds broken

bonds formed

4 C-H = 4 mol(413 kJ/mol) = 1652 kJ

3 Cl-Cl = 3 mol(243 kJ/mol) = 729 kJ

H0bonds broken = 2381

kJ

3 C-Cl = 3 mol(-339 kJ/mol) = -1017 kJ

1 C-H = 1 mol(-413 kJ/mol) = -413 kJ

H0bonds formed = -2711

kJ

3 H-Cl = 3 mol(-427 kJ/mol) = -1281 kJ

H0reaction = H0

bonds broken + H0bonds formed = 2381 kJ + (-2711 kJ) = -

330 kJ

This is just so you can see how it’s done – won’t be quiz or test.

PracticePractice

Do problem 37 in chp 9: Use the table Do problem 37 in chp 9: Use the table of bond energies to calculate of bond energies to calculate HHrxnrxn for:for:

COCO2(g)2(g) + 2 NH + 2 NH3(g)3(g) NH NH22CONHCONH2(l)2(l) + H + H22OO(l)(l)

YOU MUST draw their Lewis structures YOU MUST draw their Lewis structures and count the number of bonds of and count the number of bonds of each type!each type!

Summary of symbols used in Summary of symbols used in Chp 6:Chp 6:

HHoorxnrxn - any reaction at standard - any reaction at standard

conditionsconditions

HHooff - formation of compound from - formation of compound from

elements, standard conditionselements, standard conditions

HHoovapvap - vaporization of 1 mol at constant - vaporization of 1 mol at constant

P & TP & T

HHoofusfus - melting of 1 mol at constant P & T - melting of 1 mol at constant P & T

HHoocombcomb - combustion reaction - combustion reaction