Thermochemistry The study of heat transfer in chemical rxns.

Thermochemistry

The study of heat transfer in

chemical rxns

ReadingChapter 13 pages 498 - 506Chapter 15 page 591- 602

HW due Friday November 10th

Chapter 13 p 535: #43, 47, 49, 51Chapter 15 p 640: #59

• HW For tonight:

System

That part of nature upon which

attention is focused

Surroundings

That part of nature around the part upon

which we focus

System + Surroundings = The Universe!

Reaction Coordinate

graph of energy change vs. time in a chemical reaction

Exothermic reaction

ReleasesGives off

Loses heat

Endothermic Reaction

AbsorbsTakes in

Gainsheat

EnthalpyH= q = mcT

Heat flow/change

in a system

m is for mass!

c is for specific heat!

ΔT is for change in temp!

Grammar of Thermochemistry

Exothermic condensation reaction

H2O (g) H2O (l) + 44kJ

H2O (g) H2O (l) ΔHo = -44 kJ

Endothermic evaporation reaction

2 H2O (l) + 88 kJ 2 H2O (g)

2 H2O (l) 2 H2O (g) ΔHo = +88 kJ

Specific Heat (capacity)

cAbility of a specific quantity (1g) of a substance to store heat as its temp rises by 1oC

units Jg * oC

Heat Capacity C

Ability of a thing to store heat as its temperature rises

units JoC

Calorimeter• Device that measures Δ heat• It tries to be an adiabatic

system• In real life, gives

experimental yield

Adiabatic SystemDoes not lose heat to or take heat from surroundings

Hsystem = 0

CalorimetryHsystem = 0

Hsys = Hcal + Hrxn

Hrxn = -Hcal

Hrxn = mcTcal

• 3.358 kJ of heat added to the 50.0 g water inside a calorimeter. Twater increases from 22.34oC to 36.74oC. What is the heat capacity of the calorimeter in J/oC?

• cwater = 4.180 J/g * oC• ΔT = (36.74oC – 22.34oC) = 14.40oC• 50.00g * (4.184 J/g * oC) * 14.40oC =

3.012 x 103 J• 3.012 kJ goes into water• 3.358 kJ – 3.012 kJ = .346 kJ absorbed by

calorimeter• .346 kJ = 346 J ÷ 14.40oC = 24.0 J/oC

• 100.0 g of water at 50.0oC is added to a calorimeter that already contains 100.0g of water at 30.0oC. The final temperature is 39oC. What is the heat capacity of the calorimeter?

• cwater = 4.184 J/g * oC

• ΔTadded water = (50.0oC – 39.0oC) = 11oC• 100.0g * (4.184 J/g * oC) * 11oC =

4.60 x 103 J• ΔTcalorimeter water = (39.0oC – 30.0oC) = 9oC• 100.0g * (4.184 J/g * oC) * 9oC =

3.76 x 103 J• 4.60 kJ – 3.76 kJ = .834 kJ absorbed by calorimeter• .834 kJ = 834 J ÷ 9.0oC = 93 J/oC

Thermodynamics and Δ of state

Heat of Fusion Hf

Heat req’d to melt 1g of a substance at its MP

units J/g or J/kg

Heat of VaporizationHv

Heat req’d to boil 1g of a substance at its normal BP

units J/g or J/kg

• Calculate the amount of heat that must be absorbed by 50.0 grams of ice at -12.0oC to convert it to water at 20.0oC.

• cice = 2.09 J/g * oC Hf for ice = 334 J/g

• cwater = 4.184 J/g * oC

• Step 1 – warm the ice to 0oC requires:– (50.0 g) (2.09 J/g * oC) (0oC – (-12oC)) = 0.125 x 104 J

• Step 2 – melt the ice with no Δ in temp:– 50.0 g * 334J/g = 1.67 x 104 J

• Step 3 – warm the liquid to 20.0oC requires:– 50.0 g * 4.18 J/g * oC * (20 oC - 0 oC) = .418 x 104 J

Answer = 2.21 x 104 J = 22.1 kJ must be absorbed

Homework

• Extra Credit Homework AssignmentDue Monday November 13th – # 55 page 535, chapter 13

• Homework due Tuesday November 14th

– Chapter 15, page 641 # 61, 63, 67, 69

A certain calorimeter absorbs 20 J/oC. If 50.0 g of 50oC water is mixed with 50.0 g of 20oC water inside the calorimeter, what will be the final temperature of the mixture?

Heat lost by the hot water will be gained by the cold water and the calorimeter:

ΔHhot water = ΔHcool water + ΔHcalorimeter

ΔHhot water = (50.0 g) (4.180 J/oC*g) (50oC – x)

= 209J/oC (50oC – x)

ΔHcool water = (50.0 g) (4.180 J/oC*g) (x – 20oC)

= 209 J/oC (x – 20oC)

ΔHcalorimeter= 20 J/oC (x – 20oC)

Solve algebraically:

• 209 (50 – x) = 209 (x – 20) + 24 (x – 20)

• 209 (50 – x) = 235 (x – 20)

• 0.889 (50 – x) = x – 20

• 44 – 0.889x = x – 20

• 64 = 1.889x

• x = 33.9oC = 30oC

A certain calorimeter absorbs 24 J/oC. If 50.0 g of 52.7oC water is mixed with 50.0 g of 22.3oC water inside the calorimeter, what will be the final temperature of the mixture?

Heat lost by the hot water will be gained by the cold water and the calorimeter:

ΔHhot water = ΔHcool water + ΔHcalorimeter

ΔHhot water = (50.0 g) (4.180 J/oC*g) (52.7oC – x)

= 209J/oC (52.7oC – x)

ΔHcool water = (50.0 g) (4.180 J/oC*g) (x – 22.3oC)

= 209 J/oC (x – 22.3oC)

ΔHcalorimeter= 24 J/oC (x – 22.3oC)

Solve algebraically:

• 209 (52.7 – x) = 209 (x – 22.3) + 24 (x – 22.3)

• 209 (52.7 – x) = 235 (x – 22.3)

• 0.889 (52.7 – x) = x – 22.3

• 46.87 – 0.889x = x – 22.3

• 69.17 = 1.889x

• x = 36.6oC = 37oC

Heat of ReactionHrxn

Heat/enthalpy change of a chemical reaction

Units J or kJ

Sometimes, units J/mol rxn

Mole of reaction• Depends on how it is given in the problem

(or how you balance your reaction)

• Can say that

O2 (g) + 2 H2 (g) 2 H2O (g) + 45 kJ

• ΔHrxn = 45 kJ/mol rxn

• You can use the following conversion factors:

1 mol O2 2 mol H2 2 mol H2O 1 mol rxn 45 kJ 45 kJ 45 kJ 45 kJ

When X reacts with water the temp in a 1.5 kg

calorimeter containing 2.5 kg water went from 22.5oC to 26.5oC. Calculate Hrxn.

cwater = 4.18 J/g oC ccalorimeter = 2.00 J/g oC

ΔHrxn = ΔHwater + ΔHcalorimeter

Δ T = 26.5oC – 22.5oC = 4oC

Heat absorbed by water: Δ Hwater = mc ΔT– 2.5 kg = 2,500 g – (2,500 g)(4.18J/g*oC)(4oC) = 41,800 J = 41.8 kJ

Heat absorbed by calorimeter: Δ Hcalorimeter = mc ΔT– 1.5 kg = 1,500 g– (1,500 g)(2.00 J/g*oC)(4oC) = 12,000 J = 12 kJ

Total heat added to system = 41.8 + 12 = 53.8 kJ54 kJ

Heat of Solution

Hsoln

•The heat or enthalpy change when a substance is dissolved

• 80 g NaOH is dissolved with 1.40 L of 0.7 M HCl in a calorimeter. HCl solution has a mass of 1.4 kg or 1,400g.

• Ccalorimeter = 20 J/oC water = 10oC

• cHCl same as cwater = 4.18 J/g*oC

• What is the heat released by the solution

• What is the Hsolution for the reaction:

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

• Heat absorbed by calorimeter:– 20 J/oC * 10oC = 200 J

• Heat absorbed by HCl solution:– 1,400 g * (4.18 J/g*oC) * (10oC) = 58,520 J

• Hsolution = Hcalorimeter + HHCl solution

• 200 J + 58,520 J = 58,720 J• Heat released by solution = 58,720 J = 59 kJGo back and see how many moles of NaOH & HCl

reacted:80 g NaOH is 2 moles – therefore you have 2 moles

rxn Hsolution = 59 kJ/2 mol rxn = 30 kJ/mol rxn

Change! To the HW Due Wednesday

• Chapter 15, page 637: 13 & 15

Due Thursday November 16th• Chapter 15, page 637 – 8: 25, 27, 29, 31

• When 2.61 g of C2H6O is burned at constant pressure,82.5 kJ of heat is given off. What is ΔH for the reaction:

C2H6O (l)+ O2 (g) 2 CO2 (g) + 3 H2O (l)

• 82.5 kJ 46.0 g C2H6O 1 mol C2H6O 2.61 g C2H6O mol C2H6O mol rxn

• ΔH for the reaction = -1450 kJ/mol rxn

• When Al metal is exposed to O2 it is oxidized to form Al2O3. How much heat is released by the complete oxidation of 24.2 g of Al at 25oC and 1 atm?

4 Al (s) + 3 O2 (g) 2 Al2O3 (s) ΔH = -3352 kJ/mol rxn

• 24.2 g Al 1 mol Al 1 mol rxn -3352 kJ 27 g Al 4 mol Al mol rxn

• -751 kJ = 751 kJ of heat are released

Heat of Combustion Hcombustion

•The heat or enthalpy change when a substance is burned

Heat of Formation

Hfo

• The heat req’d to form 1 mol of a compound from pure elements

units kJ/mole

Gibb’s Free EnergyGo

•Energy of a system that can be converted to work

•Determines spontaneity

Energy of Formation Gf

o

The energy req’d to form 1 mol of a compound from pure elements

units kJ/mole

Exergonic Reaction•A reaction in which free energy is given off

G < 0

Endergonic Reaction

•A reaction in which free energy is absorbed

G > 0

Reaction at Equilibrium

G = 0

Entropy•A measure of disorder

So

Entropy of Formation•The entropy change when one mole of a substance is formed

•Sfo (J/moleoK)

Thermochemical Equation

•An equation that shows changes in heat, energy, etc

Thermochemical Equation

Ho

rxn Hf

o

productsHf

o

reactants

Thermochemical Equation

Go

rxn Gf

o

productsGf

o

reactants

Thermochemical Equation

So

rxnSf

o

productsSf

o

reactants

Thermochemical Equation

•Stoichiometry of heat change

•Solves theoretical yield

Interrelating Equation

GH S

Calculate H, G, & S when 19.7 kg of BaCO3 is decomposed into BaO + CO2

Cmpd BaCO3 CO2 . BaO

Hf

o -1216.3 -393.5 -553.5

Gf

o -1137.6 -394.4 -525.1

Sf

o 112.1 213.6 70.4

Calculate H, G, & S when 13.6 g of CaSO4 is changed

into CaO + SO2 + O2 at 27oC

Cmpd CaSO4 SO2 CaO

Hf

o -1434.1 -296.8 -635.1

Gf

o -1321.8 -300.2 -604.0

Lab Results: Cup H2O NaOH Thermo

5.0 g 50.0 g 4.0 g 15.0 g Ti = 22.0

oC Tf = 27.0

oC

Cmpd NaOH Na+ OH-

Hf

o -425.6 -240.1 -230.0

Determine: theoretical and experimental heat changes, &

Calculate the potential H, G, & S for the reaction & Sf

o for

O2 when burning 8.8 kg of C3H8 Cpd C3H8 CO2 H2O

Hf

o-103.8 -393.5 -241.8

Gf

o- 23.5 -394.4 -228.6

Sf

o269.9 213.6 188.7

Bond Energy•The energy change

when one mole of bonds are broken

Ho

bond

Bond Equation

Hbondo

rxnHbond

o

products

Hbondo

reactants

Bond Energies (kJ/mole)

C-C 347

C-H 414

O-H 464

C=O 715

Calculate H, G, & S in the production of 831ML

ammonia at 227oC under 125

kPa pressureCompd NH3

Hfo -46.1

Gfo -16.5

1st Law Thermodynamics

•Total energy change = heat + work

E = q + W

Work•W = Fd

•P = F/A

•V = Ad

•W = PV = nRT

2nd Law Thermodynamics

•Total entropy in a system always increases assuming no energy is added to the system

Thermodynamic Rxns are State

Rxns

State Reaction•Reactions that are independent of the

path; thus not dependent on intermediates

Calculate Ho, Go, & S when A + BC AC2 + Bat -23oC & Teq

Compd BC AC2

Hfo(kJ/mole) -150 -250

Gfo(kJ/mole) -125 -225

Write TE for the process2 A + B C + D

C + A H

D + B 2 K

H + K M + B

K + M Product

Write TE for the process2 A + 2 B C + D

C + A 2 H

D + B 2 K

H + K P + B

Write TE for the process2 A + 2 B C + D

C + A 3 H

D + B 2 K

H + K P + Q

Review

Calculate Htotal, when 40.0 g of

H2O is changed from - 25oC to

125oC. FPw = 0.0oC

BPw = 100.0 oC Hv = 2260 J/gCice = 2.06 (J/g K) Hf = 330 J/gCwater = 4.18 (J/g K)Csteam = 2.02 (J/g K)

Calculate Ho, G

o, & S for

AD2 + BC AC2 + BD

at (-23oC)

Cpd BC AD2 AC2 BD

Hf

o -150 -250 -300 -175

Gf

o -125 -225 -250 -150

Sf

o 75 50 80 ?

Determine Sf

o

BD

Calculate Ho, Go, & So for PbO2 + CO CO2 + Pb

Cpd PbO2 CO CO2 Hf

o -277.4 -110.5 -393.5

Gfo -217.4 -137.2 -394.4

Calculate: Teq & H of 48 g PbO2