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Unit 3: Thermochemistry. Chemistry 3202. Unit Outline. Temperature and Kinetic Energy Heat/Enthalpy Calculation Temperature changes (q = mc∆T) Phase changes (q = n∆H) Heating and Cooling Curves Calorimetry (q = C∆T & above formulas). Unit Outline. Chemical Reactions

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Unit 3 thermochemistry

Unit 3: Thermochemistry

Chemistry 3202


Unit outline
Unit Outline

  • Temperature and Kinetic Energy

  • Heat/Enthalpy Calculation

    • Temperature changes (q = mc∆T)

    • Phase changes (q = n∆H)

    • Heating and Cooling Curves

    • Calorimetry (q = C∆T & above formulas)


Unit outline1
Unit Outline

  • Chemical Reactions

    • PE Diagrams

    • Thermochemical Equations

    • Hess’s Law

    • Bond Energy

  • STSE: What Fuels You?


Temperature and kinetic energy
Temperature and Kinetic Energy

Thermochemistryis the study of energy changes in chemical and physical changes

eg. dissolving

burning

phase changes


Temperature - a measure of the average kinetic energy of particles in a substance

- a change in temperature means particles are moving at different speeds

- measured in either Celsius degrees or degrees Kelvin

Kelvin = Celsius + 273.15


The Celsius scale is based on the freezing and boiling point of water

The Kelvin scale is based on absolute zero- the temperature at which particles in a substance have zero kinetic energy.



300 K

# of particles

500 K

Kinetic Energy


Heat enthalpy calculations
Heat/Enthalpy Calculations

system - the part of the universe being studied and observed

surroundings - everything else in the universe

open system - a system that can exchange matter and energy with the surroundings

eg. an open beaker of water

a candle burning

closed system - allows energy transfer but is closed to the flow of matter.


isolated system – a system completely closed to the flow of matter and energy

heat - refers to the transfer of kinetic energy from a system of higher temperature to a system of lower temperature.

- the symbol for heat is q

WorkSheet: Thermochemistry #1




Heat enthalpy calculations1
Heat/Enthalpy Calculations

specific heat capacity – the energy , in Joules (J), needed to change the temperature of one gram (g) of a substance by one degree Celsius (°C).

  • The symbol for specific heat capacity is a lowercase c


  • A substance with a large value of c can absorb or release more energy than a substance with a small value of c.

    ie. For two substances, the substance with the larger c will undergo a smaller temperature change with the same loss or gain of heat.


Formula

q = mc∆T

q = heat (J)

m = mass (g)

c = specific heat capacity

∆T = temperature change

= T2 – T1

= Tf – Ti

FORMULA


eg. How much heat is needed to raise the temperature of 500.0 g of water from 20.0 °C to 45.0 °C?

Solve q = m c ∆T

for c, m, ∆T, T2 & T1

  • p. 634 #’s 1 – 4

  • p. 636 #’s 5 – 8

    WorkSheet: Thermochemistry #2


heat capacity 500.0 g of water from 20.0 °C to 45.0 °C?- the quantity of energy , in Joules (J), needed to change the temperature of a substance by one degree Celsius (°C)

  • The symbol for heat capacity is uppercase C

  • The unit is J/ °C or kJ/ °C


Formula1

C = mc 500.0 g of water from 20.0 °C to 45.0 °C?

q = C ∆T

C = heat capacity

c = specific heat capacity

m = mass

∆T = T2 – T1

FORMULA

Your Turn p.637 #’s 11-14

WorkSheet: Thermochemistry #3


Enthalpy changes
Enthalpy Changes 500.0 g of water from 20.0 °C to 45.0 °C?

enthalpy change - the difference between the potential energy of the reactants and the products during a physical or chemical change

AKA: Heat of Reaction or ∆H


Products 500.0 g of water from 20.0 °C to 45.0 °C?

∆H

Reactants

PE

Endothermic Reaction

Reaction Progress


Products 500.0 g of water from 20.0 °C to 45.0 °C?

∆H

Reactants

PE

Endothermic Reaction

Enthalpy

∆H

Reaction Progress


Products 500.0 g of water from 20.0 °C to 45.0 °C?

Reactants

Enthalpy

∆H is +

Endothermic


products 500.0 g of water from 20.0 °C to 45.0 °C?

∆H is -

reactants

Enthalpy

Exothermic


Enthalpy changes in reactions
Enthalpy Changes in Reactions 500.0 g of water from 20.0 °C to 45.0 °C?

  • All chemical reactions require bond breaking in reactants followed by bond making to form products

  • Bond breaking requires energy (endothermic) while bond formation releases energy (exothermic)

see p. 639


Enthalpy changes in reactions1
Enthalpy Changes in Reactions 500.0 g of water from 20.0 °C to 45.0 °C?

endothermic reaction - the energy required to break bonds is greater than the energy released when bonds form.

ie. energy is absorbed

exothermic reaction - the energy required to break bonds is less than the energy released when bonds form.

ie. energy is produced


Enthalpy changes in reactions2
Enthalpy Changes in Reactions 500.0 g of water from 20.0 °C to 45.0 °C?

∆H can represent the enthalpy change for a number of processes

  • Chemical reactions

    ∆Hrxn – enthalpy of reaction

    ∆Hcomb – enthalpy of combustion

    (see p. 643)


  • Formation of compounds from elements 500.0 g of water from 20.0 °C to 45.0 °C?

    ∆Hof– standard enthalpy of formation

    The standard molar enthalpy of formation is the energy released or absorbed when one mole of a compound is formed directly from the elements in their standard states. (see p. 642)

    eg.

    C(s) + ½ O2(g) → CO(g) ΔHfo = -110.5 kJ/mol


Use the equations below to determine the h f o for ch 3 oh l and caco 3 s
Use the equations below to determine the ΔH 500.0 g of water from 20.0 °C to 45.0 °C?fo for CH3OH(l) and CaCO3(s)

2 C(s) + 4 H2(g) + O2(g) → 2 CH3OH(l) + 477.2 kJ

2 CaCO3(s) + 2413.8kJ → 2 Ca(s) + 2 C(s) + 3 O2(g)


  • Phase Changes 500.0 g of water from 20.0 °C to 45.0 °C?(p.647)

    ∆Hvap – enthalpy of vaporization

    ∆Hfus – enthalpy of melting

    ∆Hcond – enthalpy of condensation

    ∆Hfre – enthalpy of freezing

    eg. H2O(l)  H2O(g) ΔHvap =

    Hg(l)  Hg(s) ΔHfre =

+40.7 kJ/mol

-23.4 kJ/mol


  • Solution Formation 500.0 g of water from 20.0 °C to 45.0 °C?(p.647, 648)

    ∆Hsoln– enthalpy of solution

  • eg.

  • ΔHsoln, of ammonium nitrate is +25.7 kJ/mol.

  • NH4NO3(s) + 25.7 kJ → NH4NO3(aq)

  • ΔHsoln, of calcium chloride is −82.8 kJ/mol.

  • CaCl2(s) → CaCl2(aq) + 82.8 kJ


Three ways to represent an enthalpy change: 500.0 g of water from 20.0 °C to 45.0 °C?

1. thermochemical equation - the energy term written into the equation.

2. enthalpy term is written as a separate expression beside the equation.

3. enthalpy diagram.


eg. the formation of water from the elements produces 285.8 kJ of energy.

1. H2(g) + ½ O2(g) → H2O(l) + 285.8 kJ

2. H2(g) + ½ O2(g) → H2O(l) ∆Hf = -285.8 kJ/mol

thermochemical equation


H kJ of energy.2(g) + ½ O2(g)

∆Hf = -285.8 kJ/mol

H2O(l)

enthalpy diagram

3.

Enthalpy

(H)

examples: pp. 641-643

questions p. 643 #’s 15-18

WorkSheet: Thermochemistry #4


Calculating enthalpy changes

FORMULA: kJ of energy.

q = n∆H

q = heat (kJ)

n = # of moles

∆H = molar enthalpy

(kJ/mol)

Calculating Enthalpy Changes


eg. How much heat is released when 50.0 g of CH kJ of energy.4forms from C and H ?

(p. 642)

q = nΔH

= (3.115 mol)(-74.6 kJ/mol)

= -232 kJ


eg. How much heat is released when 50.00 g of CH kJ of energy.4 undergoes complete combustion?

(p. 643)

q = nΔH

= (3.115 mol)(-965.1 kJ/mol)

= -3006 kJ


eg. How much energy is needed to change 20.0 g of H kJ of energy.2O(l) at 100 °C to steam at 100 °C ?

Mwater = 18.02 g/mol ΔHvap = +40.7 kJ/mol

q = nΔH

= (1.110 mol)(+40.7 kJ/mol)

= +45.2 kJ


∆H kJ of energy.fre and ∆Hcond have the opposite sign of the above values.


eg. The molar enthalpy of solution for ammonium nitrate is +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?

q = nΔH

= (0.4996 mol)(+25.7 kJ/mol)

= +12.8 kJ


What mass of ethane, C +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?2H6, must be burned to produce 405 kJ of heat?

ΔH = -1250.9 kJ

q = - 405 kJ

q = nΔH

n = 0.3238 mol

m = n x M

= (0.3238 mol)(30.08 g/mol)

= 9.74 g


Complete: p. 643 #’s 15 - 18 +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?

p. 645; #’s 19 – 23

pp. 648 – 649; #’s 24 – 29

19. (a) -8.468 kJ (b) -7.165 kJ

20. -1.37 x103 kJ

21. (a) -2.896 x 103 kJ

21. (b) -6.81 x104 kJ

21. (c) -1.186 x 106 kJ

22. -0.230 kJ

23. 3.14 x103 g


24. 2.74 kJ +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?

25.(a) 33.4 kJ (b) 33.4 kJ

26.(a) absorbed (b) 0.096 kJ

27.(a) NaCl(s) + 3.9 kJ/mol → NaCl(aq)

(b) 1.69 kJ

(c) cool; heat absorbed from water

28. 819.2 g

29. 3.10 x 104 kJ


p. 638 #’ 4 – 8 +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?

pp. 649, 650 #’s 3 – 8

p. 657, 658 #’s 9 - 18

WorkSheet: Thermochemistry #5


Heating and cooling curves
Heating and Cooling Curves +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?

Demo: Cooling of p-dichlorobenzene


Cooling curve for p dichlorobenzene

KE +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?

PE

KE

Cooling curve for p-dichlorobenzene

80

Temp.

(°C )

liquid

50

freezing

solid

20

Time


Heating curve for p dichlorobenzene

KE +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?

KE

PE

Heating curve for p-dichlorobenzene

80

Temp.

(°C )

50

20

Time


What did we learn from this demo?? +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?

  • During a phase change temperature remains constant and PE changes

  • Changes in temperature during heating or cooling means the KE of particles is changing


p. 651 +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?


p. 652 +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?


p. 656 +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?


Heating curve for h 2 0 s to h 2 o g
Heating Curve for H +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?20(s) to H2O(g)

A 40.0 g sample of ice at -40 °C is heated until it changes to steam and is heated to 140 °C.

  • Sketch the heating curve for this change.

  • Calculate the total energy required for this transition.


q = mc∆T +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?

140

100

q = n∆H

Temp.

(°C )

q = mc∆T

q = n∆H

0

q = mc∆T

-40

Time


Data: +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?

cice = 2.01 J/g.°C

cwater = 4.184 J/g.°C

csteam = 2.01 J/g.°C

ΔHfus = +6.02 kJ/mol

ΔHvap= +40.7 kJ/mol


warming ice: +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?

q = mc∆T

= (40.0)(2.01)(0 - -40)

= 3216 J

warming water:

q = mc∆T

= (40.0)(4.184)(100 – 0)

= 16736 J

warming steam:

q = mc∆T

= (40.0)(2.01)(140 -100)

= 3216 J


melting ice: +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?

q = n∆H

= (2.22 mol)(6.02 kJ/mol)

= 13.364 kJ

boiling water:

q = n∆H

= (2.22 mol)(40.7 kJ/mol)

= 90.354 kJ

n = 40.0 g

18.02 g/mol

= 2.22 mol

moles of water:


Total energy

90.354 kJ +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?

13.364 kJ

3216 J

3216 J

16736 J

127 kJ

Total Energy


Practice
Practice +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?

WorkSheet:

Thermochemistry #6

p. 655: #’s 30 – 34

pp. 656: #’s 1 - 9

p. 657 #’s 2, 9

p. 658 #’s 10, 16 – 20

30.(b) 3.73 x103 kJ

31.(b) 279 kJ

32.(b) -1.84 x10-3 kJ

33.(b) -19.7 kJ -48.77 kJ

34. -606 kJ


Law of conservation of energy p 627
Law of Conservation of Energy (p. 627) +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?

The total energy of the universe is constant

∆Euniverse = 0

Universe = system + surroundings

∆Euniverse = ∆Esystem + ∆Esurroundings

∆Euniverse = ∆Esystem + ∆Esurroundings = 0

OR ∆Esystem = -∆Esurroundings

OR qsystem = -qsurroundings

First Law of Thermodynamics


Calorimetry p 661
Calorimetry +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?(p. 661)

calorimetry - the measurement of heat changes during chemical or physical processes

calorimeter - a device used to measure changes in energy

2 types of calorimeters

1. constant pressure or simple calorimeter (coffee-cup calorimeter)

2. constant volume or bomb calorimeter.


Simple calorimeter
Simple +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves?Calorimeter

p.661




eg. change we apply the first law of thermodynamics:

A simple calorimeter contains 150.0 g of water. A 5.20 g piece of aluminum alloy at 525 °C is dropped into the calorimeter causing the temperature of the calorimeter water to increase from 19.30°C to 22.68°C.

Calculate the specific heat capacity of the alloy.


eg. change we apply the first law of thermodynamics:

The temperature in a simple calorimeter with a heat capacity of 1.05 kJ/°C changes from 25.0 °C to 23.94 °C when a very cold 12.8 g piece of copper was added to it.

Calculate the initial temperature of the copper. (c for Cu = 0.385 J/g.°C)


Homework
Homework change we apply the first law of thermodynamics:

  • p. 664, 665 #’s 1b), 2b), 3 & 4

  • p. 667, #’s 5 - 7


p. 665 # 4.b) change we apply the first law of thermodynamics:

(60.4)(0.444)(T2 – 98.0) = -(125.2)(4.184)(T2 – 22.3)

26.818(T2 – 98.0) = -523.84(T2 – 22.3)

26.818T2 - 2628.2 = -523.84T2 + 11681

550.66T2 = 14309.2

T2 = 26.0 °C


Calorimeter change we apply the first law of thermodynamics:

v = 100 ml

so m = 100 g

c = 4.184

T2 = 40.7

T1 = 20.4

6. System (Mg)

m = 0.50 g

= 0.02057 mol

Find ΔH

qMg = -qcal

nΔH = -mcΔT

7. System

ΔH = -53.4 kJ/mol

n = CV

= (0.0550L)(1.30 mol/L)

= 0.0715 mol

Calorimeter

v = 110 ml

so m = 110 g

c = 4.184

T1 = 21.4

Find T2


Bomb Calorimeter change we apply the first law of thermodynamics:


Bomb calorimeter
Bomb Calorimeter change we apply the first law of thermodynamics:

  • used to accurately measure enthalpy changes in combustion reactions

  • the inner metal chamber or bomb contains the sample and pure oxygen

  • an electric coil ignites the sample

  • temperature changes in the water surrounding the inner “bomb” are used to calculate ΔH


  • to accurately measure ΔH you need to change we apply the first law of thermodynamics:know the heat capacity (kJ/°C) of the calorimeter.

  • must account for all parts of the calorimeter that absorb heat

    Ctotal = Cwater +Cthermom.+ Cstirrer + Ccontainer

    NOTE: C is provided for all bomb

    calorimetry calculations


eg. A technician burned 11.0 g of octane in a steel bomb calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.

What is the enthalpy of combustion for octane?


eg. calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.

1.26 g of benzoic acid, C6H5COOH(s), is burned in a bomb calorimeter. The temperature of the calorimeter and contents increases from 23.62 °C to 27.14 °C. Calculate the heat capacity of the calorimeter. (∆Hcomb = -3225 kJ/mol)

Homework

p. 675 #’s 8 – 10

WorkSheet: Thermochemistry #7


Hess s law of heat summation
Hess’s Law of Heat Summation calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.

  • the enthalpy change (∆H) of a physical or chemical process depends only on the beginning conditions (reactants) and the end conditions (products)

  • ∆H is independent of the pathway and/or the number of steps in the process

  • ∆H is the sum of the enthalpy changes of all the steps in the process


Eg production of carbon dioxide
eg. calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.production of carbon dioxide

Pathway #1: 2-step mechanism

C(s) + ½ O2(g) → CO(g)∆H = -110.5 kJ

CO(g) + ½ O2(g) → CO2(g)∆H = -283.0 kJ

C(s) + O2(g) → CO2(g) ∆H = -393.5 kJ


Eg production of carbon dioxide1
eg. calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.production of carbon dioxide

Pathway #2: formation from the elements

C(s) + O2(g) → CO2(g) ∆H = -393.5 kJ


Using hess s law
Using Hess’s Law calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.

  • We can manipulate equations with known ΔH to determine the enthalpy change for other reactions.

    NOTE:

  • Reversing an equation changes the sign of ΔH.

  • If we multiply the coefficients we must also multiply the ΔH value.


eg. calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.

Determine the ΔH value for:

H2O(g) + C(s) → CO(g) + H2(g)

using the equations below.

C(s) + ½ O2(g) → CO(g) ΔH = -110.5 kJ

H2(g) + ½ O2(g) → H2O(g) ΔH = -241.8 kJ


Switch calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.

eg.

Determine the ΔH value for:

4 C(s) + 5 H2(g) → C4H10(g)

using the equations below.

ΔH (kJ)

C4H10(g) + 6½ O2(g) → 4 CO2(g) + 5 H2O(g) -110.5

H2(g) + ½ O2(g) → H2O(g) -241.8

C(s) + O2(g) → CO2(g) -393.5

Multiply by 5

Multiply by 4


4 CO calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.2(g) + 5 H2O(g) →C4H10(g) + 6½ O2(g)+110.5

5(H2(g) + ½ O2(g) → H2O(g) -241.8)

4(C(s) + O2(g) → CO2(g) -393.5)

4 CO2(g) + 5 H2O(g) →C4H10(g) + 6½ O2(g)+110.5

5 H2(g) + 2½ O2(g) → 5 H2O(g) -1209.0

4C(s) + 4 O2(g) → 4 CO2(g) -1574.0

Ans: -2672.5 kJ


Practice1
Practice calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.

WorkSheet:

Thermochemistry #8

pg. 681 #’s 11-14


Review
Review calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.

∆Hof(p. 642, 684, & 848)

The standard molar enthalpy of formation is the energy released or absorbed when one mole of a substance is formed directly from the elements in their standard states.

∆Hof = 0 kJ/mol

for elements in the standard state

The more negative the ∆Hof, the more stable the compound


Using hess s law and h f
Using Hess’s Law and calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.ΔHf

Use the formation equations below to determine the ΔH value for:

C4H10(g) + 6½ O2(g) → 4 CO2(g) + 5 H2O(g)

ΔHf (kJ/mol)

4 C(s) + 5 H2(g) → C4H10(g) -2672.5

H2(g) + ½ O2(g) → H2O(g) -241.8

C(s) + O2(g) → CO2(g) -393.5


Using hess s law and h f1
Using Hess’s Law and calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.ΔHf

ΔHrxn = ∑ΔHf(products) - ∑ΔHf(reactants)

eg. Use ΔHf , to calculate the enthalpy of reaction for the combustion of glucose.

C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g)


Use the molar enthalpy’s of formation to calculate calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.ΔH for the reaction below

Fe2O3(s) + 3 CO(g)→ 3 CO2(g) + 2 Fe(s)

p. 688 #’s 21 & 22


Eg. calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.

The combustion of phenol is represented by the equation below:

C6H5OH(s) + 7 O2(g) →6 CO2(g) + 3 H2O(g)

If ΔHcomb = -3059 kJ/mol, calculate the heat of formation for phenol.


Bond energy calculations p 688
Bond Energy Calculations calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.(p. 688)

  • The energy required to break a bond is known as the bond energy.

  • Each type of bond has a specific bond energy (BE).

    (table p. 847)

  • Bond Energies may be used to estimate the enthalpy of a reaction.


Bond energy calculations p 6881
Bond Energy Calculations calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.(p. 688)

ΔHrxn = ∑BE(reactants) - ∑BE (products)

eg. Estimate the enthalpy of reaction for the combustion of ethane using BE.

2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g)

Hint: Drawing the structural formulas for all reactants and products will be useful here.


calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.4 O=C=O + 6 H-O-H

2

+ 7 O = O

C

C

[2(347) + 2(6)(338) + 7(498)]

- [4(2)(745) + 6(2)(460)]

8236 - 11480

= -3244 kJ

p. 690 #’s 23,24,& 26

p. 691 #’s 3, 4, 5, & 7


Energy comparisons
Energy Comparisons calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.

  • Phase changes involve the least amount of energy with vaporization usually requiring more energy than melting.

  • Chemical changes involve more energy than phase changes but much less than nuclear changes.

  • Nuclear reactions produce the largest ΔH

    • eg. nuclear power, reactions in the sun


STSE calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.

  • What fuels you? (Handout)


aluminum alloy calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.water

m = 5.20 g m = 150.0 g

T1 = 525 ºC T1 = 19.30 ºC

T2 = ºC T2 = 22.68 ºC

FIND c for Al c = 4.184 J/g.ºC

qsys = - qcal

mcT = - mc T

(5.20)(c)(22.68 - 525 ) = -(150.0)(4.184)(22.68 – 19.30)

-2612 c = -2121

c = 0.812 J/g.°C

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calorimeter calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C.

C = 1.05 kJ/°C

T1 = 25.00 ºC

T2 = 23.94 ºC

copper

m = 12.8 g

T2 = ºC

c = 0.385 J/g.°C

FIND T1 for Cu

qsys = - qcal

mcT = - CT

(12.8)(0.385)(23.94 – T1) = -(1050)(23.94 – 25.0)

4.928 (23.94 – T1) = 1113

23.94 – T1= 1113/4.928

23.94 – T1= 225.9

T1= -202 ºC

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