Unit 3: Thermochemistry

Unit 3: Thermochemistry

Chemistry 3202

1

May 11

Unit Outline

Temperature and Kinetic Energy Heat/Enthalpy Calculation

Temperature changes (q = mc∆T)Phase changes (q = n∆H) Heating and Cooling CurvesCalorimetry (q = C∆T & above

formulas)

2

May 11

Unit Outline

Chemical ReactionsPE DiagramsThermochemical EquationsHess’s LawBond Energy

STSE: What Fuels You?

3

May 11

Temperature and Kinetic Energy

Thermochemistry is the study of energy changes in chemical and physical changes

eg. dissolving

burning

phase changes

4

May 11

Temperature , T, measures 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

5

May 11

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.

6

May 11

p. 628

7

May 11

K 50.15 450.15

°C 48 -200

8

May 11

# of

par

ticle

s

500 K

300 K

Kinetic Energy

9

May 11

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.

10

May 11

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

11

May 11

Part A: Thought Lab (p. 631)

12

May 12

Part B: Thought Lab (p. 631)

13

May 12

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

Heat/Enthalpy Calculations

14

May 12

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.

15

May 12

FORMULA

q = mc∆T

q = heat (J)

m = mass (g)

c = specific heat capacity

∆T = temperature change

= T2 – T1

= Tf – Ti

16

May 12

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


17

May 12

heat capacity - 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

18

May 13

FORMULA

C = mc

q = C ∆T

C = heat capacity

c = specific heat capacity

m = mass

∆T = T2 – T1

Your Turn p.637 #’s 11-14


May 13

Enthalpy Changes

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

20

May 18

Reaction Progress

PE

Reactants

Products

∆H

Endothermic Reaction

21

May 18

Reaction Progress

PE

Reactants

Products

∆H ∆HEnthalpy

Endothermic Reaction

22

May 18

∆H is +Enthalpy

Reactants

Products

Endothermic

23

May 18

Enthalpy

products

∆H is -

Exothermic

reactants

24

May 18

Enthalpy Changes in Reactions 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. 63925

May 18

26

May 18

Enthalpy Changes in Reactions

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 produced27

May 18

Enthalpy Changes in Reactions

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

1. Chemical reactions

∆Hrxn – enthalpy of reaction

∆Hcomb – enthalpy of combustion

(see p. 643)

28

May 18

2. Formation of compounds from elements

∆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

29

May 18

Use the equation below to determine the ΔHfo

for CH3OH(l)

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

1 C(s) + 2 H2(g) + ½ O2(g) → 1 CH3OH(l) + 238.6 kJ

∆H = -238.6 kJ/mol

30

May 18

Use the equation below to determine the ΔHfo

for CaCO3(s)

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

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

1 Ca(s) + 1 C(s) + 1.5 O2(g) → 1 CaCO3(s) + 1206.9 kJ

∆H = -1206.9 kJ/mol

31

May 18

Use the equation below to determine the ΔHf

o for PH3(g)

4 PH3(g) → P4(s) + 6 H2(g) + 21.6 kJ

a) +21.6 kJ/mol

b) -21.6 kJ/mol

c) +5.4 kJ/mol

d) -5.4 kJ/mol

32

May 18

3. Phase Changes (p.647)∆Hvap – enthalpy of vaporization (l → g)

∆Hfus – enthalpy of melting (fusion: s → l)

∆Hcond – enthalpy of condensation (g → l)

∆Hfre – enthalpy of freezing (l → s)

eg. H2O(l) H2O(g) ΔHvap =

Hg(l) Hg(s) ΔHfre =33

+40.7 kJ/mol

-23.4 kJ/mol

May 18

34

4. Solution Formation (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

May 18

Three ways to represent an enthalpy change:

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.

35

May 18

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

36

May 18

3.

H2O(l)

H2(g) + ½ O2(g)

∆Hf = -285.8 kJ/molEnthalpy (H)

enthalpy diagram

examples: pp. 641-643questions p. 643 #’s 15-18

WorkSheet: Thermochemistry #4 37

May 18

Calculating Enthalpy Changes

FORMULA:

q = n∆H

q = heat (kJ)

n = # of moles

∆H = molar enthalpy

(kJ/mol)

38

M

mn

May 24

eg. How much heat is released when 50.0 g of CH4 forms from C and H ?

(p. 642)n

50 .0 g

16.05 g / m ol

q = nΔH = (3.115 mol)(-74.6 kJ/mol) = -232 kJ

39

3 .115 m ol

May 24

eg. How much heat is released when 50.00 g of CH4 undergoes complete combustion?

(p. 643)

n 50.0 g

16.05 g / m ol

3 .115 m ol

q = nΔH = (3.115 mol)(-965.1 kJ/mol) = -3006 kJ

40

May 24

eg. How much energy is needed to change 20.0 g of H2O(l) at 100 °C to steam at 100 °C ?

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

n 20.0 g

18.02 g / m ol

1 .110 m ol

q = nΔH = (1.110 mol)(+40.7 kJ/mol) = +45.2 kJ

41

May 24

∆Hfre and ∆Hcond have the opposite sign of the above values.

42

May 24

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?

n 40.0 g

80.06 g / m ol

0 .4996 m ol

q = nΔH = (0.4996 mol)(+25.7 kJ/mol) = +12.8 kJ

43

May 24

What mass of ethane, C2H6, must be burned to produce 405 kJ of heat?

ΔH = -1250.9 kJ

q = - 405 kJ

m = ?

H

q n

q = nΔH

kJ 1250.9

kJ 405- n

n = 0.3238 mol

m = n x M = (0.3238 mol)(30.08 g/mol) = 9.74 g

44

May 24

Complete: p. 645; #’s 19 – 23

pp. 648 – 649; #’s 24 – 29

p. 638 #’ 4 – 8

pp. 649, 650 #’s 3 – 8

p. 657, 658 #’s 9 - 18

45


May 24

19. (a) -8.468 kJ (b) -7.165 kJ 20. -1.37 x103 kJ

21. (a) -2.896 x 103 kJ (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.(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 kJ46

Heating and Cooling Curves

Demo: Cooling of p-dichlorobenzene Time (s) Temperature

(°C)Time (s) Temperature (°C)

47

May 25

Cooling curve for p-dichlorobenzene

KE

PE

KE

solidfreezing

liquidTemp. (°C )

50

80

Time

20

48

May 25

Heating curve for p-dichlorobenzene

Temp. (°C )

50

20

80

KE

KE

PE

Time

49

May 25

What did we learn from this demo??

During a phase change temperature remains constant and PE changes

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

50

May 25

p. 651

51

May 25

p. 652

52

q = n∆H

q = mc∆T

May 25

p. 656

53

q = n∆H

q = mc∆T

May 25

54

May 25

Heating Curve for H20(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.

1. Sketch the heating curve for this change.

2. Calculate the total energy required for this transition.

55

May 25

Time

Temp. (°C )

-40

0

100

140

q = mc∆T

q = n∆H

q = mc∆T

q = mc∆T

q = n∆H

56

May 25

Data:

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

57

May 25

warming ice: (from -40 ºC to 0 ºC)

q = mc∆T

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

= 3216 J

warming water: (from 0 ºC to 100 ºC)

q = mc∆T

= (40.0)(4.184)(100 – 0)

= 16736 J58

May 26

warming steam: (from 100 ºC to 140 ºC)

q = mc∆T

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

= 3216 J

59

n = 40.0 g 18.02 g/mol

= 2.22 mol

moles of water:

May 26

melting ice: (fusion)

q = n∆H

= (2.22 mol)(6.02 kJ/mol)

= 13.364 kJ

boiling water: (vaporization)

q = n∆H

= (2.22 mol)(40.7 kJ/mol)

= 90.354 kJ

60

May 26

Total Energy

90.354 kJ

13.364 kJ

3216 J

3216 J

16736 J

127 kJ

61

May 27

Practicep. 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


62

May 27

Law of Conservation of Energy (p. 627)

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

63

May 30

Calorimetry (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. 64

May 30

SimpleCalorimeter

p.661

65

May 30

a simple calorimeter consists of an insulated container, a thermometer, and a known amount of water

simple calorimeters are used to measure heat changes associated with heating, cooling, phase changes, solution formation, and chemical reactions that occur in aqueous solution

66

May 30

to calculate heat lost or gained by a chemical or physical change we apply the first law of thermodynamics:

qsystem = -qcalorimeter

Assumptions:- the system is isolated- c (specific heat capacity) for water is not

affected by solutes- heat exchange with calorimeter can be

ignored67

May 30

eg.

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.

68

May 30

aluminum alloy 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

mcΔT = - mc ΔT

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

-2612 c = -2121

c = 0.812 J/g.°C69

May 31

22.68

eg. 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)

70

May 31

copper

m = 12.8 g

T2 = ºC

c = 0.385 J/g.°C

FIND T1 for Cu

qsys = - qcal

mcT = - CT(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

calorimeter

C = 1.05 kJ/°C

T1 = 25.00 ºC

T2 = 23.94 ºC

71

May 31

23.94

Homework

p. 664, 665 #’s 1b), 2b), 3 & 4 p. 667, #’s 5 - 7

72

May 31

p. 665 # 4.b)

(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

73

6. System (Mg)m = 0.50 g = 0.02057 molFind ΔH

74

Calorimeter v = 100 ml

so m = 100 gc = 4.184T2 = 40.7T1 = 20.4

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 gc = 4.184T1 = 21.4Find T2

qMg = -qcal

nΔH = -mcΔT

Bomb Calorimeter

75

June 1

Bomb Calorimeter

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

76

to accurately measure ΔH you need to 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

77

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?

78

system (octane) calorimeter

m = 11.0 g

T2 = 39.6 ºC

T1 = 20.0 ºC

qsys = - qcal

n ΔH = -CΔT

(0.09627) ΔH = - (28.0)(39.6 – 20.0)

ΔH = -5700 kJ/mol

79

n = 11.0 g 114.26 g/mol = 0.09627 mol

C = 28.0 kJ/ºC

Find ΔHcomb

May 31

eg.

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)

80

benzoic acid calorimeter

T1 = 23.62 ºC

T2 = 27.14 ºC

Find C

m = 1.26 g

ΔHcomb = -3225 kJ/mol

May 31

Homework

p. 675 #’s 8 – 10


81

qsys = - qcal

n ΔH = -CΔT

(0.01032) ΔH = - (C)(27.14 – 23.62)

C = 9.45 kJ/ ºC

n = 1.26 g 122.13 g/mol

= 0.01032 mol

May 31

Hess’s Law of Heat Summation 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

82

June 2

eg. 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 kJ83

June 2

eg. production of carbon dioxide

Pathway #2: formation from the elements

C(s) + O2(g) → CO2(g) ∆H = -393.5 kJ

84

June 2

Using Hess’s Law We can manipulate equations with

known ΔH to determine an unknown enthalpy change.

NOTE: Reversing an equation changes the

sign of ΔH. If we multiply the coefficients we must

also multiply the ΔH value.

85

June 2

eg.

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

86

reverse ? multiply ?June 2

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

Switch

Multiply by 5

Multiply by 4 87

June 2

4 CO2(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)

Ans: -2672.5 kJ

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

88

June 2

Practice

pg. 681 #’s 11-14


89

June 2

Review∆Ho

f (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 compound90

June 3

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 ΔHf

91

June 3

Using Hess’s Law and 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)

92

June 3

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

ΔHf

CO2(g) -393.5 kJ/mol

H2O(g) -241.8 kJ/mol

C6H12O6(s) -1274.5 kJ/mol

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

ΔHrxn = [6(-393.5) + 6(-241.8)] – [1(-1274.5)+ 6(0)]

= [-2361 + -1450.8] - [-1274.5 + 0]

= - 2537.3 kJ93

June 3

Use the molar enthalpy of formation to calculate ΔH for this reaction:

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

ΔHrxn = [3(-393.5) + 2(0) ] – [3(-110.5)+ 1(-824.2)]

= [-1180.5 + 0] - [-331.5 + -824.20]

= - 24.8 kJ

p. 688 #’s 21 & 2294

−824.2 kJ/mol−110.5 kJ/mol −393.5 kJ/mol

June 3

Eg.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.

95

June 7

−393.5 kJ/mol

−241.8 kJ/mol

ΔHcomb = -27.4 kJ/mol

Bond Energy Calculations (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.

96

Bond Energy Calculations (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.

97

C-C = 347C-H = 338O=O = 498

C=O = 745H-O = 460

p. 690 #’s 23,24,& 26p. 691 #’s 3, 4, 5, & 7

= -3244 kJ

98

[2(347) + 2(6)(338) + 7(498)] - [4(2)(745) + 6(2)(460)]

+ 7 O = OCC → 4 O=C=O + 6 H-O-H2

8236 - 11480

Energy Comparisons 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 ΔHeg. nuclear power, reactions in the sun

99

STSE

What fuels you? (Handout)

100

aluminum alloy 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

mcT = - mc T

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

-2612 c = -2121

c = 0.812 J/g.°C101

copper

m = 12.8 g

T2 = ºC

c = 0.385 J/g.°C

FIND T1 for Cu

qsys = - qcal

mcT = - CT(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

calorimeter

C = 1.05 kJ/°C

T1 = 25.00 ºC

T2 = 23.94 ºC

102

q heat J or kJ

c Specific heat capacity

J/g.ºC

C Heat capacity kJ/ ºC or J/ ºC

ΔH Molar heat or molar enthalpy

kJ/mol

103

Documents

Unit 3: Thermochemistry