1

Thermochemistry The study of energy

Chapter 6

2

Kinetic Energy Ek

The energy of an object in motion

Ek= ½ mv2

Mass need to be in kg

A car moving at 40 mph has a greater kinetic energy than a car moving at 20 mph

3

Types of kinetic energy

• Work: energy used to cause an object with mass to move

• Heat: energy used to cause the temperature of an object to increase

4

Potential Energy

• Stored energy in an object by virtue of its position.

• Units of energy:

Joule J,

1 Calorie Cal = 4.184 J

5

5. 1 The Nature of Energy

• Energy is capacity to do work or to produce heat.

• Thermochemistry is the study of heat change in chemical reactions.

7

Heat is the transfer of thermal energy between two bodies that are at different temperatures.

Energy Changes in Chemical Reactions

Temperature is a measure of the thermal energy

Said another way temperature is a measure of random motion (KE) of particles.

6.2

Temperature (K) = Thermal Energy (J)

8

Physics vs. Chemistry

Chemical energy: is the energy stored within the bonds of chemical substances (potential)

Thermal energy: is the energy associated with the random motion of atoms and molecules (Kinetic molecular theory) (kinetic)

In this chapter we will discuss the transfer of these types of energy.

9

Energy Systems

System: portion we single out to study typically the chemical

Ex: reactants products.

Surroundings: everything else

Ex: Reaction vessel, environment

10

11

Types of Systems

12

A

B

C

13

Exothermic

A reaction that results in the evolution of heat. Thus heat flows out of the system

Exo = out = heat loss from system = energy loss = -heat

14

An Exothermic Reaction

2H2 (g) + O2 (g) 2H2O (l) + energy

H2O (g) H2O (l) + energy

15

16

Endothermic

A reaction that absorbs heat from their surroundings. Thus heat flows into a system.

Endo = In to = heat gain to system = energy gain = + heat

17

An Endothermic Reaction

energy + 2HgO (s) 2Hg (l) + O2 (g)

energy + H2O (s) H2O (l)

18

19

Work, work, work

• Recall that energy

Can be transferred in the form of motion and heat

Work = force x distance

Units

f = Newton, N

d = meter, m

W = N*m or Joule, J

20

• Force is any kind of push or pull exerted on an object. – Ex: gravity, mechanical, electrostatic

• Distance is how far the object moved as a result of the force applied.

21

Work in Biology

• Think about water moving from the ground up the trunk of a tree. What part of the system if any undergoes a change in potential energy? Is work done in the process?

22

Guided Questions

What is changing location ?

Does this change involve potential energy?

23

• Water moves up the trunk against the force of gravity. Thus the potential energy of the water does change

• Work is movement of a mass over a distance against an opposing force.

24

25

Homework

Chang: pg 255

1, 9, 10

****BL:Pg 188

• 1, 2, 3, 6, 9, 11,

26

5.2 First law of thermodynamics

Aka: The law of conservation of energy

Energy is neither created nor destroyed, thus energy is converted from one form to another.

Thus the energy of the universe is constant

Universe = System + Surroundings

27

Chemical energy lost by combustion = Energy gained by the surroundingssystem surroundings

29

Internal Energy

• The energy (E) of a system can be defined as the sum of the kinetic and potential energies of all of the particles in a system.

• Internal energy can be changed by a flow of HEAT, WORK, or BOTH

30

E = Efinal - Einitial

E > 0 = system gained energy

E < 0 = system lost energy to the surroundings

Enthalpy can not be negative, think about it like a bank account (EX pg 235)

Change in a system energy

31

Relating heat to work

E = q + w

E = change in system’s internal energy

q = heat

w = work

32

Signs signs every where are signs

33

Write a figure description for this figure so that a student reading this text would understand.

Identify key terms in your description.

34

Work (w)

W > 0 = work is done ON the system ( + W)

W < 0 = work is done BY the system on the surroundings.

(-W)

35

Prove it….

• Write an example that illustrates work being done by a system

Then write a sentence illustrating work being done on a system

36

Heat (q)

q > 0 = heat is added to the system. (+ q endothermic)

q < 0 = heat is released from the system (-q exothermic)

37

Prove it ….

• Write an example that illustrates heat being added to a system

Then write a sentence illustrating heat being released form a system

38

Example:

Calculate E for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done by the system.

39

Answer

q = + 15.6 (endothermic)

W = - 1.4 kJ (work is done by the system)

E = q + w

E = 15.6 + (- 1.4) = 14.2

40

State Function

A Property of a function that is determined by specifying its condition or its present state

The value of a state function depends only on the present state of the system - not how it arrived there

Energy IS A STATE FUNCTIONEnthalpy IS A STATE FUNCTIONTemperature IS A STATE FUNCTIONHeat IS NOT A STATE FUNCTIONWork IS NOT A STATE FUNCTION

41

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

42

Example

It does not mater whether I heated the water (C B) or cooled that water (A B) to get it to the temperature. Internal energy at point B would be the same regardless.

A B C

43

Change in energy ΔE

• ΔE is a state function because it is independent of pathway.

(all we care about is energy at Ei and energy at Ef.

Heat (q) and work (w) are not state functions!!!!!

44

Example

45

5.2 Homework

Chang: pg 255

11,12,17,18

***BL: Pg 189

17, 19, 21, 22, 25, 26

46

5.3 Enthalpy

Enthalpy (H): Accounts for the heat flow in chemical

changes occurring at constant pressure when no forms of work other than P, V work

*** state function****

Results from a change in P (atm) or V (L) of the system.

47

What is Pressure volume work

Build up of gas that causes a piston to lift against the force of gravity.

Work = -P ΔV

48

49

HEAT

System

Surroundings

ΔH >0

Endothermic

HEAT

System

Surroundings

ΔH <0

Exothermic

Enthalpy : H = E + PV

E = internal energy

50

Enthalpy = H = E + PV

E = H PV

H = E + PV

Don’t forget you can manipulate these equations

52

Example:

• Calculate the work associated with the expansion of gas from 46 L to

64 L at a constant external pressure of 15 atm.

W = -PΔV

53

Answer

w = -PΔV

P = 15 atm V = 64L – 46L = 18L

w = -15atm (18L) = -270 L*atmTo convert to J 101.3 J/1 L*atm

55

5.3 Homework

Chang: 15, 16,

BL: Pg 190 • #’s 27 A and C only, 29 skip A do B, 30• HINT ON #30 write a balance chemical

reaction first, don’t over think it.

56

5.4 Enthalpy Changes

• Enthalpy: The heat content of a chemical system is called the enthalpy (symbol: H)

• The enthalpy change (Δ H) is the amount of heat released or absorbed when a chemical reaction occurs at constant pressure.

57

More enthalpy

• Enthalpy, along with the pressure and volume of a system, is a state function (property of a system that depends only on its state, not how it arrived at its present state).

58

Enthalpy of reactions

calculate the enthalpy of a reaction we need to subtract the final enthalpy from the initial enthalpy or

Hrxn = Hproducts – Hreactants

59

What is the enthalpy of a rxn?

• The heat of the reaction ΔH or ΔHrxn that occurs from the change of products to reactants.

Balanced equations that show an enthalpy change are called thermochemical equations.

2H2 (g) + O2 (g) 2H2O (g) ΔH = - 483.6 kJ

60

Enthalpy Diagrams

• We can use diagrams to illustrate the change in enthalpy of a reaction.

Exothermic = - ΔH value

Endothermic = + ΔH value

61

Thermochemical Equations

H2O (s) H2O (l) H = 6.01 kJ

Is H negative or positive?

System absorbs heat

Endothermic

H > 0

6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.

6.4

62

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) H = -890.4 kJ

Is H negative or positive?

System gives off heat

Exothermic

H < 0

890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.

6.4

64Reaction time Reaction time

65

66

Guidelines for thermochemical reactions

1. Enthalpy is an extensive property Ex: combustions of 1 mol = ΔH -40 kJ combustions of 2 mol = ΔH -80 kJ

2. Enthalpy of a forward reaction is equal to (but opposite in sign) of enthalpy of the reverse reaction

2H2 (g) + O2 (g) 2H2O (g) ΔH = - 483.6 kJ

2H2O (g) 2H2 (g) + O2 (g) ΔH = + 483.6 kJ

67

3. Enthalpy of the reaction depends on the state of the reactants and the products.

Producing H2O (l) or H2O (g) would change the enthalpy value for a thermochemical equation

H2O (s) H2O (l) H = 6.01 kJ

H2O (l) H2O (g) H = 44.0 kJ

69

Steps

CH4 (g) + 2O2(g) CO2 (g) + 2H2O (l) ΔH = -890 kJ

1. Use mole ratios to establish a proportion for every 1mol CH4 (g) burned -890 kJ of heat is

produced

2. Use dimensional analysis to convert g to mol to heat. (since we know heat production from 1 mole of CH4)

4.50g CH4 1mol CH4 -890kJ = -250kJ 16.0g CH4 1mol CH4

70

Homework

Chang:

BL: Pg: 190

#’s 31, 33, 36, 37, 40

Read 5.5 Calorimetry

71

Calorimetry

• Measures heat flow

• Calorimeter: used to measure the exchange of heat that accompanies chemical reactions.

72

How it works• Reaction using known quantities of

reactants is conducted in an insulated vessel that is submerge in a known qty of water.

• The heat created from the reaction will increase the temperature of the water surrounding the vessel.

• The amount of heat emitted can be calculated using the total heat capacity of the calorimeter and its contents.

73

Heat Capacity • The temperature change experienced by an object when it

absorbs a certain amount of heat energy.

• The heat required to raise the temperature of a substance by 1K or 1ºC

q = CΔT

q = heat energy released or absorbed by the rxn (J)C = heat capacity (J/K)ΔT = change in temperature (K)

NOTE: when a sample gains heat (+q) ΔT is positive when a sample loses heat (-q) ΔT is negative

74

A note on temperature scale

A temperature change in Kelvin is equal in magnitude to the temperature change in degrees Celsius.

In the following calculations we are looking at the value for the quantity of the ΔT not an actual Temp reading.

Thus 1K = 1ºC

75

Molar Heat Capacity

• The heat capacity of one mole of a substance is called its molar heat capacity.

• If the molar heat capacity for 1 mol =20 J/mol*K

Then the molar heat capacity for 10 moles = 20(10)

200 J/mol*K (20 J/mol*K for each mole present in the reaction)

76

Specific Heat • The heat capacity of 1 gram of a substance.• The heat required to raise 1 gram of a

substance by 1K or 1ºC

q = sm ΔT

q= heat energy released or absorbed by the rxn (J)

s = specific heat (J/g*K)

m = mass of solution (g)

ΔT = change in temperature in (K)

77

Enthalpy and Specific Heat

• FYI: When reactions occur at constant pressure

ΔH = q

Thus enthalpy change (ΔH) is equal to heat (q)

78

How much heat is given off when an 869 g iron bar cools from 940C to 50C?

s of Fe = 0.444 J/g • 0C

t = tfinal – tinitial = 50C – 940C = -890C

q = mst = 869 g x 0.444 J/g • 0C x –890C = -34,000 J

6.5

79

Specific Heat

• Specific heat of a substance can be determined by measuring the temperature change that a known mass of a substances undergoes when it gains or loses a specific quantity of heat.

Specific heat = qty of heat transferred(g of substance) x (temp

change)

80

Example

• 209J is required to increase the temp of 50.0 g of water by 1.00K.

• Thus the specific heat of water is 4.18 J/gK

s = 209 J (50.0g) (1.00 K)

81

Example:

• How much heat in kJ is required to increase the temperature if 150 g of water from 25°C to 42°C. Specific heat of water is 4.18 J/g°C. (Memorize specific heat water)

82

Answer

q = msΔT

q = 150 (4.18)(42-25)

q = 10.659 kJ

83

Molar Heat Capacity Example

• If you have 1 mol of water with a specific heat of 4.18 J/g*K, the molar heat capacity of water is.

• 4.18 J/g*K (18.0g) = 75.2J/mol*K

• 1mol

• What if you have 26 moles?

85

• IF we assume that no heat is lost. then the heat gained by the solution is lost from the heated metal ball.

Heat lost = Heat gained

qsol = (s) x (g of solution) x ΔT = -qrxn

ΔT > 0 means the rxn is exothermic qrxn < 0

Metal ball

86

• When a student mixes 50 mL of 1.0M HCl and 50 mL of 1.0 M NaOH in a coffee cup calorimeter the temperature of the resulting 100 mL solution increases from 21ºC to 27.5 ºC. Calculate the enthalpy change for the reaction assuming that the calorimeter loses only a negligible amount of heat. The density of the solution is 1.0 g/mL with a specific heat of

4.18 J/g*K

Example

87

q = smΔT

m = 100mL (1.0 g) = 100 g 1 mL

q = (4.18 J/gK)(100 g)(6.5) = 2.7 x103 J Now we must go back to the problem and determine if

the heat (q) is endothermic (+) or exothermic (-)Because the temperature increases heat is being

released and the reaction is exothermic

FINAL ANSWER q = = - 2.7 x103 J

50 mL 1.0 M HCl

+ 50 mL 1.0 M NaOH

=100 mL

ΔT = 27.5-21.0 = 6.5

S = 4.18 J/g*K

D = 1.0 g/mL

88

One step further

We solved for how much energy was produced from 100g of solution…

q = = - 2.7 x103 J

What if the question asked for heat produced from 1 mol HCl?

(0.05L HCl)(1.0 mol HCl) = 0.05 mol HCl in solution 1 L HCl

-2.7 103J/ (0.050 mol HCl) = -54000 J/mol of HCl

89

A coffee cup calorimeter initially contains 125 g of water at 24.2°C. Potassium bromide (10.5g) also at 24.2°C is added to the water, and after the KBr dissolves, the final temp is 21.1°C. Calculate the enthalpy change for dissolving the salt in J. Assume that the specific heat capacity of the solution is 4.184 J/°Cg and that no heat transferred to the surroundings or to the calorimeter.

90

q = msΔT (think m total)

m = 125g H2O + 10.5KBr = 135.5g total

q= 135.5 (4.184)(21.1- 24.2)

q = -1757.4 J

91

*******Example*****

A 46.2 g sample of copper is heated to

95.4 ºC and then placed into a calorimeter containing 75.0 g of water at 19.6 ºC. The final temp of the metal and water was

21.8 ºC. Calculate the specific heat of copper, assuming that all the heat lost by copper is gained by water.

92

Assume:

Heat lost by copper = heat gained by the water

qlost = 46.2g(s)(95.4-21.8 ºC )

= 3400.32(s)

qgained = 75.0 g(4.18)(21.8-19.6 ºC )

= 689.7

3400.32(s) = 689.7

S = 0.2 J/ºC *g

93

Bomb Calorimetry

• Used to study combustion reactions

qrxn = -Ccal x ΔT

94

Homework

Brown Lemay Pg 191-2

#’s

Sp heat/ capacity 43, 44, 45, 46, 47,

Calorimetry 49, 50

95

6.3 Hess’s Law

“Heat of summation”

states that if reactions are carried out in a series of steps, ΔH for the reaction will be equal to the sum of the enthalpy changes by the individual steps.

96

Reactants Products

The change in enthalpy is the same whether the reaction takes place in one step or a series of steps.

Goal: manipulate the steps of the equation (multiple, reverse) to cancel out terms to isolate the final reaction and calculate the new enthalpy (ΔH )

97

98

1. If a reaction is reversed, H is also reversed.

N2(g) + O2(g) 2NO(g) H = 180 kJ

2NO(g) N2(g) + O2(g) H = 180 kJ

2. If the coefficients of a reaction are multiplied by an integer, H is multiplied by that same integer.

2NO(g) N2(g) + O2(g) H = 180 kJ

6NO(g) 3N2(g) + 3O2(g) H = 540 kJ

-180 x 3 = -540

Calculations via Hess’s LawCalculations via Hess’s Law

99

Given the following data, calculate the ΔH for the reaction:

H2S(g) + 2O2(g) H2SO4(l) ΔH =-706.5KJ

H2SO4(l) SO3(g) + H2O(g) ΔH =184.5KJ

H2O(g) H2O(l) ΔH=-99KJ______________________________________

SO3(g) + H2O(l) H2S(g) + 2O2(g) ΔH=

Solving Hess’s Law Problems

100

Solving Hess’s Law ProblemsGiven the following data, calculate the ΔH for the reaction:

C2H2 (g)+ 5/2 O2 ( g) 2CO2 (g) + H2O (l) ΔH = -1300kJ

C (s) + O2 (g) CO2 (g) ΔH = -394 kJ

H2 (g) + ½ O2 (g) H2O (l) ΔH = -286 kJ

2C (s) + H2 (g) C2H2 (g)

101

C2H2 (g)+ 5/2 O2 2CO2 (g) + H2O (l) ΔH = -1300kJ

C (s) + O2 (g) CO2 (g) ΔH = -394 kJ

H2 (g) + ½ O2 (g) H2O (l) ΔH = -286 kJ

__________________________________

2C (s) + H2 (g) C2H2 (g) ΔH =

102

Homework

• BL packet

• #’s 57, 59, 60, 61, 62

103

5.7 Enthalpy of Formation

• Change in enthalpy due to the formation of substances from stable forms of its component elements.

Standard enthalpy of formation (ΔHfº ) is the change in enthalpy for the reaction that forms 1 mol of the compound from its elements in their standard states.

105

Where to Find ΔHfº Values

• Appendix 3 pg A-8

• NOTE THE STATE OF YOUR MOLECULE!!!

106

ΔHrxn = H (products) – H (reactants)

multiply any ∆H by its coefficient

107

Example:

Using enthalpies of formation, calculate the standard change in enthalpy for the following reaction. Then determine if the reaction is endothermic or exothermic:

2Al (s) + Fe2O3 (s) Al2O3 (s) + 2Fe (s)

108

ΔHrxn = H (products) – H (reactants)

2Al (s) + Fe2O3 (s) Al2O3 (s) + 2Fe (s)

H = 0 + -822 -1669 + 0

(-1669 + 0) - (0 + -822)

ΔHrxn = -847 kJ VERY exothermic

109

Homework

• #’s 67, 71, 72

(You will need to look up H in BL table)