ThermochemistryChapter 5 (p 165-181)

Chapter 11 (11.4 and 11.6)

1

ENERGY• Energy is the ability to do work or

transfer heat.– Energy used to cause an object that has

mass to move is called work.– Energy used to cause the temperature of

an object to rise is called heat.

2

Matter and EnergyPotential energy is energy an object possesses by virtue of its position or chemical composition.

Kinetic energy is energy an object possesses by virtue of its motion. Potential energy converted to

kinetic energy

Ek = mv212

m = mass, v = speed 3

Units of EnergyThe SI unit of energy is the joule (J).

Older, non-SI Unit: the calorie (cal).

1 cal = 4.184 J (exactly)

1 J = 1 kg m2

s2

1 cal is amount of energy required to raise the temperature of 1 g of water from 14.5°C to 15.5°C.

Nutritional Calorie: 1 Cal = 1000 cal = 1 kcal

1 J is small, often need kilojoules (kJ), 1 kJ = 1000 J

4

Definitions:System and Surroundings

• The system includes the molecules we want to study (here, the hydrogen and oxygen molecules).

• The surroundings are everything else (here, the cylinder and piston).

5

Energy Transfer: Work and HeatEnergy transferred from racquet to ball:

Energy used to move an object over some distance is work.

w = F dwhere w is work, F is the force,

and d is the distance over which the force is exerted.

• Energy can also be transferred in form of heat.

• Heat flows from warmer objects to cooler objects.

• If differences in temperature between system and surroundings, heat is transferred.

6

First Law of Thermodynamics

• Energy is neither created nor destroyed.• In other words, the total energy of the universe is

a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.

7

Internal Energy

Internal Energy E = the sum of all kinetic and potential energies of all components of the system

In cylinder: internal energy of system includes motion of H2 and O2 molecules through space, rotations, internal vibrations, energies of nuclei of each atom and their electrons.

By definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system:

E = Efinal − Einitial

8

Internal Energy Change

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

Energy lost or gained?By system (reactants) or surrounding?Reverse reaction?

9

• When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w).

• Heat gained by system and work done on system increase internal energy E, thus energy change E >0.

ΔE related to heat (q) and work (w)

E = q + w

10

E, q, w, and Their Signs

E = q + wCalculate the change in internal energy of the system:a) A balloon is heated by adding 850 J of heat. It expands, doing 382 J of work on the atmosphere.

b) A 50 g sample of water is cooled from 30C to 15C, thereby losing about 3140 J of heat.

c) A system releases 57.5 kJ of heat while doing 22.5 kJ of work on the surroundings.

q = + 850 J, w = - 382 J ΔE = 850 J – 382 J = 468 J

q = - 3140 J, w = 0 ΔE = -3140 J

q = - 57.5 kJ, w = - 22.5 kJ ΔE = -57.5 kJ – 22.5 kJ = -80.0 kJ

11

Work in a chemical reactionWe can measure the work done by a gas if the reaction is done in a vessel that has been fitted with a piston.

w = -PV

Zn (s) + 2 H+ (aq) Zn2+ (aq) + H2 (g)

Work involved in expansion or compression of gas is called:Pressure-volume work.P = pressureΔV = Vfinal – Vinitial

w done by system on surroundings <0

12

Enthalpy (greek = to warm)

• If a process takes place at constant pressure (most processes we study do) and the only work done is this pressure-volume work, we can account for heat flow during the process by measuring the enthalpy of the system.

• Enthalpy is the internal energy plus the product of pressure and volume of the system:

H = E + PV

13

Enthalpy• When the system changes at constant pressure,

the change in enthalpy, H, is

H = (E + PV)• Or: H = E + PV

14

Endothermic and Exothermic

• A process is endothermic when H is positive, the system absorbs/gains heat from surroundings

• A process is exothermic when H is negative, the system releases/loses heat to the surroundings.

15

Enthalpy of ReactionThe change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:

H = Hproducts − Hreactants

H is called the enthalpy of reaction, or the heat of reaction.

16

Enthalpies of formation

17

Section 5.7Enthalpy of formation (ΔHf) = enthalpy change associated with formation of compound from its constituent elements.

Magnitude depends on conditions of temperature, pressure, state of reactants and products

Standard set of conditions for comparison:Pure form of a compound/substance, 1 atm, 25°C = 298K

Standard enthalpy of formation (ΔH°f) = enthalpy change for reaction that forms 1 mol of compound from its elements with all substances in their standard statesElements (in std states) compound (in std state) ΔH°f

By definition: standard enthalpy of formation (ΔH°f) of most stable form of any element is zero (ΔH°f = 0)

ΔH°f and enthalpies of reactions: ΔH°rxn

18

ΔH°rxn = Σn ΔH°f (products) - Σn ΔH°f (reactants)

n = coefficients of chemical reactions of each reactant and product (varies)

Example: combustion of ethanol

C2H5OH (l) + 3 O2 (g) 2 CO2 (g) + 3 H2O (l)

Calculate the enthalpy change for combustion reaction of 1 mol of ethanol:ΔH°rxn = [2x ΔH°f (CO2) + 3 ΔH°f (H2O)] – [ΔH°f (C2H5OH) + 3x ΔH°f (O2)]Table 5.3, p 189

ΔH°f C2H5OH(l) = -277.7 kJ/molΔH°f CO2(g) = -393.5 kJ/molΔH°f H2O(l) = -285.8kJ/molΔH°f O2(g) = 0 (by definition)

ΔH°rxn = [2x(-393.5) + 3x(-285.8)] – [-277.7 + 0] = -1367 kJ

Chemical Reactions – Thermochemical Equations

19

2 H2 (g) + O2 (g) 2 H2O (g) ΔH = -483.6 kJΔH is negative reaction/system releases/loses heat to surroundings, is exothermic.

ΔH at end of balanced chemical equation, no specific amounts of reactants listed.BUT:Coefficients here represent the moles of reactants (and products) producing the associated enthalpy change,

i.e. 2 mols of H2 reacting with 1 mol O2 to produce 2 mols H2O results in an enthalpy change of -483.6 kJ.

Facts about Enthalpy

1. Enthalpy is an extensive property, depends on amount of material present.

2. H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction.

3. H for a reaction depends on the state of the products and the state of the reactants.

20

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

Practice

21

Consider the following reaction:

CH3OH (g) CO (g) + 2 H2 (g) ΔH = +90.7 kJ

a) Is heat absorbed or released in the course of this reaction?

b) Calculate the amount of heat transferred when 45.0 g of CH3OH (g) is

decomposed by this reaction.

c) For a given sample of CH3OH, the enthalpy change on reaction is 25.8 kJ.

How many grams of hydrogen gas are produced?

d) What is the value of ΔH for the reverse of the reaction?

e) How many kJ of heat are released when 50.9 g of CO(g) reacts completely

with H2(g) to form CH3OH (g) at constant pressure?

ΔH >0,heat is absorbed

45.0g CH3OH x molar massx90.7 kJ/1 mol

25.8 kJ x 2molH2/90.7 kJx molar mass H2

Sign reversed, magnitude same: -90.7 kJ

50.9 g CO(g) x molar mass CO x (-90.7 kJ/mol)

Enthalpy vs Internal Energy in reactions

22

Enthalpy (H) and internal energy (E) are both “state functions” (value depends only on state or condition of system, not on details on how it came to be in that state.But:Using E to describe energy changes in reactions requires measurement of both heat flow and work.H focuses only on the heat flow which is easier to measure.

Calorimetry

Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow.

23

Heat Capacity and Specific HeatThe amount of energy required to raise the temperature of a substance by 1 K (1C) is its heat capacity.

We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K.

24

Heat Capacity and Specific HeatSpecific heat, quantity of heat transferred, mass of

substance, and temperature change:

Specific heat =heat transferred

mass temperature change

Cs =q

m T

25

BTU – British thermal unit

1 Btu = amount of heat required to raise the temperature of 1 lb of water by 1°F.

How many Joules in 1 Btu?

Specific heat of water: Cs=4.18 J/g-K

Have mass of water: m = 1 lb = 454 g

Need ΔT in K (°C) instead of 1°F (ΔT scale:100°C units = 180°F units thus: ΔT of °F = 5/9°C)

So: 4.18 J/g-K x 454 g x 5/9 = 1054 J = 1 Btu

26

Practice

27

The specific heat of iron metal is 0.450 J/g-K. How many Joules of heat are necessary to raise the temperature of a 1.05 kg block of iron from 25°C to 88.5°C?

Specific heat x mass x temperature change = heat required in J.

0.450 J/g-K x 1.05 kg x kg g x dT (88.5-25.0) =30004 J (3 SF) = 3.00 x 104 J = 30.0 kJ

Phase Changes (11.4)

28

Endothermic? Exothermic?

Phase Changes – Energy Changes

29

• Solid melts liquid: units (molecules) are free to move, average separation increases

• Melting process called: “fusion”• Requires input of energy: heat of fusion,

or enthalpy of fusion (ΔHfus)

• ΔHfus for ice: 6.01 kJ/mol

Phase Changes – Energy Changes

30

• Temperature increases• Liquids gas: units (molecules) are

widely separated• Requires input of energy: heat of

vaporization, or enthalpy of vaporization (ΔHvap)

• ΔHvap for water: 40.7 kJ/mol

Heat of fusion and vaporization

The heat of fusion is the energy required to change a solid at its melting point to a liquid.

The heat of vaporization is defined as the energy required to change a liquid at its boiling point to a gas.

ΔHvap > ΔHfus , all intermolecular interactions are severed.

31

Phase Change: solid to gasSublimation

32

Enthalpy change for solid to gas transition = heat of sublimation (ΔHsub)

ΔHsub= ΔHfus + ΔHvap

Energy Changes Associated with Changes of State

• The heat added to the system at the melting and boiling points goes into pulling the molecules farther apart from each other.

• The temperature of the substance does not rise during a phase change.

33

Practice

34

What is the enthalpy change during the process in which 100.0 g of water at 50.0°C is cooled to ice at -30.0°C?

Cs(ice) = -2.03 J/g-K

ΔHfus= -6.01 kJ/mol

Cs(water) = -4.18 J/g-KAB

Heating curve for waterCD

BC

100.0 g water (H2O) = ? mol H2O

Phase DiagramsPhase diagrams display the state of a substance at various pressures and temperatures and the places where equilibria exist between phases.

35

Phase Diagrams• The circled line is the liquid-vapor interface.• It starts at the triple point (T), the point at

which all three states are in equilibrium.

36

Phase DiagramsIt ends at the critical point (C); above this critical temperature and critical pressure the liquid and vapor are indistinguishable from each other.

37

Phase DiagramsEach point along this line is the boiling point of the substance at that pressure.

38

Phase Diagrams• The circled line in the diagram below is the

interface between liquid and solid.• The melting point at each pressure can be found

along this line.

39

Phase Diagrams• Below the triple point the substance cannot exist

in the liquid state.• Along the circled line the solid and gas phases

are in equilibrium; the sublimation point at each pressure is along this line.

40

Phase Diagram of Water

• Note the high critical temperature and critical pressure.– These are due to the

strong van der Waals forces between water molecules.

41

Phase Diagram of Water• The slope of the solid-

liquid line is negative.– This means that as the

pressure is increased at a temperature just below the melting point, water goes from a solid to a liquid.

42

Phase Diagram of Carbon DioxideCarbon dioxide cannot exist in the liquid state at pressures below 5.11 atm; CO2 sublimes (solid to gas) at normal pressures (1 atm), (dry ice!).

43

Phase diagrams

44

Describe any changes in the phases of water when: a) temperature is kept at 0°C, while pressure is decreased from that at point 5 to the pressure at point 1.

b) Pressure is kept at 1 atm, while temperature is decreased from that at point 9 to the temperature at point 6.

Practice

45

Describe all the phase changes that would occur when:a)Water vapor originally at 0.005 atm and -0.5°C is slowly compressed at constant temperature until the final pressure is 20 atm.b)Water originally at 100.0C and 0.5 atm is cooled at constant pressure until the temperature is -10°C.

Practice

46

Describe the phase changes (and the appr. temperatures at which they occur) when CO2 is heated from -80°C to -20°C at:

a) a constant pressure of 3 atm

b) a constant pressure of 6 atm