thermochem

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6-1

1 LITER OF GASOLINE CONTAINS 8000 CALORIES.


6-2


6-3

A PERSON USES AN AVERAGE OF

2000 CALORIES IN A DAY.


6-4

EXCESS CALORIESIS STORED BY THE BODY

AS FAT.


6-5

SO WHAT?


6-6

WHAT DOES CALORIES MEAN ANYWAY?


6-7

NO. CALORIES IS NOT FAT.IT IS THE ENERGY

FROM THE FOOD WE EAT THAT IS MEASURED IN

CALORIES.


6-8


6-9

SO HOW DO THEY KNOW HOW MUCH CALORIES THERE ARE

IN STUFF?


6-10

THEY KNOW THROUGHTHERMOCHEMISTRY

ANDCALORIMETRY


6-11

Thermochemistry: Energy Flow and Chemical Change

6.1 Forms of Energy and Their Interconversion

6.2 Enthalpy: Heats of Reaction and Chemical Change

6.3 Calorimetry: Laboratory Measurement of Heats of Reaction

6.4 Stoichiometry of Thermochemical Equations

6.5 Hess’s Law of Heat Summation

6.6 Standard Heats of Reaction (H0rxn)


6-12

Thermochemistry is a branch of thermodynamics that deals withthe heat involved with chemical and physical changes. (Duhh!!)

Thermodynamics is the study of heat and its transformations.

Fundamental premise

When energy is transferred from one object to another, it appears as work and/or as heat.

For our work we must define a system to study; everything elsethen becomes the surroundings.

The system is composed of particles with their own internal energies (E or U).Therefore the system has an internal energy. When a change occurs, theinternal energy changes.


6-13

A boundary separates a system from the surrounding. It is arbitrary in character, real or imaginary.

Types of System 1. Open system – separated by an imaginary boundary that allows matter

and energy exchange between the system and surroundings 2. Closed system – separated by a real diathermal and non-permeable boundary that only allows energy exchange to and from the surroundings 3. Isolated system – a system enclosed by a real adiabatic, non- permeable boundary where no matter and energy exchange occur

State and Non-state PropertiesState properties – observed when a system is in an equilibrium state

e.g. P, T, V, U, H, G, SNon-state properties – path dependent properties that are measured or

observed during a change in statee.g. q, w,


6-14

Thermodynamics

State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.

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

energy , pressure, volume, temperature

6.7


6-15

Thermochemistry is the study of heat change in chemical reactions.

The system is the specific part of the universe that is of interest in the study.

open

mass & energyExchange:

closed

energy

isolated

nothing

SYSTEMSURROUNDINGS

6.2


6-16

E = Efinal - Einitial = Eproducts - Ereactants

Figure 6.2 Energy diagrams for the transfer of internal energy (E) between a system and its surroundings.


6-17

1st Law of Thermodynamics

• This is a generalization of the Conservation of Energy Law and it also serves to define the change in internal energy of a system.

• Energy cannot be created nor destroyed. Energy can only be transferred and or transformed.


6-18

1st Law of Thermodynamics (cont.)

Equation:

Q = W + ΔU

U = internal energy

Q = heat energy (=heat transfer)

W = work done• The equation states that heat supplied to a system goes

to increase the internal energy of the system and perform mechanical work on the surroundings.


6-19

Internal Energy & Heat

Internal Energy:• Is the total molecular energy of a system

= U

Heat:• Is the movement of energy = transfer of energy = flow

of energy = Q


6-20

Isothermal Process

• Any process in which the temperature of a system remains constant.


6-21

Isothermal Process (cont.)

1st Law as applied to an Isothermal Process:

Q = W


6-22

Isobaric Process

• Is any process in which pressure of the system remains constant.


6-23

Isobaric Process (cont.)

1st Law as applied to an Isobaric Process:

W = P (Vf – Vi)


6-24

Adiabatic Process

• Is any process in which no heat enters or leaves the system.


6-25

Adiabatic Process (cont.)

1st Law as applied to an Adiabatic Process:

ΔU = - W


6-26

Thermodynamics

6.7

E = q + w

E is the change in internal energy of a system

q is the heat exchange between the system and the surroundings

w is the work done on (or by) the system

w = -PV when a gas expands against a constant external pressure


6-27

Enthalpy and the First Law of Thermodynamics

6.7

E = q + w

At constant pressure, q = H and w = -PV

E = H - PV

H = E + PV


6-28

Figure 6.3 A system transferring energy as heat only.

Figure 6.4 A system losing energy as work only.

Energy, E

Zn(s) + 2H+(aq) + 2Cl-(aq)

H2(g) + Zn2+(aq) + 2Cl-(aq)

E<0work done onsurroundings


6-29

Table 6.1 The Sign Conventions* for q, w and E

q w+ = E

+

+

-

-

-

-

+

+

+

-

depends on sizes of q and w

depends on sizes of q and w

* For q: + means system gains heat; - means system loses heat.

* For w: + means word done on system; - means work done by system.


6-30

Limitations of the First Law of Thermodynamics

E = q + w

Euniverse = Esystem + Esurroundings

Esystem = -Esurroundings

The total energy-mass of the universe is constant.

However, this does not tell us anything about the direction of change in the universe.

Esystem + Esurroundings = 0 = Euniverse

Euniverse = Esystem + Esurroundings

Units of Energy

Joule (J)

Calorie (cal)

British Thermal Unit

1 cal = 4.18J

1 J = 1 kg*m2/s2

1 Btu = 1055 J


6-31

Sample Problem 6.1 Determining the Change in Internal Energy of a System

PROBLEM: When gasoline burns in a car engine, the heat released causes the products CO2 and H2O to expand, which pushes the pistons outward. Excess heat is removed by the car’s cooling system. If the expanding gases do 451 J of work on the pistons and the system loses 325 J to the surroundings as heat, calculate the change in energy (E) in J, kJ, and kcal.


6-32

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.

H = H (products) – H (reactants)

H = heat given off or absorbed during a reaction at constant pressure

Hproducts < Hreactants

H < 0Hproducts > Hreactants

H > 0 6.3


6-33

The Meaning of Enthalpy

w = - PV

H = E + PV

qp = E + PV = H

H ≈ E in

1. Reactions that do not involve gases.

2. Reactions in which the number of moles of gas does not change.

3. Reactions in which the number of moles of gas does change but q is >>> PV.

H = E + PV

where H is enthalpy


6-34

Figure 6.8

Enthalpy diagrams for exothermic and endothermic processes.

Ent

halp

y, H

Ent

halp

y, H

CH4 + 2O2

CO2 + 2H2O

Hinitial

HinitialHfinal

Hfinal

H2O(l)

H2O(g)

heat out heat inH < 0 H > 0

A Exothermic process B Endothermic process

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H2O(l) H2O(g)


6-35

Sample Problem 6.2 Drawing Enthalpy Diagrams and Determining the Sign of H

PROBLEM: In each of the following cases, determine the sign of H, state whether the reaction is exothermic or endothermic, and draw and enthalpy diagram.

SOLUTION:

PLAN: Determine whether heat is a reactant or a product. As a reactant, the products are at a higher energy and the reaction is endothermic. The opposite is true for an exothermic reaction

(a) H2(g) + 1/2O2(g) H2O(l) + 285.8kJ

(b) 40.7kJ + H2O(l) H2O(g)

(a) The reaction is exothermic.

H2(g) + 1/2O2(g) (reactants)

H2O(l) (products)

EXOTHERMIC

(products)H2O(g)

(reactants)H2O(l)

H = -285.8kJ H = +40.7kJENDOTHERMIC

(b) The reaction is endothermic.


6-36

Some Important Types of Enthalpy Change

heat of combustion (Hcomb)

heat of formation (Hf)

heat of fusion (Hfus)

heat of vaporization (Hvap)

C4H10(l) + 13/2O2(g) 4CO2(g) + 5H2O(g)

K(s) + 1/2Br2(l) KBr(s)

NaCl(s) NaCl(l)

C6H6(l) C6H6(g)


6-37

Constant-Pressure Calorimetry

No heat enters or leaves!

quniv = qwater + qcal + qmetal

quniv = 0qmetal = - (qwater + qcal)

qwater = mct

qcal = Ccalt

6.4

Reaction at Constant PH = qmetal

A piece of heated metal is dropped in an insulated cup with 75.0 grams of water. The temperature of the water increased from 29degC to 32degC. How much heat was lost by the piece of metal?


6-38

The specific heat (c) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.

C = mc

Heat (q) absorbed or released:

q = mct

q = Ct

t = tfinal - tinitial

6.4


6-39

Table 6.2 Specific Heat Capacities of Some Elements, Compounds, and Materials

Specific Heat Capacity (J/g*K)

SubstanceSpecific Heat Capacity (J/g*K)

Substance

Compounds

water, H2O(l)

ethyl alcohol, C2H5OH(l)

ethylene glycol, (CH2OH)2(l)

carbon tetrachloride, CCl4(l)

4.184

2.46

2.42

0.864

Elements

aluminum, Al

graphite,C

iron, Fe

copper, Cu

gold, Au

0.900

0.711

0.450

0.387

0.129

wood

cement

glass

granite

steel

Materials

1.76

0.88

0.84

0.79

0.45


6-40

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

6.4


6-41

Sample Problem 6.3 Finding the Quantity of Heat from Specific Heat Capacity

PROBLEM: A layer of copper welded to the bottom of a skillet weighs 125 g. How much heat is needed to raise the temperature of the copper layer from 250C to 300.0C? The specific heat capacity (c) of Cu is 0.387 J/g*K.

Sample Problem 6.4 Determining the Heat of a Reaction

PROBLEM: You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at 25.000C and carefully add 25.0 mL of 0.500 M HCl, also at 25.000C. After stirring, the final temperature is 27.210C. Calculate qsoln (in J) and Hrxn (in J/mol). (Assume the total volume is the sum of the individual volumes and that the final solution has the same density and specfic heat capacity as water: d = 1.00 g/mL and c = 4.184 J/g*K)


6-42

Sample Problem 6.5 Calculating the Heat of Combustion

PROBLEM: A manufacturer claims that its new dietetic dessert has “fewer than 10 Calories per serving.” To test the claim, a chemist at the Department of Consumer Affairs places one serving in a bomb calorimeter and burns it in O2(the heat capacity of the calorimeter = 8.15 kJ/K). The temperature increases 4.9370C. Is the manufacturer’s claim correct?


6-43

PROBLEM: A balloon is filled with 3 liters of helium (atomic mass=4.0) at 1 atm pressure. It initially had a temperature of 25°C but when left under the sun the temperature changed to 28°C and the balloon expanded. The specific heat of helium is 5.1932 J/gK.

a. What is the amount of heat absorbed?

b. What is the amount of work done?

c. What is the amount of change in internal energy?


6-44

Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest?

Establish an arbitrary scale with the standard enthalpy of formation (H0) as a reference point for all enthalpy expressions.f

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm.

f

The standard enthalpy of formation of any element in its most stable form is zero.

H0 (O2) = 0f

H0 (O3) = 142 kJ/molf

H0 (C, graphite) = 0f

H0 (C, diamond) = 1.90 kJ/molf6.5


6-45

Calculate the standard enthalpy of formation of CS2 (l) given that:C(graphite) + O2 (g) CO2 (g) H0 = -393.5 kJRxn/f

S(rhombic) + O2 (g) SO2 (g) H0 = -296.1 kJRxn/f

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1. Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)

2. Add the given rxns so that the result is the desired rxn.

rxnC(graphite) + O2 (g) CO2 (g) H0 = -393.5 kJ

2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -296.1x2 kJrxn

CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)

H0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJrxn6.5


6-46

Table 6.5 Selected Standard Heats of Formation at 250C(298K)

Formula H0f(kJ/mol)

calciumCa(s)CaO(s)CaCO3(s)

carbonC(graphite)C(diamond)CO(g)CO2(g)CH4(g)CH3OH(l)HCN(g)CSs(l)

chlorineCl(g)

0-635.1

-1206.9

01.9

-110.5-393.5-74.9

-238.6135

87.9

121.0

hydrogen

nitrogen

oxygen

Formula H0f(kJ/mol)

H(g)H2(g)

N2(g)NH3(g)NO(g)

O2(g)O3(g)H2O(g)

H2O(l)

Cl2(g)

HCl(g)

0

0

0

-92.30

218

-45.990.3

143-241.8

-285.8

107.8

Formula H0f(kJ/mol)

silverAg(s)AgCl(s)

sodium

Na(s)Na(g)NaCl(s)

sulfurS8(rhombic)S8(monoclinic)SO2(g)

SO3(g)

0

0

0

-127.0

-411.1

2-296.8

-396.0


6-47

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atm.

rxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn nH0 (products)f= mH0 (reactants)f-

6.5

Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)


6-48 6.5


6-49

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn nH0 (products)f= mH0 (reactants)f-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ

-5946 kJ2 mol

= - 2973 kJ/mol C6H6

6.5


6-50

Sample Problem 6.7 Using Hess’s Law to Calculate an Unknown H

PROBLEM: Two gaseous pollutants that form in auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following equation:

CO(g) + NO(g) CO2(g) + 1/2N2(g) H = ?

Given the following information, calculate the unknown H:

Equation A: CO(g) + 1/2O2(g) CO2(g) HA = -283.0 kJ

Equation B: N2(g) + O2(g) 2NO(g) HB = 180.6 kJ


6-51

Sample Problem 6.9 Calculating the Heat of Reaction from Heats of Formation

PROBLEM: Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia:

Calculate H0rxn from H0

f values.

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)


6-52

Sample Problem 6.6 Using the Heat of Reaction (Hrxn) to Find Amounts

PROBLEM: The major source of aluminum in the world is bauxite (mostly aluminum oxide). Its thermal decomposition can be represented by

If aluminum is produced this way, how many grams of aluminum can form when 1.000x103 kJ of heat is transferred?

Al2O3(s) 2Al(s) + 3/2O2(g) Hrxn = 1676 kJ


6-53

Entropy

• Is a measure of a system’s organizational order.

• A system that has some recognizable order has a low entropy value; the more DISORDER the system, the HIGHER its entropy value.

• Symbol: ΔS

• Units: Joules/Kelvin


6-54

Entropy (cont.)

2nd Law as applied to any process:– The entropy of an isolated system will always tend

to increase.– It is very easy to destroy (=high entropy) than to

build (=low entropy)


6-55

Entropy (cont.)

Equation:Δ S ≥ 0S = entropyΔS = change in entropyFor an isothermal process where a change in entropy occurs, the entropy change is given by:

ΔS = ΔQ / T


6-56

Entropy (cont.)

Ex: Phase change: going from ice (solid) to water (liquid), the increase in entropy value,

ΔS = ΔQ / T

T = melting or boiling point (in Kelvin)


6-57

Entropy (cont.)

2nd Law as applied to Entropy:

ΔS > 0 every system tends to get more disordered, hence the entropy increases


6-58

Entropy - Calculations

• Is also based on probability concepts. According to Boltzman, the entropy of a system taking place is given by:

S = K ln W

K = Boltzman constant

W = Probability of occurrence of an event

ln = natural log


6-59

Entropy (cont.)

Sample space for throwing 3 coins simultaneously:


6-60

Entropy (cont.)

Conclusion:

S (= entropy) for all heads is far less compared to S for only two heads. So the probability of obtaining 2 heads far exceeds all heads.


6-61

Entropy (cont.)

Relevance:

Can be used to determine direction of any process or invention.

thermochem

Technology

heat energy

h heat

hp v h

h h ein

h enthalpy

energy diagrams

transfer of energy

system w