Chapter 6

Thermochemistry: Energy Flow and Chemical Change

Thermochemistry involves the monitoring of energy transformations that occur with a chemical reaction.

CH4(g) + 2 O2 ==> CO2 + 2 H2O + HEAT

NH4NO3 + H2O + HEAT ==> NH4+ + NO3-

• Reaction gives off heat with rise in temperature in the flask • Reactions that evolve heat = EXOTHERMIC • Heat is written as a product.

• Reaction absorbs heat with decrease in temperature • Reactions that evolve heat = ENDOTHERMIC • Heat is written as a reactant.

Energy is the capacity to do work or produce heat. It takes many faces:

• Radiant energy comes from the sun and is earth’s primary energy source (nuclear fusion)

• Thermal energy is the energy associated with the vibration and rotational motion of atoms and molecules.

• Chemical energy is the energy stored within the chemical bonds of substances.

• Nuclear energy is the energy stored within the neutrons and protons in the atom (E = mc2)

• Kinetic energy: energy associated with moving mass

• Potential energy: energy available by virtue of an object’s position or height above a reference height.

The SI Unit of energy is the Joule (kg m2/s2).

KE = 1/2 m ·v2 = 1/2 (kg)(m/s)2 = kg·m2/s2 = 1 joule

PE = mgh = (kg)(m/s2)(m) = kg·m2/s2 = 1 jouleg = acceleration gravity (9.8 m/s2), and h is object’s height.

Work = F x d = (ma) x d = (kg)(m/s2) x (m) = kg·m2/s2 = 1 joule

1 calorie (cal) = 4.18J

1 British Thermal Unit (BTU) = 1055 Joules

1000 calorie (cal) = 1 Cal (food calorie)

What is the kinetic energy (in kilojoules) of a 2300 lb car moving at 55 mi/h?

• KE = 1/2 mv2 =?• 1 Joule = 1 kg⋅m2/s2

• 1 mile = 1.609 km = 1609 m• 1 lb = 0.454 kg

A single Fritos chip snack has a energy content of 5.0 Cal. How many joules is this?

E = 5 Cal ! 1000 cal

1 Cal! 4.18J

1 cal= 2.1! 104J

Energy can be transformed from one form to another but it can not be destroyed. (1st Law)

Potential Energy = mghKinetic Energy = 1/2 mv2 = 0

PE and KE exist here

High PE

No PE

KE + PE = constant

Potential Energy = 0

Kinetic Energy = 1/2 mv2

We can describe behavior of physical and chemical phenomena as a tendency to seek lower potentials:

High PE

No PE

High Temp

Low Temp

H2(g) + O2(g)

2H2O

Acid + Base

Water + salt

Ent

halp

y, H

Hinitial

Hfinal

H2O(l)

H2O(g)

heat inΔH > 0

Ent

halp

y, H

CH4 + 2O2

CO2 + 2H2O

Hinitial

Hfinal

heat outΔH < 0

A Exothermic process B Endothermic process

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

Energy is also transformed in chemical reactions.

In 1843, James Joule finds that heat and work can be interconverted and are manifestations of the same thing, energy.

Joule found 1 cal heat = 4.1868 J work

amount of heat required to raise the temperature of 1g of water by 1˚C

W = Force x distance

• Energy is the capacity to do work and exchange heat-- it can take on many many forms.

• Work = Energy = Force x Distance.

• Thermal Energy is the kinetic energy of molecules in motion (translational, rotational, and vibrational).

• Heat is thermal energy transferred between two objects at different temperatures.

• Chemical Energy is energy stored in a chemical bond.

Thermodynamics uses a specialized vocabulary that we need to get used to.

• Reactants and products are the system; everything else is the surroundings.

• We track the energy flow into and out of the system.

• Loss of energy from the system to the surroundings has a negative sign.

• Increase of the system energy from the surroundings has a positive sign.

Thermodynamic Formalism: A system has a bulk property of matter that we call “internal energy”.

-

energy in +

surroundings

system

energy out

• Translational kinetic energy.• Molecular rotation.• Bond vibration.• Intermolecular attractions.• Chemical bonds.• Electrons.• Nuclear • Electrochemical

The internal energy, E, of a system is a bulk property of matter that on a microscopic level represents all energy within the system.

On a microscopic scale the internal energy is:

The internal energy of a system can only change by exchange of heat (q) and work (w) with the surroundings.

System

Surroundings

Internal Energy

Rotation, Vibrations, Electronic, translational energy of molecules

+q = heat added to the system

+w = work done on the system

-q = heat lost from the system

-w = work done by the system

ΔESys = q + w

WorkHeatInternal energy

Surroundings

There are 3-Types of “Thermodynamic Systems”

1.Open System2. Closed System 3. Isolated System

No matter exchangedNo heat or work exchanged

Matter exchangedHeat or work exchanged

Matter not exchangedHeat or work exchanged

Matter

Mechanical work can be only be done by a system or on a system by the system being mechanically linked to the environment through P-V work.

• When pistons are attached there could be work done by the system or on the system.

• We call this: pressure-volume work or PV work.

2KClO3(s) ==> 2KCl(s) + 3 O2(g)

w = Force!Distance

w =Force Area

Area! height

w = P ! (Area! height)

w = P !!V

w = P !!V

height

Note PV has units of energy! 101.3 J = 1 L atm

A system can do work on the surroundings or the surroundings can do mechanical work on the system.

work = w = -PΔV

Reworking our Internal Energy Equation, some points of clarification on confusing stuff.

Definition of Work

ΔEsys = q + w = q - PΔV = q - P(Vf - Vi)

ΔESys = q + wWorkHeat

Substituting

Definition of internal energy, ∆E

P = 5.0 atm ΔV = (14.5-12.0) L = 2.5 L w = –PΔVw = -(5.0 atm)(2.5 L) = -12.5 L·atm(-12.5 L·atm)(101 J/L·atm) = -1.3 × 103 J = -1.3 kJ

Calculate the work (in kilojoules) done during a reaction in which the volume expands from 12.0 L to 14.5 L against an external pressure of 5.0 atm. Watch the conversion of PV units to joules.

A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm?

w = -P ΔV

(a) ΔV = 5.4 L – 1.6 L = 3.8 LP = 0 atm

W = -0 atm x 3.8 L = 0 L•atm = 0 joules

(b) ΔV = 5.4 L – 1.6 L = 3.8 LP = 3.7 atm

w = -3.7 atm x 3.8 L = -14.1 L•atm

w = -14.1 L•atm x 101.3 J1L•atm

= -1430 J

The 1st Law of Thermodynamics that energy is neither created nor destroyed in any process, it is just transformed between the two boundaries.

ΔEsystem + ΔEsurroundings = 0

ΔEsystem = -ΔEsurroundings

“The total energy of a system and its surroundings remains constant.”

We can represent the change in internal energy, ΔE of a chemical reaction in a Energy Reaction Diagram.

Energy, E

ΔE<0

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

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

PΔV work done onsurroundings

EIntial

EFinal

A system does work on its surroundings. The energy is lost by the system.

We can not measure the “absolute” internal energy of a system......we can only measure changes in internal energy---we call this a “change in state”.

1. Mount Everest is king if measured from sea level (8,848 meters)2. Mauna Kea is when measured from the depths of the Pacific Ocean floor. It rises 10,203 meters. 3. Mt. Chimborazo in the Andes triumphs. Although it stands but 6,267 meters above sea level, its peak is the farthest from the earth's core.

Measurements are relative to something. What’s the tallest mountain in the world?

ΔESYS = q + wWorkHeatChange in

Internal Energy

A state function is any math function that depends only on the initial and final states and not the path taken to between the two states.

The potential energy of hiker 1 and hiker 2 is the same even though they took very different paths to get to the mountain top. PE only depends on the initial and final heights only.

PE = mghh

ΔE = Efinal - Einitial

ΔP = Pfinal - Pinitial

ΔV = Vfinal - Vinitial

ΔT = Tfinal - Tinitial

ΔH = Hfinal - Hinitial

All are state functions!

0

Final State

Initial State

Why Do We Care About State Functions?• A change in a state function tells us that the function

has one unique value between any two arbitrary states. This fact is why thermodynamics is so powerful and useful.

mgh2

h = 1

Δh

Initial State

Final State

Arbitray State

A chemical system in equilibrium posses a state function called internal energy. A chemical reaction causes a change in internal energy equal to the difference in heat transfer into or from the system and the work done by or on the system.

Initial State Final State

+/-q

+/-w

SOLUTION:

q = - 325 J and w = - 451 J

ΔE = q + w = -325 J + (-451 J) = -776 J

-776J 103JkJ = -0.776kJ -0.776k4.18k

kcal= -0.185 kcal

Determining A Change in Internal Energy

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 (ΔE) in J, kJ, and kcal.

Chemists don’t use internal energy to monitor energy changes in the lab. Instead, we define and use a related parameter called enthalpy.

ΔEsys = q + w = q - PΔV

ΔE = qV - PΔV

At constant volume: ΔV=0

ΔE = qV

CASE 2

ΔE = qP - PΔV

At constant pressure:

qP = ΔE + PΔV

CASE 1

Chemists measure the enthalpies of nearly all chemical reactions and give special names to them.

An Exothermic reaction releases heat to the surroundings, ΔH < 0.

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

H2O (g) H2O (l) + heatΔH = <0

heat written as a product (given off)

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

heat + H2O (s) H2O (l)

An Endothermic reaction absorbs heat from the surroundings into the system, ΔH > 0.

ΔH = >0

heat written as a reactant (consumed)

I. In systems containing solids and liquids; there is no volume change (or it is small); PΔV = 0 and ΔH = ΔE

2. In reactions where the number of moles of gas does not change from reactants to products; no work is done.

N2(g) + O2(g) ==> 2NO(g) PΔV = 0 and ΔH = ΔE

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

ΔH = ΔE + PΔV

There are 3-Cases When ΔH = ΔE

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔH = -2043 kJ

2) 2043 kJ are released and can be written as productC3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) + 2043 kJ

A thermochemical equation shows the enthalpy change (ΔH) of a chemical reaction and also the reactions stoichimetric factors.

1 mol C3H8(g) = -2043 kJ 5O2(g) = -2043 kJ

3CO2(g) = -2043 kJ 4H2O(l) = -2043 kJ

Stoichiometeric factors

3) more useful stoichiometric factors for calculations

1) ΔH <0 = REACTION IS EXOTHERMIC--gives off heat

Using the Heat of Reaction (ΔHrxn) to Find AmountsThe major source of aluminum in the world is bauxite which is mostly aluminum oxide. Its thermal decomposition can be represented by:

If aluminum is produced this way, how many grams of aluminum can form when 1.000 x 103 kJ of heat is is used to decompose the oxide?

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

SOLUTION:

g Al = 1000. kJ! 2 mol Al1676 kJ

! 26.98 g Al1 mol Al

= 32.2 g Al

Using the thermochemical equation calculate how much heat is evolved (or what is the enthalpy) when 266 g of white phosphorus (P4) is combusted in air? The molar mass of P4 is 123.9 g/mol.

P4 (s) + 5O2 (g) P4O10 (s) ΔHrxn = -3013 kJ/mol

ΔH = H (products) – H (reactants)

!Hrxn = 266. g P4 !1 mol P4

123.9 g P4

! "3013 kJ

1 mol P4

= "6470 kJ

In order to apply Hess’s Law we must learn to interpret and manipulate thermochemical equations. The rules are:

• 1. The stoichiometric coefficients always refer to the number of moles of a substance---it’s extensive!

• 2. If you reverse a reaction, the sign of ΔH changes

• 3. If you multiply both sides of the equation by a factor n, then ΔH must change by the same factor n.

2H2O (s) 2H2O (l) ΔH = 2 x 6.01 = 12.0 kJ

H2O (l) H2O (s) ΔH = -6.01 kJ

H2O (s) H2O (l) ΔH = 6.01 kJ

• 4. The physical states of all reactants and products must be specified in thermochemical equations.

H2O (s) H2O (l) ΔH = 6.01 kJ

H2O (l) H2O (g) ΔH = 44.0 kJ

Ent

halp

y, H

CH4 + 2O2

CO2 + 2H2O

Hinitial

Hfinal

heat outΔH < 0

A Exothermic process

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

Ent

halp

y, H

H2O(l)

H2O(g)

heat inΔH > 0

B Endothermic process

H2O(l) H2O(g)

Hfinal

Hinitial

An energy level diagram summarizes the enthalpy changes from an initial state to a final state in a chemical reaction.

q = C ΔT = s m ΔT

1. Law of Conservation of Energyn!

i=1

qi = 0

2. Physics of heat transfer: the heat absorbed by a substance depends on the material’s specific heat capacity, mass and temperature change.

Enthalpies (ΔH) of chemical reactions can be determined using the physics of heat transfer a lab method called calorimetry.

Principle 1: The amount of heat, q, transferred from an object at higher temperature to an object at lower temperature is proportional to the difference in temperature of the two objects!

C = heat capacity

q = C ΔT

q ∝ ΔT

ThermometerThermometer

T1T1 T2

Object 1 Object 2Heat Q

Heat transferred = q ∝ (Tfinal - Tinitial) q = C ΔT

C has units of Energy per degree T (J/°C, J/K, cal/°C, cal/K)

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a quantity of the substance by one degree Celsius or Kelvin (units of J/˚C or J/K or cal/˚C).

material dependent constant “heat capacity”

temperature difference between objects

ΔTC =q

The heat capacity, C, of any material is found to be proportional to mass of a substance under question.

C ∝ m

C = m s

Units of s: J/g °C, J/g K

specific heat

mass

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

The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. (How thermally sensitive a substance is to the addition of energy!)

Heat (q) absorbed or released:

q = s m ΔT

q = C ΔT C = m s

Substituting (2) into 1 gives:

(1) (2)

The molar heat capacity of a substance is the amount of heat (q) required to raise the temperature of one mole of the substance by one degree Celsius.

Heat (q) absorbed or released:

q = s m ΔT

Units: J/mol °C or J/mol K Units: moles

Units: ˚C or K

Heat (q) absorbed or released:

q = s m ΔT

How much heat, q, is required to raise the temperature of 1000. kg of iron and 1000 kg of water from 25 to 75˚C? Note that the specific heat capacity of iron is 0.45 J/g ˚C

qH2O = 1.00! 103 kg ! 1000 g

1 kg! (75! " 25!)! 4.18J

!C g= 2.09! 108J

qFe = 1.00! 103 kg ! 1000 g

1 kg! (75! " 25!)! 0.450J

!C g= 2.25! 107J

Determine the specific heat capacity of lead when 150.0 grams of lead pellets at 100.0˚C are added to 50.0 ml of water held at 22.0˚C. The final temperature of the insulated calorimeter is 28.8˚C. Assume the container and thermometer do not absorb heat.

1. 150.0 g Pb @ 100.0˚C

2. Add pellets to 50.0 ml H2O @ 22.0˚C

3. What’s Up?

28.8˚C

sp. heat ?

qwater = mcΔT = (50.0 g)(4.184 J/g °C)(28.8 - 22.0)°C = 1.40 x103 J

qlead = mcΔT = (150.0 g)(c)(28.8 - 100.0)°C = -1.40 x103 J

clead = 0.13 Jg-1°C-1

1. Use the law of conservation of energy pertains to ALL OF THE PARTICIPANTS!

mPb cPb ΔTPb = -mwat cwat ΔTwat

clead = mwat cwat ΔTwat / mlead ΔTlead

qPb + qwater + qcontainer = 0

qPb = - qwater

n!

i=1

qi = 0

1. Conservation of energy is the key fundamental physics!

qsystem + qsurroundings = 0

Heat (q) absorbed depends on specific heat capacity, mass and temperature change.

q = s m ΔT = C ΔT

1. Law of Conservation of Energyn!

i=1

qi = 0

2. Specific Heat Capacity of Substances

Enthalpies (ΔH) of chemical reactions can be determined using a lab method called calorimetry.

Understand two concepts and you can nail this to the wall!

• Constant Pressure Calorimetry: A constant pressure calorimeter measures the heat change at constant pressure such that qp = ΔH = ΔE - P ΔV

• Constant Volume Bomb Calorimetry: A bomb calorimeter measures the heat change at constant volume such that qv = ΔE. (SKIP THIS IN BOOK)

There are two ways we can do calorimetry, (1) at constant pressure calorimetry; 2) at constant volume and

Skip the constant volume calorimetery in the textbook!

Coffee Cup Calorimeter• A simple cheap calorimeter---pressure is constant--

so we are measuring: ΔH = qP = ΔE + PΔV

No heat enters or leaves!

– Well insulated and therefore isolated.– Calorimeter and other parts are assumed

to not absorb heat (not true but assumed)– heat will come from a chemical reaction,

and is denoted: qrxt

– Energy Conservation is key: Sum the heats

qcal + qtherm + qsolution + qrxn = 0

qi = Ci ΔT = mi si ΔT

n!

i=1

qi = 0

Chemistry in Action

Fuel Values of Foods and Other Substances

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) ΔH = -2801 kJ/mol

1 cal = 4.184 J

1 Cal = 1000 cal = 4184 J

All determined using a bomb calorimeter!

Little calorie = chemistry

Big Calorie = Food

Coffee-Cup Calorimetry A student mixes 50.0 mL of 1.00 M HCl and 50.0 mL of 1 M NaOH in a

coffee-cup calorimeter and finds the mixture goes from 21.00 °C to 27.50 °C. Calculate the Heat of Neutralization for the reaction, assuming that the calorimeter and thermometer absorbs a negligible quantity of heat and that the total volume of the solution is additive; that the water density is 1.000 g/mL and its specific heat capacity is that of water = 4.18 J/g K.

HCl(aq) + NaOH(aq) → H2O(l ) + NaCl(aq) ΔH = ? kJ

A student mixes 50.0 mL of 1.00 M HCl and 50.0 mL of 1 M NaOH in a coffee-cup calorimeter and finds the mixture goes from 21.00 °C to 27.50 °C. Calculate the change Heat of Neutralization per mole acid for the reaction, assuming that the calorimeter loses (or absorbs) only a negligible quantity of heat, that the total volume of the solution is 100.0 mL, that its density is 1.000 g/mL and its specific heat is that of water = 4.18 J/g K.

HCl(aq) + NaOH(aq) → H2O(l ) + NaCl(aq) ΔH = ? kJ/mol

Δ

Δ

and each qi = mi si ΔT

n!

i=1

qi = 0 = qrxt + qsoln = 0

qrxn

qrxn

qrxn -qcal

• Hess’s Law: The enthalpy change for any reaction is equal to the sum of the enthalpy changes for any individual step in the reaction.

CH4 (g) + 2O2 ==> CO2 + 2H2O ΔH1 = ? kJ

CO(g) + 2H2O + 1/2 O2(g) ==> CO2 + 2H2O ΔH3 = - 283 kJ

CH4 (g) + 2O2 ==> CO + 2H2O + 1/2 O2 ΔH2 = - 607 kJ

From ATable

• The heat of reaction for any reaction is a constant and depends only on the initial and final states (state function).

CH4 (g) + 2O2 ==> CO2 + 2H2O ΔH3 = ? kJ

CO(g) + 2H2O + 1/2 O2(g) ==> CO2 + 2H2O ΔH2 = - 283 kJ

CH4 (g) + 2O2 ==> CO + 2H2O + 1/2 O2 ΔH1 = - 607 kJ

CH4 (g) + 2O2

CO(g) + 2H2O + 1/2 O2(g)

CO2 + 2H2O

∆H1 = - 607 kJ

∆H2 = - 283 kJ

∆H3 = ?

Because enthalpy is a state function, its value for any reaction only depends on the final and initial state. This gives us a powerful way to predict enthalpy of many chemical reactions.

Energy level diagrams are useful pictorials of the thermodynamic transitions that take place during a chemical reaction.

CH4 (g) + 2O2 ==> CO2 + 2H2O ΔH3 = -890 kJCO(g) + 2H2O + 1/2 O2(g) ==> CO2 + 2H2O ΔH2 = - 283 kJ CH4 (g) + 2O2 ==> CO + 2H2O + 1/2 O2 ΔH1 = - 607 kJ

Enth

alpy

, HCH4 (g) + 2O2

CO2 + 2H2O

∆H1 = - 607 kJ

∆H2 = - 283 kJ

∆H3 = -890 kJCO(g) + 2H2O + 1/2 O2(g)

NO2(g) → NO(g) + 1/2O2(g) ΔH = +57.07 kJ

1/2N2(g) + O2(g) → NO2(g) ΔH = +33.18 kJ

½N2(g) + O2(g) → NO + 1/2O2(g) ΔH = 90.25 kJ

Calculate the enthalpy of the following reaction and draw and energy level diagram that summarize the thermodynamic enthalpies involved in this reaction.

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) ΔH = ?

Given the following information, calculate the unknown ΔH:

CO(g) + 1/2O2(g) CO2(g) ΔHA = -283.0 kJ

N2(g) + O2(g) 2NO(g) ΔHB = 180.6 kJ

Multiply Equation B by 1/2 and reverse it.

ΔHB = -90.3 kJCO(g) + 1/2O2(g) CO2(g) ΔHA = -283.0 kJNO(g) 1/2N2(g) + 1/2O2(g)

ΔHrxn = -373.3 kJCO(g) + NO(g) CO2(g) + 1/2N2(g)

Physicists and chemists define a standard thermodynamic “standard state” and denote with a superscript degree sign, ΔH˚

• 1 atmosphere pressure• 25˚C = 298.15K• 1 Molar concentration for solutions containing solutes

Thermodynamic Standard State Conditions

ΔHrxn ΔH˚rxn

The superscript indicates thermodynamic standard state: P = 1 atm, T = 298, [1M]

Conditions of P, T, [] are given in problem and stated

What has changed, not much other than the conditions of pressure, temperature and concentration are implicit when you see the superscript.

N2 (g) + 3H2 (g) -------> 2NH3 (g) ΔH˚ = -92.38 kJ

1 mol of N2 reacts with 3 mol H2 to produce 3NH3 and evolves -92.38 kJ of heat at 1 atm and 25˚C

2N2 (g) + 6H2 (g) -------> 4NH3 (g) ΔH˚ = 2 X -92.38 kJ

1/3N2 (g) + H2 (g) ----> 2/3NH3 (g) ΔH˚ = 1/3 X -92.38 kJ

Example of Reaction

Types of Heats of Reaction, ΔHrxn

ΔH˚

The Standard Enthalpy of Formation, ΔHf°, is the

enthalpy change associated with the formation of 1 mole of product from its naturally occurring elements under standard state conditions.

H2(g) + O2(g) → H2O(l ) ΔHf° = -285.8 kJ/mol

3C(s) + 4H2(g) → C3H8(g) ΔHf° = -103.85 kJ/mol

Ag(s) + 1/2Cl2(g) → AgCl(s) ΔHf° = -127.0 kJ/mol

•Oxygen exists as O2 gas at 25 °C•Carbon exists as solid graphite (C) at 25 °C.•Sulfur exits as S8 as a solid at 25˚C

•Water is H2O(l ) in its standard state (not ice or water vapor).

•Examples:Note: 1 mol product!

Note: fractional coefficients allowed

It is also not possible to measure the “absolute enthalpy”. We define an arbitrary scale with the standard enthalpy of formation (ΔHf

°) as a reference point.

1) The standard enthalpy of formation, ΔHf˚ of any element in its most stable form is zero.

The standard enthalpy of formation, ΔHf˚ of many compounds is tabulated in many handbooks. This gives us predictive capability and is someone useful in the real world.

ΔH0 (O2) = 0f

ΔH0 (O3) = 142 kJ/molf

ΔH0 (C, graphite) = 0f

ΔH0 (C, diamond) = 1.90 kJ/molf

ΔH0 (S8, rhombic) = 0 kJ/molf

Writing Formation EquationsWrite balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include ΔH0

f.

PLAN:

(a) Silver chloride, AgCl, a solid at standard conditions.

(b) Calcium carbonate, CaCO3, a solid at standard conditions.

Use the table of heats of formation for values. Remember 1 mole of product is formed and fractions are allowed as coefficients.

(c) Hydrogen sulfide, H2S, a gas at standard conditions.

Writing Formation EquationsWrite balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include ΔH0

f.

PLAN:

(a) Silver chloride, AgCl, a solid at standard conditions.(b) Calcium carbonate, CaCO3, a solid at standard conditions.

Use the table of heats of formation for values.

(c) Hydrogen cyanide, HCN, a gas at standard conditions.

SOLUTION:

ΔH0f = -127.0 kJ(a) Ag(s) + 1/2Cl2(g) AgCl(s)

ΔH0f = -1206.9 kJ(b) Ca(s) + C(graphite) + 3/2O2(g) CaCO3(s)

ΔH0f = 33.9 kJ(c) O2(g) + 1/2N2(g) NO2(g)

Equations where each enthalpy data is provided

and you use Hess’s Law to calculate enthalpy of

another reaction.

Use ΔH˚f data from a table and use the

equation:

2-Problem Types To Expect For Enthalpy Calculations

ΔH°rxn

= Σ ni ΔHif°(products) - Σ mi ΔHjf

°(reactants)

We can use heats of formation ΔHf°, to determine

enthapies of reactions, ΔHrxn°.

ΔH˚rxn = Σ mΔH˚

f(products) - Σ nΔH˚f(reactants)

Energy level diagrams are useful pictorials of the thermodynamic transitions that take place during a chemical reaction.

Ent

halp

y, H

Elements

Reactants

Products

deco

mpo

sitio

n-ΔH0

f

ΔH0f

form

atio

nΔH0

rxn

where Σ means “the sum of”

ni is the respective stoich coefficient for ith product mi is the respective stoich coefficient for each ith reactant

The enthalpy of any reaction, ΔH°rxn can be obtained by

using standard heats of formations and the following equation:

ΔH°rxn

= Σ ni ΔHif°(products) - Σ mi ΔHjf

°(reactants)

ΔH°rx = [cΔH°f (C) + dΔH°f (D)] – [aΔH°f (A) + bΔH°f (B)]

ΔH°rx = ΔH°f (Products) – ΔH°f (Reactants)

aA + bB cC + dDExample: Suppose ΔH°rx = ?

Calculating ΔHf° from tabulated data

Calculate the ΔHcomb for the combustion of 2 moles of C6H6(l ) from enthalpies of formation data table.

C(s) + O2(g) → CO2(g) ΔHf° = -393.5

H2(g) + O2(g) → H2O(l ) ΔHf° = -285.8 kJ

6C(s) + 3H2(g) → C6H6(l ) ΔHf° = 49.04 kJ

ΔHf° Data From Table In Back of the Textbook (or

Handbook of Physics and Chemistry

Calculating ΔHf° from tabulated data

2 C6H6(l ) +15 O2(g) → 12 CO2(g) + 6 H2O(l ) ΔHcomb = 2 X -3267 kJ/mol

that’s for one mole of benzene only!

ΔHcomb° = [6(-393.5 kJ) + 3(-285.8 kJ)] - [(-49.04 kJ)] = -3,267.4 kJ

6C(s) + 6O2(g) → 6CO2(g) ΔHf° = 6(-393.5 kJ)

3H2(g) + O2(g) → 3H2O(l ) ΔHf° = 3(-285.8 kJ)

C6H6(l ) → 6C(s) + 3H2(g) ΔHf° = (-49.04 kJ)

C6H6(l ) + O2(g) → 6CO2(g) + 3H2O(l) ΔHcomb° = ?

ΔH°rxn

= Σ ni ΔHif°(products) - Σ mi ΔHjf

°(reactants)

Calculate the heat of decomposition for the following process using standard enthalpies of formation foundin Appendix. What kind of reaction is this? Is energy given off or needed to make this reaction proceed?

CaCO3 (s) CaO (s) + CO2 (g)

]ΔH0rxn ΔH0 (CO2)fΔH0 (CaO)f= [ + ] - ΔH0 (CaCO3)f[

ΔH0rxn = [ -635.6 + -393.5 ] – [ -1206.9 ] = 179 kJ

ΔH°rxn

= Σ ni ΔHif°(products) - Σ mi ΔHif

°(reactants)

Reaction is decomposition and endothermic. Energy must be absorbed to proceed.

Calculate the standard enthalpy of formation of CS2 (l) given that:C(graphite) + O2 (g) CO2 (g) ΔH0 = -393.5 kJrxn

S(rhombic) + O2 (g) SO2 (g) ΔH0 = -296.1 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) ΔH0 = -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) ΔH0 = -393.5 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) ΔH0 = -296.1x2 kJrxn

CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) ΔH0 = +1072 kJrxn+

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

ΔH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJrxn

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)

ΔH0rxn nΔH0 (products)f= Σ mΔH0 (reactants)f

Σ-

ΔH0rxn 6ΔH0 (H2O)f12ΔH0 (CO2)f= [ + ] - 2ΔH0 (C6H6)f[ ]

ΔH0rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ

-5946 kJ2 mol

= - 2973 kJ/mol C6H6

PLAN:

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 ΔH0rxn from ΔH0

f values.

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

Look up the ΔH0f values in Appendix and use Hess’s Law to

find ΔHrxn.

ΔH0f NH3(g) = -46.3 kJ/mol

ΔH0f H2O(g) = -241.8 kJ/mol ΔH0

f NO(g) = 90.4 kJ/mol

ΔH0f O2(g) = 0

Calculate ΔH0rxn from ΔH0

f values.

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

SOLUTION: ΔHrxn = Σ mΔH0f (products) - Σ nΔH0

f (reactants)

ΔHrxn = [4(ΔH0f NO(g) + 6(ΔH0

f H2O(g)] - [4(ΔH0f NH3(g) + 5(ΔH0

f O2(g)]

= (4 mol)(90.4 kJ/mol) + (6 mol)(-241.8 kJ/mol) -

[(4 mol)(-46.3 kJ/mol) + (5 mol)(0 kJ/mol)]

ΔHrxn = -904 kJ

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:

Heats of Reaction From ΔHf°