PRINCIPLES OF CHEMISTRY I CHEM 1211 CHAPTER 5 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state.

PRINCIPLES OF CHEMISTRY I

CHEM 1211

CHAPTER 5

DR. AUGUSTINE OFORI AGYEMANAssistant professor of chemistryDepartment of natural sciences

Clayton state university

CHAPTER 5

THERMOCHEMISTRY

- The study of the relationship between heat and chemical reactions

THERMOCHEMISTRY

- The ability to do work or to transfer heat

- Energy is necessary for life: humans, plants, animals, cars

- Forms of energy are interconvertible

Two major categories- Kinetic energy

- Potential energy

ENERGY

Kinetic Energy (Ek)

- Energy of motion- Atoms and molecules possess kinetic energy since they have

mass and are in motion

m = mass of the object (kg)v = velocity (speed) of the object (m/s)

- Units: The joule (J), 1 J = 1 kg-m2/s2, 4.184 J = 1 cal

ENERGY

2k mv

2

1E

ENERGY

Calculate the kinetic energy of an object of mass 38 g, whichis moving with a constant velocity of 54 m/s.

(a) in joules and (b) in calories

J55m/s) )(54g 1000

kg 1g)( (38

2

1E 2

k

cal13)J4.184

cal1(m/s) )(54

g 1000

kg 1g)( (38

2

1E 2

k

Potential Energy (Ep)

- Energy by virtue of its position relative to other objects- Arises when there is a force operating on an object

Ep = mgh

m = mass of the object (kg)g = gravitational constant (9.8 m/s2)

h = height of the object relative to a reference point (m)

- Units: The joule (J), 1 J = 1 kg-m2/s2, 4.184 J = 1 cal

ENERGY

ENERGY

A 25-g marble is thrown upward and travels through a vertical distance of 10.0 m from the ground. Calculate the potential energy

of the marble at a height of 5.0 m from the ground.

J1.2m))(5.0m/s)(9.8g1000

kg1g)((25E 2

p

ENERGY

Electrostatic Potential Energy (Eel):

- Due to interactions between charged particles

d

QκQE 21

el

κ = constant of proportionality (8.99 x 109 J-m/C2)C = coulomb, a unit of electrical charge

Q1 and Q2 = electrical charges on two interacting objectsd = distance between the two objects

ENERGY

Electrostatic Potential Energy (Eel):

- Due to interactions between charged particles

- For molecular-level objects, Q1 and Q2 are on the order of magnitude of electron charge (1.60 x 10-19 C)

- Same sign Q1 and Q2 (both positive or both negative) causes repulsion and Eel is positive

- Opposite signs Q1 and Q2 (one positive and one negative) cause attraction and Eel is negative

- The lower the energy of a system, the more stable the system- Opposite charges interact more strongly and the system is

more stable

Work (w)- The energy transferred when a force moves an object

- The product of force (F) and distance (d) through which the object moves

w = F x d

ENERGY

Force- Any kind of push or pull exerted on an object

Heat (q)- Energy used to cause the temperature of an object to change

- A form of energy necessary to change the temperature of a substance

Chemical Energy- Potential energy resulting from forces that hold atoms together

ENERGY

SYSTEM AND SURROUNDINGS

System- The limited and well-defined portion of the

universe under study

Surroundings - Everything else in the universe

Studying energy changes in a chemical reaction- The reactants and products make up the system

- The reaction container and everything else make up the surroundings

SYSTEM AND SURROUNDINGS

Open System- Matter and energy can be exchanged with the surroundings

(water boiling on a stove without a lid)

Closed System- Energy but not matter can be exchanged with the surroundings (two reactants in a closed cylinder reacting to produce energy)

Isolated System- Neither matter nor energy can be exchanged with the surroundings

(insulated flask containing hot tea)

- Sum of all potential and kinetic energies of all components

- Change in internal energy = final energy minus initial energy

E = Efinal - Einitial

- Energy can neither be created nor destroyed

- Energy is conserved

INTERNAL ENERGY (E)

E = Efinal - Einitial

If Efinal > Einitial

E is positive and system has gained energy from its surroundings

If Efinal < Einitial

E is negative and system has lost energy to its surroundings

INTERNAL ENERGY (E)

Energy Diagram

E < 0 E > 0

Inte

rnal

ene

rgy,

E Einitial

EfinalIn

tern

al e

nerg

y, E

Einitial

Efinal

Energy lostto surroundings

Energy gained from surroundings

E of system decreases E of system increases

INTERNAL ENERGY (E)

E = q + wq = heat added to or liberated from a system

w = work done on or by a system

Internal energy of a system increases when - Heat is added to the system from surroundings (positive q)- Work is done on the system by surroundings (positive w)

+w +q

system

INTERNAL ENERGY (E)

E = q + wq = heat added to or liberated from a system

w = work done on or by a system

Internal energy of a system decreases when - Heat is lost by the system to the surroundings (negative q)

- Work is done by the system on the surroundings (negative w)

-w -q

system

INTERNAL ENERGY (E)

Endothermic Process

- Process in which system absorbs heat (endo- means ‘into’)

- Heat flows into system from its surroundings (melting of ice - the reason why it feels cold)

- Heat is a reactant and E is positive

N2(g) + O2(g) + heat → 2NO(g)

INTERNAL ENERGY (E)

Exothermic Process

- Process in which system loses heat

- Heat flows out of the system (exo- means ‘out of’) (combustion of gasoline)

- Heat is a product and E is negative

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

INTERNAL ENERGY (E)

State Function

- Property that depends on initial and final states of the system

- Does not depend on path or how a change occurs

- Internal Energy depends on initial and final states

- Internal Energy is a state function

- q and w, on the other hand, are not state functions

INTERNAL ENERGY (E)

Internal energy is influenced by

- Temperature

- Pressure

- Total quantity of matter

- Internal Energy is an extensive property

INTERNAL ENERGY (E)

Calculate E for a system absorbing 58 kJ of heat from itssurroundings while doing 19 kJ of work on the surroundings.State whether it is an endothermic or an exothermic process

q = +58 kJ (heat is added to the system from surroundings)w = -19 kJ (work is done by the system on the surroundings)

E = q + wE = (+ 58 – 19) kJ = +39 kJ

Endothermic

INTERNAL ENERGY (E)

ENTHALPY (H)

- Heat flow in processes occurring at constant pressure- Only work-pressure (P-V work) are performed

At constant pressurew = - PV

w = workp = pressure

V = change in volume = Vfinal - Vinitial

ENTHALPY (H)

Expansion of Volume- V is a positive quantity and w is a negative quantity

- Energy leaves the system as work- Work is done by the system on the surroundings

Compression of Volume- V is a negative quantity and w is a positive quantity

- Energy enters the system as work- Work is done on the system by the surroundings

ENTHALPY (H)

Calculate the work associated with the expansion of a gas from32 L to 58 L at a constant pressure of 12 atm

w = - PV

w = - (12 atm)(58 L - 32 L) = - 310 L.atm

Gas expands hence work is done by system on surroundings

ENTHALPY (H)

Calculate the work associated with the compression of a gas from58 L to 32 L at a constant pressure of 12 atm

w = - PV

w = - (12 atm)(32 L - 58 L) = 310 L.atm

Gas compresses hence work is done on system by surroundings

ENTHALPY (H)

H = E + PV

H, E, P, and V are all state functions

Change in Enthalpy

H = (E + PV)

ENTHALPY CHANGE (H)

Change in Enthalpy at Constant Pressure

H = E + PV

H = (qp + w) - w = qp

qp = heat at constant pressure

E = q + w

PV = - w

H = qp

Change in enthalpy = heat gained or lost at constant pressure

Positive H - System gains heat from the surroundings

- Endothermic process

Negative H - System releases heat to the surroundings

- Exothermic process

ENTHALPY CHANGE (H)

H = Hfinal - Hinitial

- Enthalpy change is a state function

- Enthalpy is an extensive property

ENTHALPY CHANGE (H)

Standard Enthalpy Change (Ho) - When all reactants and products are in their standard

states

Standard State - Pure form of a substance at

standard temperature and pressure (STP)

Conditions of STP- Standard temperature: 273 K or 0 oC

- Standard pressure: 1.00 atm (101.325 kPa or 100 kPa)

ENTHALPY CHANGE (H)

Enthalpy of Reaction (Hrxn) H accompanying a chemical reaction

Enthalpy of Formation (Hf) - H for forming a substance from its component elements

Enthalpy of Vaporization- H for converting liquids to gases

Enthalpy of Fusion- H for melting solids

Enthalpy of Combustion- H for combusting a substance in oxygen

ENTHALPY CHANGE (H)

ENTHALPY OF REACTION (Hrxn)

- Heat of reaction

- Enthalpy change accompanying a chemical reaction

Hrxn = Hproducts – Hreactants

Horxn = standard enthalpy of reaction

Horxn = Ho

products – Horeactants

Thermochemical Equation- A chemical equation for which H is given

Consider combustion of methane

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

H is an extensive property (depends on amount)

From balanced equation1 mol CH4 reacts with 2 mol O2 to release 890 kJ of heat

Similarly3 mol CH4 reacts with 6 mol O2 to release (3 x 890 kJ) of heat

ENTHALPY OF REACTION (Hrxn)

For the reverse reactionH is equal in magnitude but opposite in sign

CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) H = +890 kJ

H depends on state of the reactants and products

For instance, enthalpy of H2O(g) > enthalpy of H2O(l)Heat is absorbed when converting from liquid to gas

ENTHALPY OF REACTION (Hrxn)

Consider the following reactionCH4(g) + 2O2(g) → CO2(g) + 2H2O(l) H = -890 kJ/mol

Calculate the enthalpy change if 2.0 g methane is burned in excess oxygen

When 1 mol of methane is burned, 890 kJ of heat is released

44

44 CHmol0.12

CHg16.05

CHmol1xCHg2.0

kJ110-CHmol1

kJ 890 -xCHmol 0.12H

44

ENTHALPY OF REACTION (Hrxn)

CALORIMETRY

- Measurement of heat flow

- Device used to measure heat flow is the calorimeter

CALORIMETRY

Heat Capacity - Amount of heat required to raise the temperature of a substance

by 1 K (or 1 oC)

etemperaturinincrease

absorbedheatcapacityheat

Units: cal/K (cal/oC) or J/K (J/oC) 1 cal = 4.184 J

cal = calorie and J = Joule

- Greater heat capacity requires greater amount of heat to produce a given temperature increase

Specific Heat Capacity (Cs)- The quantity of heat energy necessary to raise the temperature

of 1 gram of a substance by 1 K (or 1 oC) Units: cal/g.K (cal/g.oC) or J/g.K (J/g.oC)

Calorie - The amount of heat energy needed to raise the temperature of

1 gram of water by 1 Kelvin (or 1 degree Celsius)

Heat = [mass(g) of solution] x [specific heat of solution] x [T]

q = mCs T

CALORIMETRY

Constant-Pressure

- Heat lost by reaction (qrxn) = heat gained by solution (qsoln)

- qrxn = qsoln

qsoln = [mass(g) of solution] x [specific heat of solution] x [T]

- Specific heat of dilute aqueous solutions are approximately equal to that of water (4.18 J/g.K)

CALORIMETRY

Constant-Volume (Bomb Calorimetry)

- Mostly used to study combustion reactions

- Heat capacity of the calorimeter (Ccal) is first determined using a substance that releases a known quantity of heat

qrxn = - Ccal x T

CALORIMETRY

qsoln = mCs T

T = (72 - 28) oC = 44 oC = 44 K

q = (4.18 J/g.K)(140 g)(44 K) = 26,000 J

Calculate the amount of heat needed to increase the temperatureof 140 g of a solution from 28 oC to 72 oC. Take the specific heat

capacity of the solution as that of water, 4.18 J/g.K

CALORIMETRY

When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed in a constant-pressure calorimeter,

the temperature of the mixture increases from 22.20 oC to 23.11 oC as per the following reaction

AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq)

Calculate H (kJ/mol) for the reaction if the combined solution has a mass of 100.0 g and a specific heat of 4.18 J/g.oC

CALORIMETRY

T = (23.11 - 22.20) oC = 0.91 oC

- qrxn = qsoln

Solution temperature increases so qrxn is negative (exothermic)

qrxn = - (4.18 J/g.oC)(100.0 g)(0.91 oC) = - 380 J = - 0.38 kJ

mol AgNO3 = (0.100 M)(0.0500 L) = 0.00500 mol

H = (-0.38 kJ)/(0.00500 mol) = -76 kJ/mol

CALORIMETRY

When a 0.5865-g sample of lactic acid (HC3H5O3) is burned in a bomb calorimeter whose heat capacity is 4.812 kJ/oC, the

temperature increases from 23.10 oC to 24.95 oC. Calculate the heat of combustion of lactic acid (a) per gram and (b) per mole

T = (24.95 - 23.10) oC = 1.85 oC

qrxn = - Ccal x T = - (4.812 kJ/oC)(1.85 oC) = - 8.90 kJ

(a) H = (8.90 kJ)/(0.5865 g) = - 15.2 kJ/g

(b) H = (15.2 kJ/g)(90.09 g/mol) = - 1370 kJ/mol

CALORIMETRY

HESS’S LAW

If a reaction is carried out in a series of steps

- Enthalpy change for the overall reaction will equal the sum of the enthalpy changes for the individual steps

- Overall H is independent of the number of steps

- Overall H is independent of the reaction path

- Useful for calculating energy changes that are very difficult to measure

Useful Hints

- Work backward from the required reaction, manipulating reactants and products of any given reactions

- Reverse any reactions as needed

- Multiply reactions by appropriate factors as needed

- Use trial and error but allow the final reaction to guide you

HESS’S LAW

To determine H for A → 2C, given

A → 2B HA

C → B HC

Reverse second reaction and multiply by 2

A → 2B HA

2B → 2C -2HC

A → 2C H = HA + -2HC

HESS’S LAW

Calculate the H for the conversion of graphite to diamondCgraphite(s) → Cdiamond(s)

using the following combustion reactions:Cgraphite(s) + O2(g) → CO2(g) H = -394 kJCdiamond(s) + O2(g) → CO2(g) H = -396 kJ

Reverse the second reaction and sum the 2 reactions

Cgraphite(s) + O2(g) → CO2(g) H = -394 kJCO2(g) → Cdiamond(s) + O2(g) H = +396 kJ

Cgraphite(s) → Cdiamond(s) H = +2 kJ

HESS’S LAW

ENTHALPY OF FORMATION (Hf)

- H for forming a substance from its component elements

Standard Enthalpy of Formation (Hfo)

- Hf when all substances are in their standard states

- Most stable form of elements are used for elements existing in more than one form under standard conditions

(O2 is used for oxygen, H2 is used for hydrogen)

- Hfo of the most stable form of any element is zero

)(reactantsmΔ(products)nΔΔH of

of

orxn HH

n and m are stoichiometric coefficients

ENTHALPY OF FORMATION (Hf)

)(reactantsmΔ(products)nΔΔH of

of

orxn HH

Calculate the standard change in enthalpy for the thermite reaction in which a mixture of powdered aluminum and iron(III)

oxide is ignited with a magnesium fuse as shown below2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s)

Hfo for Fe2O3(s) = -826 kJ/mol

Hfo for Al2O3(s) = -1676 kJ/mol

Hfo for Al(s) = Hf

o for Fe(s) = 0

= Hfo for Al2O3(s) - Hf

o for Fe2O3(s)= -1676 - (-826 kJ) = -850 kJ

ENTHALPY OF FORMATION (Hf)

SOURCES OF ENERGY

Foods

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) Ho = -2803 kJ

Glucose from carbohydrate reacts with oxygen to produce carbon dioxide

water and

energy in our bodies

Fuels

Fossil Fuels - Carbon or hydrocarbons found in the earth’s crust

- Major sources of energy

Coal (solid) - High molecular weight hydrocarbons (most abundant)

Petroleum (liquid) - Composed of many compounds and mostly hydrocarbons

Natural Gas - Composed of gaseous hydrocarbons

SOURCES OF ENERGY

Fuels

Fossil Fuels - Carbon or hydrocarbons found in the earth’s crust

- Major sources of energy

Hydrocarbons - Compounds of carbon and hydrogen

- The greater the percentage of carbon and hydrogen in a fuel, the higher its fuel value

SOURCES OF ENERGY

Nuclear Energy- generated from splitting or fusion of the nuclei of atoms

Nonrenewable Sources of Energy- Limited resources and rate of consumption is much

greater than rate of regeneration- Fossil fuel and nuclear energy

SOURCES OF ENERGY

Renewable Sources of Energy

- Unlimited resources and are inexhaustible

Solar energy: from the sun Wind energy: from wind mills

Geothermal energy: from heat stored in the earth Biomass: from crops and biological waste

Hydroelectric energy: from rivers

SOURCES OF ENERGY