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PRINCIPLES OF CHEMISTRY I
CHEM 1211
CHAPTER 5
DR. AUGUSTINE OFORI AGYEMANAssistant professor of chemistryDepartment of natural sciences
Clayton state university
- 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 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
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