Transcript Thermochemistry

Thermochemistry
A. Phase Changes
A. Phase Changes
Evaporation
molecules at the surface gain enough
energy to overcome IMF
Volatility
measure of evaporation rate
depends on temp & IMF
A. Phase Changes
Equilibrium
trapped molecules reach a balance between
evaporation & condensation
Vapor Pressure
pressure of vapor above
a liquid at equilibrium
• depends on temp & IMF
• directly related to volatility
v.p.
A. Phase Changes
temp
A. Phase Changes
• Boiling Point
– temp at which v.p. of liquid
equals external pressure
• depends on Patm & IMF
• Normal B.P. - b.p. at 1 atm
A. Phase Changes
• Melting Point
– equal to freezing point
 Which
has a higher m.p.?
• polar or nonpolar? polar
• covalent or ionic? ionic
A. Phase Changes
• Sublimation
– solid  gas
– v.p. of solid equals
external pressure
 EX:
dry ice, mothballs,
solid air fresheners
B. Heating Curves
Gas - KE 
Boiling - PE 
Liquid - KE 
Melting - PE 
Solid - KE 
B. Heating Curves
• Temperature Change
– change in KE (molecular motion)
– depends on heat capacity
 Heat Capacity
• energy required to raise the temp of 1
gram of a substance by 1°C
B. Heating Curves
• Phase Change
– change in PE (molecular arrangement)
– temp remains constant
 Heat
of Fusion (Hfus)
• energy required to melt 1 gram of a
substance at its m.p.
B. Heating Curves
• Heat of Vaporization (Hvap)
– energy required to boil 1 gram of a
substance at its b.p.
– usually larger than Hfus…why?
 EX:
sweating,
steam burns,
the drinking bird
C. Phase Diagrams
• Show the phases of a substance at
different temps and pressures.
Thermal Energy
A. Temperature & Heat
1. Temperature is related to the
average kinetic energy of the
particles in a substance.
2. SI unit for temp. is the Kelvin
a. K = C + 273 (10C = 283K)
b. C = K – 273 (10K = -263C)
3. Thermal Energy – the total of
all the kinetic and potential energy
of all the particles in a substance.
4. Thermal energy relationships
a. As temperature increases, so
does thermal energy (because the
kinetic energy of the particles
increased).
b. Even if the temperature doesn’t
change, the thermal energy in a
more massive substance is higher
(because it is a total measure of
energy).
5. Heat
a. The flow of thermal
energy from one object to
another.
Cup gets cooler while
hand gets warmer
b. Heat always flows from
warmer to cooler objects.
Ice gets warmer
while hand gets
cooler
6. Specific Heat
a. Some things heat up or cool
down faster than others.
Land heats up and cools down faster than water
b. Specific heat is the amount of heat required to raise the
temperature of 1 kg of a material by one degree (C or K).
1) C water = 4184 J / kg C
2) C sand = 664 J / kg C
This is why land heats up quickly during the day and
cools quickly at night and why water takes longer.
Why does water have such a high specific
heat?
water
metal
Water molecules form strong bonds with each other;
therefore it takes more heat energy to break them. Metals
have weak bonds and do not need as much energy to break
them.
How to calculate changes
in thermal energy
q = m x cxT
q = change in thermal energy
m = mass of substance
T = change in temperature (Tf – Ti)
c = specific heat of substance
Heat loss (released) = heat gained (absorbed)
-q = + q
c. A calorimeter is used to help
measure the specific heat of a
substance.
First, mass and temperature
of water
measured
Knowing
its are
Q value,
its mass,
and its T, its Cp can be
calculated
Then
sample
put
Thisheated
gives the
heatislost
insidebyand
flows into
T is measured for water to
theheat
substance
water
help get its heat gain
HAPPY Monday!
• PRE-AP
Take out your notebooks!!
Turn in any late work/check your boxes for returned quiz and
returned redox homework.
Heat Capacity Homework DUE WEDNESDAY
REGULAR
– Take out your Enthalpy Homework – we are going to go
over some of the problems.
– Enthalpy Hmk DUE TOMORROW.
– Pre-TEST TOMORROW!!
– TEST ON THURSDAY
– REMINDERS
– Project Description and Outline DUE FRIDAY!
MAY 2nd .
STUDY FOR PRE-TEST
1. LeChatelier’s Principle (Disney ride analogy!)
1. Add too much to L, move to right
2. Take away from L, move to left
2. Phase Changes
1. Heating Curve
2. Phase Change Diagram
3. Exothermic/Endothermic
4. Vocab for phase changes (sublimation etc)
3. Basic Temperature Vocabulary
1. Kinetic Energy and Potential Energy
2. Thermal Energy
3. Intermolecular forces
4. q and heat capacity problems
1. Q = amount of heat transferred
(neg/pos)
2. C = specific heat
3. Units come from the C value
q = mc T
5. Enthalpy Problems
1. H = (Products added) –
(Reactants added)
2. Multiply by the coefficents
3. Elements in basic form are 0.
TWO Trends in Nature
• Order  Disorder


• High energy  Low energy

Exothermic process is any process that gives off heat –
transfers thermal energy from the system to the surroundings.
2H2 (g) + O2 (g)
H2O (g)
2H2O (l) + energy
H2O (l) + energy
Endothermic process is any process in which heat has to be
supplied to the system from the surroundings.
energy + 2HgO (s)
energy + H2O (s)
2Hg (l) + O2 (g)
H2O (l)
6.2
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 < 0
Hproducts > Hreactants
H > 0
6.4
Thermochemical Equations
Is H negative or positive?
System absorbs heat
Endothermic
H > 0 and is positive
6.01 kJ are absorbed for every 1 mole of ice that
melts at 00C and 1 atm.
H2O (s)
H2O (l)
H = 6.01 kJ
6.4
Thermochemical Equations
Is H negative or positive?
System gives off heat
Exothermic
H < 0 and is negative
890.4 kJ are released for every 1 mole of methane
that is combusted at 250C and 1 atm.
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (l) H = -890.4 kJ
6.4
Thermochemical Equations
•
The stoichiometric coefficients always refer to the number
of moles of a substance
H2O (s)
•
H2O (l)
ΔH = 6.01 kJ
If you reverse a reaction, the sign of H changes
H2O (l)
•
H = 6.01 kJ/mol
H2O (s)
H = -6.01 kJ
If you multiply both sides of the equation by a factor n,
then H must change by the same factor n.
2H2O (s)
2H2O (l)
H = 2 mol x 6.01 kJ/mol = 12.0 kJ
6.4
Thermochemical Equations
•
The physical states of all reactants and products must be
specified in thermochemical equations.
H2O (s)
H2O (l)
H = 6.01 kJ
H2O (l)
H2O (g)
H = 44.0 kJ
How much heat is evolved when 266 g of white
phosphorus (P4) burn in air?
P4 (s) + 5O2 (g)
266 g P4 x
P4O10 (s)
1 mol P4
123.9 g P4
x
Hreaction = -3013 kJ
3013 kJ
= 6470 kJ
1 mol P4
6.4
Standard enthalpy of formation (Hf0) is the heat
change that results when one mole of a compound is
formed from its elements at a pressure of 1 atm.
The standard enthalpy of formation of any element in its
most stable form is zero.
H0f (O2) = 0
H0f (C, graphite) = 0
H0f (O3) = 142 kJ/mol
H0f (C, diamond) = 1.90 kJ/mol
6.6
6.6
0 ) is the enthalpy of
The standard enthalpy of reaction (Hrxn
a reaction carried out at 1 atm.
aA + bB
cC + dD
H0rxn = [ cH0f (C) + dH0f (D) ] - [ aH0f (A) + bH0f (B) ]
0 (reactants)
H
S
H0rxn = S H0 (products)
f
f
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.)
6.6
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 = S H0 f(products) - S H0 f(reactants)
H0rxn = [ 12H0f (CO2) + 6H0f (H2O)] - [ 2H0f (C6H6)]
H0rxn = [ 12 × -393.5 + 6 × -285.8 ] – [ 2 × 49.04 ] = -6535 kJ
-6535 kJ
= - 3267 kJ/mol C6H6
2 mol
6.6
Calculate the standard enthalpy of formation of CS2 (l)
given that:
C(graphite) + O2 (g)
CO2 (g) H0rxn = -393.5 kJ
S(rhombic) + O2 (g)
CS2(l) + 3O2 (g)
SO2 (g)
H0rxn = -296.1 kJ
CO2 (g) + 2SO2 (g)
0 = -1072 kJ
Hrxn
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.
C(graphite) + O2 (g)
2S(rhombic) + 2O2 (g)
+ CO2(g) + 2SO2 (g)
CO2 (g) H0rxn = -393.5 kJ
2SO2 (g) H0rxn = -296.1x2 kJ
CS2 (l) + 3O2 (g)
0 = +1072 kJ
Hrxn
C(graphite) + 2S(rhombic)
CS2 (l)
0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ
H
rxn
6.6
Chemistry in Action:
Fuel Values of Foods and Other Substances
C6H12O6 (s) + 6O2 (g)
6CO2 (g) + 6H2O (l) H = -2801 kJ/mol
1 cal = 4.184 J
1 Cal = 1000 cal = 4184 J
The enthalpy of solution (Hsoln) is the heat generated or
absorbed when a certain amount of solute dissolves in a
certain amount of solvent.
Hsoln = Hsoln - Hcomponents
Which substance(s) could be
used for melting ice?
Which substance(s) could be
used for a cold pack?
6.7
The Solution Process for NaCl
Hsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol
6.7
Energy Diagrams
Exothermic
Endothermic
(a) Activation energy (Ea) for the forward reaction
50 kJ/mol
300 kJ/mol
(b) Activation energy (Ea) for the reverse reaction
150 kJ/mol
100 kJ/mol
(c) Delta H
-100 kJ/mol
+200 kJ/mol
Entropy (S) is a measure of the randomness or disorder of a
system.
order
disorder
S
S
If the change from initial to final results in an increase in randomness
S > 0
For any substance, the solid state is more ordered than the
liquid state and the liquid state is more ordered than gas state
Ssolid < Sliquid << Sgas
H2O (s)
H2O (l)
S > 0
18.3
First Law of Thermodynamics
Energy can be converted from one form to another but
energy cannot be created or destroyed.
Second Law of Thermodynamics
The entropy of the universe increases in a spontaneous
process and remains unchanged in an equilibrium process.
Spontaneous process:
Suniv = Ssys + Ssurr > 0
Equilibrium process:
Suniv = Ssys + Ssurr = 0
18.4
Entropy Changes in the System (Ssys)
The standard entropy of reaction (S0rxn ) is the entropy
change for a reaction carried out at 1 atm and 250C.
aA + bB
S0rxn =
cC + dD
[ cS0(C) + dS0(D) ] - [ aS0(A) + bS0(B) ]
S0rxn = S S0(products) - S S0(reactants)
What is the standard entropy change for the following
reaction at 250C? 2CO (g) + O2 (g)
2CO2 (g)
S0(CO) = 197.9 J/K•mol
S0(O2) = 205.0 J/K•mol
S0(CO2) = 213.6 J/K•mol
S0rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]
S0rxn = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol
18.4
Entropy Changes in the System (Ssys)
When gases are produced (or consumed)
•
If a reaction produces more gas molecules than it
consumes, S0 > 0.
•
If the total number of gas molecules diminishes,
S0 < 0.
•
If there is no net change in the total number of gas
molecules, then S0 may be positive or negative
BUT S0 will be a small number.
What is the sign of the entropy change for the following
reaction? 2Zn (s) + O2 (g)
2ZnO (s)
The total number of gas molecules goes down, S is negative.
18.4
Spontaneous Physical and Chemical Processes
• A waterfall runs downhill
• A lump of sugar dissolves in a cup of coffee
• At 1 atm, water freezes below 0 0C and ice melts above 0 0C
• Heat flows from a hotter object to a colder object
• A gas expands in an evacuated bulb
• Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
18.2
Gibbs Free Energy
Spontaneous process:
Suniv = Ssys + Ssurr > 0
Equilibrium process:
Suniv = Ssys + Ssurr = 0
For a constant-temperature process:
Gibbs free
energy (G)
G = Hsys -TSsys
G < 0
The reaction is spontaneous in the forward direction.
G > 0
The reaction is nonspontaneous as written. The
reaction is spontaneous in the reverse direction.
G = 0
The reaction is at equilibrium.
18.5
G = H - TS
18.5
The standard free-energy of reaction (G0rxn) is the freeenergy change for a reaction when it occurs under standardstate conditions.
aA + bB
cC + dD
0
Grxn
= [cG0f (C) + dG0f (D) ] - [aG0f (A) + bG0f (B) ]
0
Grxn
= S G0 f(products) - S G0 f(reactants)
Standard free energy of
formation (G0f ) is the free-energy
change that occurs when 1 mole
of the compound is formed from its
elements in their standard states.
G0f of any element in its stable
form is zero.
18.5
What is the standard free-energy change for the following
reaction at 25 0C?
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
0
Grxn
= S G0 f(products) - S G0 f (reactants)
0
Grxn
= [12G0f (CO2) + 6G0f (H2O)] - [ 2G0f (C6H6)]
0
Grxn
= [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ
Is the reaction spontaneous at 25 0C?
G0 = -6405 kJ < 0
spontaneous
18.5
Recap: Signs of Thermodynamic Values
Negative
Enthalpy (ΔH) Exothermic
Positive
Endothermic
Entropy (ΔS)
Less disorder More disorder
Gibbs Free
Energy (ΔG)
Spontaneous Not
spontaneous
The specific heat (s) [most books use lower case 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 = ms
Heat (q) absorbed or released:
q = mst
q = Ct
t = tfinal - tinitial
6.5
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 = mst = 869 g x 0.444 J/g • 0C x –890C = -34,000 J
6.5
Constant-Pressure Calorimetry
qsys = qwater + qcal + qrxn
qsys = 0
qrxn = - (qwater + qcal)
qwater = mst
qcal = Ccalt
Reaction at Constant P
H = qrxn
No heat enters or leaves!
6.5
Sample Problem
• How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C?
Step 1: Heat the ice
Q=mcΔT
Q = 36 g x 2.06 J/g deg C x 8 deg C = 593.28 J = 0.59 kJ
Step 2: Convert the solid to liquid
ΔH fusion
Q = 2.0 mol x 6.01 kJ/mol = 12 kJ
Step 3: Heat the liquid
Q=mcΔT
Q = 36g x 4.184 J/g deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem
• How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C?
Step 4: Convert the liquid to gas
Q = 2.0 mol x 44.01 kJ/mol =
Step 5: Heat the gas
ΔH vaporization
88 kJ
Q=mcΔT
Q = 36 g x 2.02 J/g deg C x 20 deg C = 1454.4 J = 1.5 kJ
Now, add all the steps together
0.59 kJ + 12 kJ + 15 kJ + 88 kJ + 1.5 kJ
= 118 kJ