Thermochemistry
Chapters 6 and 16
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.
ΔH = H (products) – H (reactants)
ΔH = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
ΔH < 0
Hproducts > Hreactants
ΔH > 0
6.4
Thermochemical Equations
Is ΔH negative or positive?
System absorbs heat
Endothermic
ΔH > 0
6.01 kJ are absorbed for every 1 mole of ice that
melts at 00C and 1 atm.
H2O (s)
H2O (l)
ΔH = 6.01 kJ
6.4
Thermochemical Equations
Is ΔH negative or positive?
System gives off heat
Exothermic
ΔH < 0
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) ΔH = -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 ΔH changes
H2O (l)
•
ΔH = 6.01 kJ/mol
H2O (s)
ΔH = -6.01 kJ
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 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)
ΔH = 6.01 kJ
H2O (l)
H2O (g)
ΔH = 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) ΔHreaction = -3013 kJ
1 mol P4
123.9 g P4
x
3013 kJ
= 6470 kJ
1 mol P4
6.4
Standard enthalpy of formation (ΔHf0) 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.
ΔH0f (O2) = 0
ΔH0f (C, graphite) = 0
ΔH0f (O3) = 142 kJ/mol
ΔH0f (C, diamond) = 1.90 kJ/mol
6.6
6.6
0 ) is the enthalpy of
The standard enthalpy of reaction (ΔHrxn
a reaction carried out at 1 atm.
aA + bB
cC + dD
0
ΔHrxn
= [cΔH0f (C) + dΔH0f (D) ] - [aΔH0f (A) + bΔH0f (B) ]
0 (reactants)
0
ΔH

ΔHrxn
= ΔH0 (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)
0
ΔHrxn
=  H0 f(products) -  H0 f(reactants)
0
ΔHrxn
= [12ΔH0f (CO2) + 6ΔH0f (H2O)] - [ 2ΔH0f (C6H6)]
0
ΔHrxn
= [ 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) ΔH0rxn = -393.5 kJ
S(rhombic) + O2 (g)
CS2(l) + 3O2 (g)
SO2 (g)
ΔH0rxn = -296.1 kJ
CO2 (g) + 2SO2 (g)
0 = -1072 kJ
ΔHrxn
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) ΔH0rxn = -393.5 kJ
2SO2 (g) ΔH0rxn = -296.1x2 kJ
CS2 (l) + 3O2 (g)
0 = +1072 kJ
ΔHrxn
C(graphite) + 2S(rhombic)
CS2 (l)
0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ
ΔH
rxn
6.6
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
The enthalpy of solution (ΔHsoln) is the heat generated or
absorbed when a certain amount of solute dissolves in a
certain amount of solvent.
ΔHsoln = 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
The lattice energy of an ionic solid is a
measure of the strength of bonds in that
ionic compound. It is given the symbol U
and is equivalent to the amount of
energy required to separate a solid ionic
compound into gaseous ions.
ΔHsoln = 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
ΔS > 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)
ΔS > 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:
ΔSuniv = ΔSsys + ΔSsurr > 0
Equilibrium process:
ΔSuniv = ΔSsys + ΔSsurr = 0
18.4
Entropy Changes in the System (ΔSsys)
0 ) is the entropy
The standard entropy of reaction (ΔSrxn
change for a reaction carried out at 1 atm and 250C.
aA + bB
ΔS0rxn =
S0
cC + dD
[ cS0(C) + dS0(D) ] - [ aS0(A) + bS0(B) ]
0(products) -  S0(reactants)
S

=
rxn
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
ΔS0rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]
ΔS0rxn = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol
18.4
Entropy Changes in the System (ΔSsys)
When gases are produced (or consumed)
•
If a reaction produces more gas molecules than it
consumes, ΔS0 > 0.
•
If the total number of gas molecules diminishes,
ΔS0 < 0.
•
If there is no net change in the total number of gas
molecules, then ΔS0 may be positive or negative
BUT ΔS0 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, ΔS 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:
ΔSuniv = ΔSsys + ΔSsurr > 0
Equilibrium process:
ΔSuniv = ΔSsys + ΔSsurr = 0
For a constant-temperature process:
Gibbs free
energy (G)
ΔG = ΔHsys -TΔSsys
ΔG < 0
The reaction is spontaneous in the forward direction.
ΔG > 0
The reaction is nonspontaneous as written. The
reaction is spontaneous in the reverse direction.
ΔG = 0
The reaction is at equilibrium.
18.5
ΔG = ΔH - TΔS
18.5
The standard free-energy of reaction (ΔG0rxn ) is the freeenergy change for a reaction when it occurs under standardstate conditions.
aA + bB
cC + dD
0
ΔGrxn
= [cΔG0f (C) + dΔG0f (D) ] - [aΔG0f (A) + bΔG0f (B) ]
0
ΔGrxn
=  ΔG0f (products) -  ΔG0 f (reactants)
Standard free energy of
formation (ΔGf0) is the free-energy
change that occurs when 1 mole
of the compound is formed from its
elements in their standard states.
ΔG0f 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
ΔGrxn
=  ΔG0f (products) -  ΔG0 f (reactants)
0
ΔGrxn
= [12ΔG0f (CO2) + 6ΔG0f (H2O)] - [ 2ΔG0f (C6H6)]
0
ΔGrxn
= [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ
Is the reaction spontaneous at 25 0C?
ΔG0 = -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
Gibbs Free Energy and Chemical Equilibrium
ΔG = ΔG0 + RT lnQ
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature (K)
Q is the reaction quotient
At Equilibrium
ΔG = 0
Q=K
0 = ΔG0 + RT lnK
ΔG0 = - RT lnK
18.6
Gibbs Free Energy and Chemical Equilibrium
Some concepts to remember
Equilibrium constant (K) and Reaction Quotient (Q)
http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Calculating_Equilibrium_Constants.htm
http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Reaction_Quotient.htm
K is the concentrations of products raised to their
coefficient over the concentrations of the reactants raised
to their coefficient. At equilibrium
ΔG = ΔGΕ + RT ln (Q)
Define terms: ΔG = free energy not at standard conditions
ΔGΕ = free energy at standard conditions
R = universal gas constant 8.3145 J/mol K
T = temp. in Kelvin
ln = natural log
Q = reaction quotient: (for gases this is the partial
pressures of the products divided by the partial pressures of the reactants—
all raised to the power of their coefficients) Q = [products]
[reactants]
“RatLink”: ΔG° = -RTlnK
Terms: basically the same as above --however, here the system is at equilibrium,
so ΔG = 0 and K represents the equilibrium constant
under standard conditions.
K = [products]
[reactants]
still raised to power of coefficient
“nFe”: ΔG° = - nFE°
remember this!!
Terms: ΔG° = just like above—standard free
energy
n = number of moles of electrons
transferred (look at ½ reactions)
F = Faraday’s constant 96,485
Coulombs/mole electrons
E° = standard voltage ** one volt =
joule/coulomb**
BIG MAMMA, verse 3: ΔG°rxn = ΣGΕ (products) – Σ GΕ
(reactants)
G0 = - RT lnK
18.6
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 = msΔT
q = CΔT
ΔT = 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
Dt = tfinal – tinitial = 50C – 940C = -890C
q = msDt = 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 = msΔt
qcal = Ccal/Δt
Cp= ΔH/ΔT
Reaction at Constant P
ΔH = qrxn
No heat enters or leaves!
6.5
6.5
Phase Changes
The boiling point is the temperature at which the
(equilibrium) vapor pressure of a liquid is equal to the
external pressure.
The normal boiling point is the temperature at which a liquid
boils when the external pressure is 1 atm.
11.8
The critical temperature (Tc) is the temperature above which
the gas cannot be made to liquefy, no matter how great the
applied pressure.
The critical pressure
(Pc) is the minimum
pressure that must be
applied to bring about
liquefaction at the
critical temperature.
11.8
Where’s Waldo?
Can you find…
The Triple Point?
Critical pressure?
Critical
temperature?
Where fusion
occurs?
Where vaporization
occurs?
Melting point
(at 1 atm)?
Carbon Dioxide
Boiling point
(at 6 atm)?
The melting point of a solid
or the freezing point of a
liquid is the temperature at
which the solid and liquid
phases coexist in equilibrium
Freezing
H2O (l)
Melting
H2O (s)
11.8
Molar heat of sublimation
(DHsub) is the energy required
to sublime 1 mole of a solid.
Deposition
H2O (g)
Sublimation
H2O (s)
DHsub = DHfus + DHvap
( Hess’s Law)
11.8
Molar heat of fusion (ΔHfus) is the energy required to melt
1 mole of a solid substance.
11.8
11.8
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