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Thermochemistry
Chapters 6 and 18
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.
DH = H (products) – H (reactants)
DH = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
DH < 0
Hproducts > Hreactants
DH > 0
6.4
Thermochemical Equations
Is DH negative or positive?
System absorbs heat
Endothermic
DH > 0
6.01 kJ are absorbed for every 1 mole of ice that
melts at 00C and 1 atm.
H2O (s)
H2O (l)
DH = 6.01 kJ
6.4
Thermochemical Equations
Is DH negative or positive?
System gives off heat
Exothermic
DH < 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) DH = -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 DH changes
H2O (l)
•
DH = 6.01 kJ/mol
H2O (s)
DH = -6.01 kJ
If you multiply both sides of the equation by a factor n,
then DH must change by the same factor n.
2H2O (s)
2H2O (l)
DH = 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)
DH = 6.01 kJ
H2O (l)
H2O (g)
DH = 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
DHreaction = -3013 kJ
3013 kJ
= 6470 kJ
1 mol P4
6.4
Standard enthalpy of formation (DHf0) 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.
DH0f (O2) = 0
DH0f (C, graphite) = 0
DH0f (O3) = 142 kJ/mol
DH0f (C, diamond) = 1.90 kJ/mol
6.6
6.6
0 ) is the enthalpy of
The standard enthalpy of reaction (DHrxn
a reaction carried out at 1 atm.
aA + bB
cC + dD
DH0rxn = [ cDH0f (C) + dDH0f (D) ] - [ aDH0f (A) + bDH0f (B) ]
0 (reactants)
DH
S
DH0rxn = S DH0 (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)
DH0rxn = S DH0 f(products) - S DH0 f(reactants)
DH0rxn = [ 12DH0f (CO2) + 6DH0f (H2O)] - [ 2DH0f (C6H6)]
DH0rxn = [ 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) DH0rxn = -393.5 kJ
S(rhombic) + O2 (g)
CS2(l) + 3O2 (g)
SO2 (g)
DH0rxn = -296.1 kJ
CO2 (g) + 2SO2 (g)
0 = -1072 kJ
DHrxn
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) DH0rxn = -393.5 kJ
2SO2 (g) DH0rxn = -296.1x2 kJ
CS2 (l) + 3O2 (g)
0 = +1072 kJ
DHrxn
C(graphite) + 2S(rhombic)
CS2 (l)
0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ
DH
rxn
6.6
Chemistry in Action:
Fuel Values of Foods and Other Substances
C6H12O6 (s) + 6O2 (g)
6CO2 (g) + 6H2O (l) DH = -2801 kJ/mol
1 cal = 4.184 J
1 Cal = 1000 cal = 4184 J
The enthalpy of solution (DHsoln) is the heat generated or
absorbed when a certain amount of solute dissolves in a
certain amount of solvent.
DHsoln = 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
DHsoln = 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
DS > 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)
DS > 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:
DSuniv = DSsys + DSsurr > 0
Equilibrium process:
DSuniv = DSsys + DSsurr = 0
18.4
Entropy Changes in the System (DSsys)
The standard entropy of reaction (DS0rxn ) is the entropy
change for a reaction carried out at 1 atm and 250C.
aA + bB
DS0rxn =
cC + dD
[ cS0(C) + dS0(D) ] - [ aS0(A) + bS0(B) ]
DS0rxn = 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
DS0rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]
DS0rxn = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol
18.4
Entropy Changes in the System (DSsys)
When gases are produced (or consumed)
•
If a reaction produces more gas molecules than it
consumes, DS0 > 0.
•
If the total number of gas molecules diminishes,
DS0 < 0.
•
If there is no net change in the total number of gas
molecules, then DS0 may be positive or negative
BUT DS0 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, DS 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:
DSuniv = DSsys + DSsurr > 0
Equilibrium process:
DSuniv = DSsys + DSsurr = 0
For a constant-temperature process:
Gibbs free
energy (G)
DG = DHsys -TDSsys
DG < 0
The reaction is spontaneous in the forward direction.
DG > 0
The reaction is nonspontaneous as written. The
reaction is spontaneous in the reverse direction.
DG = 0
The reaction is at equilibrium.
18.5
DG = DH - TDS
18.5
The standard free-energy of reaction (DG0rxn) is the freeenergy change for a reaction when it occurs under standardstate conditions.
aA + bB
cC + dD
0
DGrxn
= [cDG0f (C) + dDG0f (D) ] - [aDG0f (A) + bDG0f (B) ]
0
DGrxn
= S DG0 f(products) - S DG0 f(reactants)
Standard free energy of
formation (DG0f ) is the free-energy
change that occurs when 1 mole
of the compound is formed from its
elements in their standard states.
DG0f 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
DGrxn
= S DG0 f(products) - S DG0 f (reactants)
0
DGrxn
= [12DG0f (CO2) + 6DG0f (H2O)] - [ 2DG0f (C6H6)]
0
DGrxn
= [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ
Is the reaction spontaneous at 25 0C?
DG0 = -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
DG = DG0 + 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
DG = 0
Q=K
0 = DG0 + RT lnK
DG0 = - RT lnK
18.6
DG0 = - 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 = msDt
q = CDt
Dt = 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 = msDt
qcal = CcalDt
Reaction at Constant P
DH = 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 (DHfus) 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