Download Thermodynamics

Thermodynamics
- The study of heat changes that accompany chemical reactions and phase
changes.
- It determines three factors about the reaction:
-the direction
-the degree (extent of reaction)
-its spontaneity
- It does not determine the speed of the reaction.
-Speed is determined by kinetics.
Thermodynamic Law
- First Law:
Energy can be converted to one form or another but it can not be created
or destroyed.
Enthalpy
-Enthalpy (ΔH): The heat change of a system at constant pressure.
-It determines the ability of a reaction to produce heat.
-It is a state function.
-A state function’s value is:
1.Dependent on temperature & pressure
2.Independent on the path.
-Letter has a °as a superscript
-There are two types of heat (q) involved in a reaction:
1.Exothermic: heat is released
2.Endothermic: heat is absorbed
-Not a good indication of spontaneity since both can be spontaneous.
-Standard Conditions
Temperature:25oC or 298K Pressure: 1.00 atm Element in its stable state
Standard Molar Enthalpy of Formation
-The heat change for the formation of one mole substance from its
elements in their standard states (ΔHf).
-Units are kJ/K mol
Enthalpy
-Units are kJ/K mol
- Equation:
- Equation:
ΔHrxn = ∑nΔH°f
products
- ∑nΔH°f reactants
n = number of moles
-Elements inEnthalpy
their standard state have a ΔH°f of zero.
where
n is the number of moles
Example:
Find the ΔHf for the reaction
below
C (graphite) + O2 (g)  CO2 (g)
ΔHf for CO2= -393.5 kJ/K mol
ΔHof for the rxn =
Enthalpy
-393.5 kJ/K mole
Find the ΔH°f for the reaction
below:
2 CH3OH (l) + 3 O2 (g) 
2 CO2 (g) + 4 H2O (g)
Enthalpy
Compound
CH3OH (l)
O2 (g)
CO2 (g)
H2O (g)
Total
ΔH°f
nΔH°f
(kJ/K mol) (kJ/K mol)
-238.7
0
-477.4
0
-393.5
-241.8
-787
-967.2
n is the number of moles
Enthalpy
ΔHrxn = ∑nΔH°f products - ∑nΔH°f reactants
(-787 + -967.2) - (-477.4)
-1754.2 - (-477.4)
-1276.8 kJ/K mol
Entropy
- A measure of randomness or disorder of a system.
-For any substance, the solid state is more ordered
than the liquid state and the liquid state is more ordered than the gas
state. (ΔS is positive)
Ssolid SliquidSgas
Predict which compound has the highest S°.
A. H2O (s)
B. Na (s)
C. Br2 (l)
D. H2O (g)
Answer: C3H8 most complex molecule
E. C3H8 (g)
-Behavior in gases
If a reaction
-Produces more gas molecules than it consumes, ΔS°> 0.
Predict the sign of ΔS°
2SO2 (g) + O2(g) 2SO3(g) 2H2 (l) + O2 (g)  2H2O (s) CS2 (l)  CS2 (g) +
CaCO3 (s)  CaO (s) + CO2 (g) +
Thermodynamic Law
-Second Law:
The entropy of the universe always increases for a spontaneous process and
remains unchanged in an equilibrium process.
Spontaneous Processes
-Occur naturally
-Tendency to want to go to a lower energy state
-Increase in entropy
-Spontaneity tells the direction of the energy flow.
-It does not tell anything about the speed of the reaction.
For a spontaneous process
SUNIVERSE = SSYS + SSURR > 0
For an equilibrium process
SUNIVERSE = SSYS + SSURR = 0
-
If
ΔSuniverse > 0 spontaneous
ΔSuniverse < 0 not spontaneous
-The reverse reaction is
ΔSuniverse = 0 at equilibrium
Entropy
-The
standard
entropy
of reaction
– Neither
the forward
nor the reverse
reaction is favored
(ΔS) is the entropy change for a
Entropy
reaction carried out at
1 atm and
-The25°C.
standard entropy of reaction (ΔS) is the entropy change for a reaction
carried out at 1 atm and 25°C.
-Equation:
-Equation:
ΔSrxn entropy
= ∑nΔS°f products - ∑nΔS°f reactants
-Unit: J/K mol
Example:
Find the ΔS° for the reaction
below:
2 H2 (g) + O2 (g)  2 H2O (l)
Unit: J/K mol
entropy
Total
nΔS°
(J/K mol)
Compound
ΔS°
(J/K mol)
H2 (g)
O2 (g)
H2O(l)
131
205
262
205
69.9
139.8
Enthalpy
n is
the number of moles
ΔSrxn = ∑nΔS°products - ∑nΔS°reactants
139.8 - (262 + 205)
139.8 - 467
-327.2 J/K mol
Entropy and Enthalpy
-The entropy of the surroundings is related to the heat gained or lost by
the system.
Entropy
Equation:
Equation:
Used ONLY
for phase
changes
Used ONLY
for phase
changes
ΔSsurr = ΔHsys
T
ΔS is in J/K mol
is *All
in J/K
phase mol
changes are at equilibrium
ΔS
ΔH is in kJ/K mol
T is in Kelvin
*All phase changes are at
equilibrium
ΔH is in kJ/K mol
T is in Kelvin
Example:
Entropy
Estimate the normal boiling
point of
H2O (l) H2O (g)
H2O (l)
H2O (g)
ΔH°f
-285.8
-241.8
S°
69.9
188.7
kJ/K mol
J/K mol
Entropy
Units
ΔHsys =∑nΔH°f products-∑nΔH°f reactants
= -241.8 – (-285.8)
= 44 kJ/K mol
ΔSsurr =∑nΔS°products-∑nΔS°reactants
= 188.7 – (69.9)
Entropy
= 118.8 J/K mol
Covert kJ to J
44 kJ = 44000 J
ΔS = ΔH
T
118.8 J/K mol = 44000 J/K mol
T
T = 370.4 K
Thermodynamic Law
-Third Law: The entropy of a perfect crystal is zero at 0 K.
-Defines absolute zero.
Gibbs Free Energy (G)
-Combines ΔH & ΔS
-Represents the energy that is free (released from the system) to do work.
-Direct indicator of spontaneity.
Equation:
(Gibbs Helmholtz Equation)
ΔG = ΔH –free
T ΔS energy
Units: J/K mol
Gibbs
Can use under any conditions
What is the G for the reaction at 25°C ? N2 (g) + 3 H2 (g)  2 NH3 (g)
Compound
ΔH°f
(kJ/K mol)
ΔS°
(J/K mol)
0
N2 (g)
0
H2 (g)
Gibbs
energy
NH3 (g) free-46.3
191.5
131
193
ΔH°f =∑nΔH°f products-∑nΔH°f reactants
= (2 x -46.3)-(0)
= -92.6 kJ/K mol
ΔS°rxn =∑nΔS°products-∑nΔS°reactants
= (2x193) – (191.5)+(3x131)
Gibbs
free energy
= -198.5 J/K mol
ΔH°= -96600 J/K mol
ΔS°= -198.5 J/K mol
ΔG° = ΔH°-TΔS°
= (-96600) – (298 x -198.5)
= -37447 J/K mol
Relationship with spontaneity
-If G<0, the reaction is spontaneous (forward dir.)
-If G>0, the reaction is not spontaneous (forward dir.)
-If G=0,the reaction is at equilibrium
-A reaction is spontaneous in the forward direction only if ΔG is negative.
-Spontaneity is controlled by enthalpy and entropy.
If ΔH (-) & ΔS (+),then what is the sign of ΔG?
Spontaneous? Answer: ΔG is negative & spontaneous
Standard Free Energy (ΔG°): The free-energy change that occurs when 1
mole of the compound is formed from its elements in their standard
states.
-Occurs under standard state conditions.
Gibbs free energy
- Equation:
Equation:
ΔG°rxn = ∑nΔG°f
products
- ∑nΔG°f reactants
where n is the number of moles
-Units J/K mol
-Units
are
molstate have a ΔG°f of zero.
-Elements in
theirJ/K
standard
Gibbs Free Energy & Equilibrium
Equation:
ΔG° = -2.303RT log Keq
ΔG°= Free Energy
R = 8.31 J/K mol (Gas Constant)
T = Temperature in Kelvin Keq = Equilibrium Constant
Gibbs free energy
Reaction is..
IF
ΔG°>0 & Keq < 1 Not Spontaneous
ΔG°<0 & Keq > 1
Spontaneous
Gibbs
free energy
ΔG°=0 & K = 1 At Equilibrium
eq
Find Keq at 298 K for
CaCO3 (s)
 CaO
(s) + CO2 (g)
Gibbs
free
energy
Solve for ΔG° first
ΔG° = 130900 J
ΔG° = -2.303 RT log Keq
130900 J=(-2.303)(8.31)(298)log Keq
log Keq = -22.95
= 1.12 x 10
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