chemical thermodynamics ap chemistry chapter 19 notes
TRANSCRIPT
CHEMICAL
THERMODYNAMICS
AP Chemistry
Chapter 19 Notes
REVIEW1st Law of
Thermodynamics
Energy in the universe is conserved
REVIEW
Path Functionvalue depends on how process takes
place (i.e. q, w)
heat
q = mCpT
REVIEWState Function
value independent of path - is a defined reference or zero
point (i.e. H)
Enthalpy - Enthalpy - HHzero pointzero point
HHff00 of of elementselements in in
natural form at natural form at 252500CC
Enthalpy - H
types:
Hrxn = nHf0(prod) -
nHf0(react)
Enthalpy - H
types:
Hfus , Hvap
ENTROPY “S”ENTROPY “S”a measure of a measure of
randomness or randomness or disorder of a disorder of a
systemsystem
Entropy is NOT conserved. The universe seeks
maximum disorder.
2nd Law of Thermodynamics
for a spontaneousprocess, entropy
increases
Entropy - state functionEntropy - state function
zero reference: S = 0zero reference: S = 0
for a perfect crystal at for a perfect crystal at absolute zeroabsolute zero
Ssolid < Sliquid << Sgas
S0 = standard entropyof elements & cmpdsat P = 1 atm; T = 250C
units = J/K mol
[Appendix L, text]
Calculate S using Hess’ Law
S(rxn) = nS0(prod) - nS0(react)
Example:
Ca(s) + C(gr) + 3/2 O2 CaCO3(s)
Suniv = Ssys + Ssurr
if Suniv > 0 process spontaneous
Suniv = Ssys + Ssurr
if Suniv < 0 process ?
Suniv = Ssys + Ssurr
if Suniv = 0 process ?
to relate S and H
consider
H2O(s) H2O(l)
where the water is the system & everything
else is the surroundings
Temperature Dependence
S = J/mol (1/T)
Ssurr = -H/T
Ssys Ssurr Suniv spon
+ +
Ssys Ssurr Suniv spon
+ + + Y
- -
Ssys Ssurr Suniv spon
+ + + Y
- - - N
+ -
Ssys Ssurr Suniv spon
+ + + Y
- - - N
+ - ? D
- +
Ssys Ssurr Suniv spon
+ + + Y
- - - N
+ - ? D
- + ? D
change phase
change phase0change phase T
HS
For a phase change:
GIBB’S FREEENERGY
G
most abstract of thermodynamic state
functions
w1 w = reversible
work
w2
P
V
Definition:
G = w + PVw: reversible workPV: pressure/volume work
isothermal, reversible path
G = w + PV + VP
at constant PP = 0 so VP = 0
G = w + PV
G = w + PV + VPat constant PG = w + PV
at constant VV = 0 so PV = 0G = w = useful work
G cannot be G cannot be measuredmeasured
must measure must measure G G over a processover a process
ZERO REFERENCEG = 0 for elements in
stable form under Standard
Thermodynamic Conditions
T = 25oC ; P = 1 atm
Gf0 = standard Free
Energy of Formation from the elements
Appendix L, text
G follows Hess’ Law:
G0 (rxn) = nGf
0(p) - nGf0(r)
Summary of Laws ofThermodynamics
Zeroth Law:Heat Gain = Heat Loss
Summary of Laws ofThermodynamics
First Law:Law of Conservation
of Energy
Summary of Laws ofThermodynamics
Second Law:Defines Entropy
Summary of Laws ofThermodynamics
Third Law:Defines Absolute Zero
GIBB’S HELMHOLTZEQUATION
G H
T
combine
G = -aT + H
G, H are state functions, thus “a” must be a state function
G = -S(T) + H
Gibb’s HelmholtzEquation
G = H - TS
Units on the State Functions
HkJ
mol
SJ
K mol
GkJ
mol
G H T S
GT
HT
S
SST
Gsurr
surrST
H but
univST
G
univsurr SSS
but
thus, a process is spontaneousif and only if
G is negative
Spontaneitycontrolled by
enthalpy(minimum energy)
Sponaneitycontrolled by
enthalpyentropy
(maximum disorder)
Sponaneitycontrolled by
enthalpyentropy
both
Predict Spontaneity
IF H(-) and S(+)
G = -H - T(+S)
G < 0, => spontaneous
Predict Spontaneity
IF H(+) and S(-)
G = +H - T(-S)
G > 0, => NOT spontan
Summary of Spontaneity
H S G Spont. - + - yes + - + no + + + or - ? - - + or - ?
Uses of theGibb’s Helmholtz
Equation
1. Find the molar entropy of formation for ammonia.
2. Elemental boron, in thin fibers, can be made from a boron halide:
BCl3(g) + 3/2 H2(g) ->B(s) + 3HCl(g)
Calculate: Calculate: HH00, , SS00 and and GG00..
Spontaneous?Spontaneous?
Driving force?Driving force?
3. Using thermodynamic information, determine the boiling point of bromine.
ThermodynamicDefinition ofEquilibrium
Geq = 0
by definition
G = H - TS&
H = E + PV
thus,
G = E + PV - TS
take derivative of both sides
dG = dE + PdV + VdP - TdS - SdT
for a reversible process
TdS = q
derivative used for state function
whilepartial derivative used
for path function
if the only work is PVwork of expansion
PdV = w
First Law ofThermodynamics
E = q - w
dE = q - w = 0
thus
q = wor
TdS = PdV
by substitution
dG = 0 + PdV + VdP - PdV - SdTor
dG = VdP - SdT
let’s assume we havea gaseous system atequilibrium, therefore,examine Kp at thatconstant temperature
at constant T
SdT = 0
thusdG = VdP
dG V dP
VnRT
P
G
G
P
P
1
2
1
2
but
dG nRTdPP
G
G
P
P
1
2
1
2
dG nRTdPP
G G nRTPP
G
G
P
P
1
2
1
2
2 12
1
ln
G G RTPP
n
1 21
2
ln
if “condition 2” isat standard thermodynamic conditions, thenG2 = G0 and P2 = 1 atm
thus
G G RT Pn 0 ln
determine G for
aA + bB cC + dD
where all are gases
G = Gprod - Greact
= cGC + dGD - aGA - bGB
but
cG cG RT PC Cc 0 ln
and likewise for the others
cG dG aG bGC D A B0 0 0 0
Grxn0
and
G G RTP P
P PrxnC
cD
d
Aa
Bb
0 ln( ) ( )
( ) ( )
( ) ( )
( ) ( )?
P P
P PC
cD
d
Aa
Bb
but
Thus,
G = G0 + (RT) ln Q
But
aA + bB cC + dD
G = 0
thus
G RT Krxn P0 ln
or in general
G RT Krxn eq0 ln
THERMODYNAMICSTHERMODYNAMICS
&&
EQUILIBRIUMEQUILIBRIUM
3. Using thermodynamic information, determine
the boiling point of bromine.
FACT: Product-favored FACT: Product-favored
systems have Ksystems have Keqeq > 1. > 1.
Thermodynamics and KThermodynamics and Keqeq
Therefore, both Therefore, both
∆G˚∆G˚rxnrxn and K and Keqeq are are
related to reaction related to reaction
favorability.favorability.
Thermodynamics and KThermodynamics and Keqeq
KKeqeq is related to reaction is related to reaction
favorability and thus to ∆Gfavorability and thus to ∆Goorxnrxn..
The larger the value of K the The larger the value of K the more negative the value of more negative the value of
∆G∆Goorxnrxn
Thermodynamics and KThermodynamics and Keqeq
∆∆GGoorxnrxn= - RT lnK= - RT lnK
where R = 8.314 J/K•molwhere R = 8.314 J/K•mol
Thermodynamics and KThermodynamics and Keqeq
Calculate K for the reactionCalculate K for the reaction
NN22OO44 2 NO2 NO22
∆∆GGoorxnrxn = +4.8 kJ = +4.8 kJ
K = 0.14K = 0.14
When ∆GWhen ∆G00rxnrxn > 0, then K < 1 > 0, then K < 1
∆Gorxn = - RT lnK
Thermodynamics and Keq
∆∆G, ∆G˚, and KG, ∆G˚, and Keqeq
∆∆G is change in free energy at G is change in free energy at non-standard conditions.non-standard conditions.
∆∆G is related to ∆G˚G is related to ∆G˚∆∆G = ∆G˚ + RT ln Q G = ∆G˚ + RT ln Q
where Q = reaction quotient where Q = reaction quotient
∆∆G, ∆G˚, and KG, ∆G˚, and Keqeq
When Q < K or Q > K, reaction is When Q < K or Q > K, reaction is spontaneous.spontaneous.
When Q = K reaction is at equilibriumWhen Q = K reaction is at equilibrium When ∆G = 0 reaction is at equilibriumWhen ∆G = 0 reaction is at equilibrium Therefore, ∆G˚ = - RT ln KTherefore, ∆G˚ = - RT ln K
Product favored Product favored reactionreaction
–∆–∆GGoo and K > 1 and K > 1
In this caseIn this case
∆ ∆GGrxnrxn is < ∆G is < ∆Goorxnrxn , so , so
state with both state with both reactants and reactants and
products present products present is MORE STABLE is MORE STABLE
than complete than complete conversion.conversion.
∆∆G, ∆G˚, G, ∆G˚, andand K Keqeq
Product-favored reaction. Product-favored reaction.
2 NO2 NO22 N N22OO44
∆∆GGoorxnrxn = – 4.8 kJ = – 4.8 kJ
Here ∆GHere ∆Grxnrxn is less than ∆G is less than ∆Goorxnrxn , so the , so the
state with both reactants and state with both reactants and products present is more stable products present is more stable than complete conversion.than complete conversion.
∆G, ∆G˚, and Keq
∆∆GGoorxnrxn is the change in free is the change in free
energy when reactants energy when reactants
convert COMPLETELY to convert COMPLETELY to
products.products.
Thermodynamics and KThermodynamics and Keqeq
OverviewOverview
KKeqeq is related to reaction is related to reaction
favorability.favorability.
When ∆GWhen ∆Goorxnrxn < 0, reaction < 0, reaction
moves energetically moves energetically
“downhill”“downhill”
4. For the following reaction, calculate the temperature at which the reactants are favored.
)g(NH)g(H2
3)g(N
2
1322
THERMODYNAMICSTHERMODYNAMICSOFOF
CHEMICALCHEMICALREACTIONSREACTIONS
5. How much useful work can be obtained from an engine fueled
with 75.0 L of hydrogen at 10 C at 25 atm?
6. The reaction to split water into hydrogen and oxygen can be promoted by firstreacting silver with water.
2 Ag(s) + H2O(g) Ag2O(s) + H2(g)
Ag2O(s) 2 Ag(s) + 1/2 O2(g)
Calculate H0, S0 and G0 for each reaction.
Combine the reactions and calculate H0 and G0 for the combination.
Is the combination spontaneous?
At what temperature does the second reaction become spontaneous?
7. The conversion of coal into hydrogen for fuel is:
C(s) + H2O(g) CO(g) + H2(g)
Calculate G0 and Kp at 250C.
Is the reaction spontaneous?
At what temperature does the reaction become spontaneous?
Calculate the temperature at which K = 1.0 x 10-4.
8. The generation of nitric acid in the upper atmosphere might destroy the ozone layer by: NO(g) + O3(g) NO2(g) + O2(g)
Calculate G0 (reaction)and K at 250C.