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1
Reaction Equilibrium in Ideal Gas Mixture
2
Subtopics1.Chemical Potential in an Ideal Gas Mixture.
2.Ideal-Gas Reaction Equilibrium3.Temperature Dependence of the Equilibrium Constant
4.Ideal-Gas Equilibrium Calculations
3
1.1 Chemical Potential of a Pure Ideal GasExpression for μ of a pure gasdG=-S dT + V dPDivision by the no of moles gives:
dGm = dμ = -Sm dT + Vm dP
At constant T,dμ = Vm dP = (RT/P) dP
If the gas undergoes an isothermal change from P1 to P2:
.
μ (T, P2) - μ (T, P1) = RT ln (P2/P1)
Let P1 be the standard pressure P˚ μ (T, P2) – μ˚(T) = RT ln (P2/ P˚) μ = μ˚(T) + RT ln (P/ P˚) pure ideal gas
dPP
RTd
P
P 2
1
2
1
1
4
1.2 Chemical Potential in an Ideal Gas MixtureAn ideal gas mixture is a gas mixture having
the following properties:1) The equation of state PV=ntotRT
obeyed for all T, P & compositions. (ntot = total no. moles of gas).
2) If the mixture is separated from pure gas i by a thermally conducting rigid membrane permeable to gas i only, at equilibrium the partial pressure of gas i in the mixture is equal to the pure-gas-i system.
PxP ii At equilibrium, P*i = P i
Mole fraction of i(ni/ntot)
5
1.2 Chemical Potential in an Ideal Gas MixtureLet μi – the chemical potential of gas i in the
mixtureLet μ*i – the chemical potential of the pure
gas in equilibrium with the mixture through the membrane.
The condition for phase equilibrium:The mixture is at T & P, has mole fractions x1,
x2,….xi
The pure gas i is at temp, T & pressure, P*i.
P*i at equilibrium equals to the partial pressure of i, Pi in the mixture:
Phase equilibrium condition becomes:
gas in the mixture pure gas (ideal gas mixture)
*ii
PxP ii
iiiii PTPxTxxPT ,,,....,,, **21 At equilibrium, P*i = P i
6
1.2 Chemical Potential in an Ideal Gas MixtureThe chemical potential of a pure gas, i:
(for standard state,
)
The chemical potential of ideal gas mixture:
(for standard state,
)
00* ln, PPRTTPT iiii barP 10
00 ln)( PPRTT iii barP 10
7
2. Ideal-Gas Reaction EquilibriumAll the reactants and products are ideal gasesFor the ideal gas reaction:
the equilibrium condition:
Substituting into μA , μB ,
μC and μD :
dDcCbBaA
0i ii DCBA dcba
00 ln PPRT iii
000000 lnlnln PPcRTcPPbRTbPPaRTa CCBBAA
00 ln PPdRTd DD
8
2. Ideal-Gas Reaction Equilibrium
The equilibrium condition becomes:
where eq – emphasize that these are partial pressure at
equilibrium.
00000000 lnlnlnln
0
PPbPPaPPdPPcRTbadc BADC
G
BADC
T
beqB
a
eqA
d
eqD
c
eqC
PPPP
PPPPRTG
0,
0,
0,
0,0 ln
000000 lnlnln PPcRTcPPbRTbPPaRTa CCBBAA
00 ln PPdRTd DD
i
iii
iTmiT vGvG ,,
PK
9
2. Ideal-Gas Reaction EquilibriumDefining the standard equilibrium constant (
) for the ideal gas reaction: aA + bB cC +
dD
Thus,
beqB
a
eqA
d
eqD
c
eqCP
PPPP
PPPPK
0,
0,
0,
0,0
barP 10
00 ln PKRTG
PK
10
2. Ideal-Gas Reaction EquilibriumFor the general ideal-gas reaction:Repeat the derivation above,
Then,
Define:
Then,Standard equilibrium constant: (Standard pressure equilibrium constant)
ii iA 0
i
ieqieqi
iiT PPRTPPRTG
0,
0,
0 lnln
i
v
eqiTiPPRTG 0
,0 ln
n
n
ii aaaa .....21
1
iv
ieqiP PPK 0,
0
00 ln PKRTG RTG
P eK00
11
Example 1A mixture of 11.02 mmol of H2S & 5.48mmol of CH4 was
placed in an empty container along with a Pt catalyst & the equilibrium
was established at 7000C & 762 torr.
The reaction mixture was removed from the catalyst & rapidly cooled to room temperature, where the rates of the forward & reverse reactions are negligible.
Analysis of the equilibrium mixture found 0.711 mmol of CS2.
Find & for the reaction at 7000C.
)()(4)()(2 2242 gCSgHgCHgSH
0PK
0G
PxP ii beqB
a
eqA
d
eqD
c
eqCP
PPPP
PPPPK
0,
0,
0,
0,0 00 ln PKRTG
1bar =750torr
12
Answer (Example 1)
Mole fraction:
P = 762 torr,Partial pressure:
Standard pressure, P0 = 1bar =750torr.
mmol
mmolmmol
mmolmmolmmol
mmolmmolmmol
gCSgHgCHgSH711.0
2
84.2)711.0(4
2
77.4711.048.5
4
60.9)711.0(202.11
2 )()(4)()(2
536.0)92.1760.9(2
mmolx SH
266.0)92.1777.4(4
mmolxCH
158.02Hx
0397.02CSx
torrtorrP SH 408)762(536.02
torrtorrPCH 203)762(266.0
4
torrPH 1202
torrPCS 3.302
000331.0750203750408
7503.307501202
4
1020
10400
42
22 PPPP
PPPPK
CHSH
CSHP
13
Answer (Example 1)Use
At 7000C (973K),
00 ln PKRTG
molkJ
KmolKJG
/8.64
]000331.0ln[]973][/314.8[0
14
3. Temperature Dependence of the Equilibrium Constant
The ideal-gas equilibrium constant (Kp0) is a function of temperature only.
Differentiation with respect to T:
From
RTGKP00ln
dT
Gd
RTRT
G
dT
Kd p0
2
001ln
00
SdT
Gd
2
000
2
00ln
RT
STG
RT
S
RT
G
dT
Kd p
3. Temperature Dependence of the Equilibrium Constant
Since ,
This is the Van’t Hoff equation.The greater the | ΔH0 |, the faster changes
with temperature.Integration:
Neglect the temperature dependence of ΔH0,
15
000 STHG 2
00ln
RT
H
dT
Kd P
dT
RT
TH
TK
TKT
TP
P
2
1
2
0
10
20
ln
21
0
10
20 11
lnTTR
H
TK
TK
P
P
2
000ln
RT
STG
dT
Kd p
0PK
16
Example 2
Find at 600K for the reaction by using the approximation that ΔH0 is independent of T;
Note:
)(2)( 242 gNOgON
Substance kJ/mol
kJ/mol
NO2 (g) 33.18 51.31
N2O4 (g) 9.16 97.89
0298Gf0
298Hf
0PK
21
0
10
20 11
lnTTR
H
TK
TK
P
P i
iTfiT HH 0,
0
RTGKP00ln
17
Answer (Example 2)If ΔH0 is independent of T, then the van’t Hoff
equation gives
From
From
21
0
10
20 11
lnTTR
H
TK
TK
P
P
i
iTfiT HH 0,
0 molkJmolkJH /20.57/]16.9)18.33(2[0298
molJmolkJG /4730/]89.97)31.51(2[0298
RTGP eK
00 148.0298314.847300
298, eKP
609.11600
1
15.298
1
./314.8
/57200
148.0ln
0600,
KKKmolJ
molJKP
40600, 1063.1 xKP
RTGKP00ln
18
3. Temperature Dependence of the Equilibrium ConstantSince , the van’t Hoff equation
can be written as:
The slope of a graph of ln Kp0 vs 1/T at a particular
temperature equals –ΔH0/R at that temperature. If ΔH0 is essentially constant over the temperature
range, the graph of lnKp0 vs 1/T is a straight line.
The graph is useful to find ΔH0 if ΔfH0 of all the species are not known.
dTTTd 21)(
R
H
Td
Kd P
1
ln 0
dTRT
H
dT
Kd P2
00ln
19
Example 3Use the plot ln Kp
0 vs 1/T for
for temperature in the range of 300 to 500K
Estimate the ΔH0.
)(2)(3)( 322 gNHgHgN
0 0.001 0.002 0.003 0.004 0.005
-20
-15
-10
-5
0
5
10
15
20
25
T -1 /K -1
lnKp0
Plot of lnKp0 vs
1/T
11987.1 KmolcalR
R
H
Td
Kd P
1
ln 0
20
Answer (Example 3)T-1 = 0.0040K-1, lnKp
0 = 20.0.
T-1 = 0.0022K-1, lnKp0 = 0.0.
The slope:
From
So,
Kx
K4
11011.1
0022.00040.0
0.00.20
KxR
H
Td
Kd P 40
1011.11
ln
molkcal
KxKmolcalH
/22
)1011.1)(987.1( 4110
21
4. Ideal-Gas Equilibrium CalculationsThermodynamics enables us to find the Kp
0 for a reaction without making any measurements on an equilibrium mixture.
Kp0 - obvious value in finding the maximum
yield of product in a chemical reaction.If ΔGT
0 is highly positive for a reaction, this reaction will not be useful for producing the desired product.
If ΔGT0 is negative or only slightly positive,
the reaction may be useful.A reaction with a negative ΔGT
0 is found to proceed extremely slow - + catalyst
22
4. Ideal-Gas Equilibrium CalculationsThe equilibrium composition of an ideal gas
reaction mixture is a function of :T and P (or T and V).the initial composition (mole numbers)
n1,0,n2,0….. Of the mixture.The equilibrium composition is related to the
initial composition by the equilibrium extent of reaction (ξeq).
Our aim is to find ξeq.
eqiieqii nnn 0,,
23
4. Ideal-Gas Equilibrium CalculationsSpecific steps to find the equilibrium composition
of anideal-gas reaction mixture:1) Calculate ΔGT
0 of the reaction using and a table of ΔfGT
0 values.
2) Calculate Kp0 using
[If ΔfGT0 data at T of the reaction
are unavailable, Kp
0 at T can be estimated using
which assume ΔH0 is constant]
0,
0iTfi iT GG
00 ln PKRTG
21
0
10
20 11
lnTTR
H
TK
TK
P
P
24
4. Ideal-Gas Equilibrium Calculations3) Use the stoichiometry of the reaction to
express the equilibrium mole numbers (ni) in terms of the initial mole number (ni,0) & the equilibrium extent of reaction (ξeq), according to ni=n0+νi ξeq.
4) (a) If the reaction is run at fixed T & P, use
(if P is known) & the expression for ni from ni=n0+νi ξeq
to express each equilibrium partial pressure Pi in
term of ξeq.
(b) If the reaction is run at fixed T & V, use Pi=niRT/V (if V is known)
to express each Pi in terms of ξeq
PnnPxPi iiii
25
Ideal-Gas Equilibrium Calculations5) Substitute the Pi’s (as function of ξeq) into
the equilibrium constant expression & solve ξeq.
6) Calculate the equilibrium mole numbers from ξeq and the expressions for ni in step 3.
iv
i iP PPK 00
26
1.
2.
3. ni=n0+νi ξeq.
4.
5.
6. Get 𝜉 and find n
Example 4Suppose that a system initially contains 0.300
mol of N2O4 (g) and 0.500 mol of NO2 (g) & the equilibrium is attained at 250C and 2.00atm (1520 torr).
Find the equilibrium composition.Note:
)(2)( 242 gNOgON
Substance kJ/mol
NO2 (g) 51.31
N2O4 (g) 97.89
0298Gf
0,
0iTfi iT GG 00 ln PKRTG
PnnPxPi iiii
iv
i iP PPK 00
Answer (Example 4)
27
Get:
From
By the stoichiometry,
molkJG /73.489.97)31.51(20298
00 ln PKRTG 0ln1.298./314.8/4730 PKKKmolJmolJ 908.1ln 0 PK
148.00 PK
xmequilibriureachtoreactmolesxLet
gNOgON2
242 )(2)(
molxn ON 300.042
molxnNO 2500.02
Answer (Example 4)
28
Since T & P are fixed:
Use
Px
xPxP NONO
800.0
2500.022
Px
xPxP ONON
800.0
300.04242
iv
i iP PPK 00
10200
422
PPPPK ONNOP
02
202
300.0
800.0
800.0
2500.0148.0
PPx
x
x
PPx
02
2
500.0240.0
42250.0148.0
P
P
xx
xx
2
20
500.0240.0
42250.0148.0
xx
xx
P
P
Answer (Example 4)
29
The reaction occurs at: P=2.00atm=1520 torr & P0=1bar=750torr.
Clearing the fractions:Use quadratic formula:
So, x = -0.324 @ -0.176 Number of moles of each substance present at
equilibrium must be positive. Thus,
So,
As a result,
02325.00365.20730.4 2 xx
a
acbbx
2
42
0300.042
molxn ON
02500.02
molxnNO
300.0 x250.0 x
300.0250.0 x 176.0 x
moln ON 476.042 molnNO 148.0
2
30
Example 5Kp
0 =6.51 at 800K for the ideal gas reaction:
If 3.000 mol of A, 1.000 mol of B and 4.000 mol of C are placed in an 8000 cm3 vessel at 800K.
Find the equilibrium amounts of all species.
DCBA 2
1.
2.
3. ni=n0+νi ξeq.
4. Pi=niRT/V
5.
6. Get 𝜉 and find n
0,
0iTfi iT GG 00 ln PKRTG
iv
i iP PPK 00
1 bar=750.06 torr, 1 atm = 760 torrR=82.06 cm3 atm mol-1 K-1
Answer (Example 5)
31
Let x moles of B react to reach equilibrium, at the equilibrium:
The reaction is run at constant T and V.Using Pi=niRT/V & substituting intoWe get:
Substitute P0=1bar=750.06 torr, R=82.06 cm3 atm mol-1 K-1,
molxnmolxnmolxnmolxn DcBA
DCBA
41)23(
2
iv
i iP PPK 00
102010100 PPPPPPPPK BADCP
RT
VP
nn
nn
VRTnVRTn
PVRTnVRTnK
BA
DC
BA
DCP
0
22
00
KKmolbarcm
barcm
molxx
molxx
80014.83
8000
123
451.6
113
3
32
2
Answer (Example 5)
32
We get,By using trial and error approach, solve the cubic
equation.The requirements: nB>0 & nD>0, Hence, 0 < x <1.Guess if x=0, the left hand side = -2.250Guess if x =1, the left hand side = 0.024Guess if x=0.9, the left hand side = -0.015Therefore, 0.9 < x < 1.0.For x=0.94, the left hand side = 0.003For x=0.93, the left hand side=-0.001As a result, nA=1.14 mol, nB=0.07mol, nC=4.93mol,
nD=0.93mol.
0250.2269.5995.3 23 xxx