1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium...

32
Reaction Equilibrium in Ideal Gas Mixture 1

Transcript of 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium...

Page 1: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

1

Reaction Equilibrium in Ideal Gas Mixture

Page 2: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 3: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 4: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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)

Page 5: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 6: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 7: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 8: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 9: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 10: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 11: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 12: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 13: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

13

Answer (Example 1)Use

At 7000C (973K),

00 ln PKRTG

molkJ

KmolKJG

/8.64

]000331.0ln[]973][/314.8[0

Page 14: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 15: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 16: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 17: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 18: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 19: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 20: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 21: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 22: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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,,

Page 23: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 24: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 25: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 26: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 27: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 28: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 29: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 30: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 31: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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

Page 32: 1. Subtopics 1.Chemical Potential in an Ideal Gas Mixture. 2.Ideal-Gas Reaction Equilibrium 3.Temperature Dependence of the Equilibrium Constant 4.Ideal-Gas.

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