1. Reaction Equilibrium in Ideal Gas Mixture

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

3. 1.1 Chemical Potential of a Pure Ideal Gas
Expression for μ of a pure gas
dG=-S dT + V dP
Division 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

4. 1.2 Chemical Potential in an Ideal Gas Mixture
An ideal gas mixture is a gas mixture having the
following properties:
The equation of state PV=ntotRT obeyed for all T, P & compositions (ntot = total no. moles of gas).
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.
At equilibrium, P*i = P i
Mole fraction of i(ni/ntot)

5. 1.2 Chemical Potential in an Ideal Gas Mixture
Let μi – the chemical potential of gas i in the mixture
Let 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)
At equilibrium, P*i = P i

6. 1.2 Chemical Potential in an Ideal Gas Mixture
The chemical potential of a pure gas, i:
(for standard state, )
The chemical potential of ideal gas mixture:

7. 2. Ideal-Gas Reaction Equilibrium
All the reactants and products are ideal gases
For the ideal gas reaction:
the equilibrium condition:
Substituting into μA , μB , μC and μD :

8. 2. Ideal-Gas Reaction Equilibrium
The equilibrium condition becomes:
where eq – emphasize that these are partial pressure at
equilibrium.

9. 2. Ideal-Gas Reaction Equilibrium
Defining the standard equilibrium constant ( ) for
the ideal gas reaction: aA + bB cC + dD
Thus,

10. 2. Ideal-Gas Reaction Equilibrium
For the general ideal-gas reaction:
Repeat the derivation above,
Then,
Define:
Standard equilibrium constant:
(Standard pressure equilibrium constant)

11. Example 1
A mixture of 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 mmol of CS2.
Find & for the reaction at 7000C.
1bar =750torr

12. Answer (Example 1) Mole fraction: P = 762 torr, Partial pressure:
Standard pressure, P0 = 1bar =750torr.

13. Answer (Example 1)
Use
At 7000C (973K),

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

15. 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,

16. Example 2
Find at 600K for the reaction by using the approximation that ΔH0 is independent of T;
Note:
Substance
kJ/mol
NO2 (g)
33.18
51.31
N2O4 (g)
9.16
97.89

17. Answer (Example 2)
If ΔH0 is independent of T, then the van’t Hoff equation gives
From

18. 3. Temperature Dependence of the Equilibrium Constant
Since , 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.

19. Example 3 Use the plot ln Kp0 vs 1/T for
for temperature in the range of 300 to 500K
Estimate the ΔH0.
Plot of lnKp0 vs 1/T

20. Answer (Example 3) T-1 = 0.0040K-1, lnKp0 = 20.0.
The slope:
From
So,

21. 4. Ideal-Gas Equilibrium Calculations
Thermodynamics enables us to find the Kp0 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 ΔGT0 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 ΔGT0 is found to proceed extremely slow - + catalyst

22. 4. Ideal-Gas Equilibrium Calculations
The 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.

23. 4. Ideal-Gas Equilibrium Calculations
Specific steps to find the equilibrium composition of an
ideal-gas reaction mixture:
Calculate ΔGT0 of the reaction using and a table of ΔfGT0 values.
Calculate Kp0 using [If ΔfGT0 data at T of the reaction are unavailable,
Kp0 at T can be estimated using
which assume ΔH0 is constant]

24. 4. Ideal-Gas Equilibrium Calculations
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.
(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

25. Ideal-Gas Equilibrium Calculations
Substitute the Pi’s (as function of ξeq) into the equilibrium constant expression & solve ξeq.
Calculate the equilibrium mole numbers from ξeq and the expressions for ni in step 3.

26. Example 4
Suppose that a system initially contains mol of N2O4 (g) and mol of NO2 (g) & the equilibrium is attained at 250C and 2.00atm (1520 torr).
Find the equilibrium composition.
Note: 1. 2.
3. ni=n0+νi ξeq. 4. 5.
6. Get 𝜉 and find n
Substance
kJ/mol
NO2 (g)
51.31
N2O4 (g)
97.89

27. Answer (Example 4)
Get:
From
By the stoichiometry,

28. Answer (Example 4)
Since T & P are fixed:
Use

29. Answer (Example 4)
The reaction occurs at: P=2.00atm=1520 torr & P0=1bar=750torr.
Clearing the fractions:
Use quadratic formula:
So, x =
Number of moles of each substance present at equilibrium must be positive.
Thus,
So,
As a result,

30. Example 5 Kp0 =6.51 at 800K for the ideal gas reaction:
If mol of A, mol of B and mol of C are placed in an 8000 cm3 vessel at 800K.
Find the equilibrium amounts of all
species. 1. 2.
3. ni=n0+νi ξeq.
4. Pi=niRT/V 5.
6. Get 𝜉 and find n
1 bar= torr,
1 atm = 760 torr
R=82.06 cm3 atm mol-1 K-1

31. Answer (Example 5)
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 into
We get:
Substitute P0=1bar= torr, R=82.06 cm3 atm mol-1 K-1,

32. Answer (Example 5) 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 =
Guess if x =1, the left hand side = 0.024
Guess if x=0.9, the left hand side =
Therefore, 0.9 < x < 1.0.
For x=0.94, the left hand side = 0.003
For x=0.93, the left hand side=-0.001
As a result,
nA=1.14 mol, nB=0.07mol, nC=4.93mol, nD=0.93mol.