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Reaction Equilibrium in Ideal Gas Mixture

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Reaction Equilibrium in Ideal Gas Mixture. 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. 1.1 Chemical Potential of a Pure Ideal Gas.

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### Reaction Equilibrium in Ideal Gas Mixture

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

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
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)

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

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:

(for standard state, )

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

equilibrium.

2. Ideal-Gas Reaction Equilibrium
• Defining the standard equilibrium constant ( ) for

the ideal gas reaction: aA + bB cC + dD

• Thus,
2. Ideal-Gas Reaction Equilibrium
• For the general ideal-gas reaction:
• Repeat the derivation above,
• Then,
• Define:
• Then,
• Standard equilibrium constant:

(Standard pressure equilibrium constant)

Ideal gas reaction equilibrium

Ideal gas reaction equilibrium

Example 1
• A 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.

1bar =750torr

Answer (Example 1)

Mole fraction:

P = 762 torr,

Partial pressure:

Standard pressure, P0 = 1bar =750torr.

Answer (Example 1)

Use

At 7000C (973K),

3. Temperature Dependence of the Equilibrium Constant

Eq 6.14

• The ideal-gas equilibrium constant (Kp0) is a function of temperature only.
• Differentiation with respect to T:
• From
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,
Example 2
• Find at 600K for the reaction by using the approximation that ΔH0 is independent of T;

Note:

Answer (Example 2)

If ΔH0 is independent of T, then the van’t Hoff equation gives

From

From

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.
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

Answer (Example 3)

T-1 = 0.0040K-1, lnKp0 = 20.0.

T-1 = 0.0022K-1, lnKp0 = 0.0.

The slope:

From

So,

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 ishighly 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
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.
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]

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+νiξeq.
• (a) If the reaction is run at fixed T & P, use

(if P is known)

& the expression for ni from ni=n0+νiξeqto 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

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.
Example 4
• Suppose 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:

1.

2.

3. ni=n0+νiξeq.

4.

5.

6. Get 𝜉 and find n

Answer (Example 4)
• Get:
• From
• By the stoichiometry,
Answer (Example 4)
• Since T & P are fixed:
• Use
Answer (Example 4)
• 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,
Example 5
• Kp0 =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.

1.

2.

3. ni=n0+νiξeq.

4. Pi=niRT/V

5.

6. Get 𝜉 and find n

1 bar=750.06 torr,

1 atm = 760 torr

R=82.06 cm3 atm mol-1 K-1

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=750.06 torr, R=82.06 cm3 atm mol-1 K-1,
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 = -2.250
• Guess if x =1, the left hand side = 0.024
• Guess if x=0.9, the left hand side = -0.015
• 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.