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

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Reaction equilibrium in ideal gas mixture

Reaction Equilibrium in Ideal Gas Mixture


Subtopics

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

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

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 mixture1

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 mixture2

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

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 equilibrium1

2. Ideal-Gas Reaction Equilibrium

  • The equilibrium condition becomes:

  • where eq – emphasize that these are partial pressure at

    equilibrium.


2 ideal gas reaction equilibrium2

2. Ideal-Gas Reaction Equilibrium

  • Defining the standard equilibrium constant ( ) for

    the ideal gas reaction: aA + bB cC + dD

  • Thus,


2 ideal gas reaction equilibrium3

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

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

Answer (Example 1)

Mole fraction:

P = 762 torr,

Partial pressure:

Standard pressure, P0 = 1bar =750torr.


Answer example 11

Answer (Example 1)

Use

At 7000C (973K),


3 temperature dependence of the equilibrium constant

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 constant1

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

Example 2

  • Find at 600K for the reaction by using the approximation that ΔH0 is independent of T;

    Note:


Answer example 2

Answer (Example 2)

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

From

From


3 temperature dependence of the equilibrium constant2

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

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

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

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 calculations1

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 calculations2

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 calculations3

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

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

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

Answer (Example 4)

  • Get:

  • From

  • By the stoichiometry,


Answer example 41

Answer (Example 4)

  • Since T & P are fixed:

  • Use


Answer example 42

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

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

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 51

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.


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