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## PowerPoint Slideshow about ' Reaction Equilibrium in Ideal Gas Mixture' - preston-lindsey

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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.

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:

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

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.

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