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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.

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Presentation on theme: "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."— Presentation transcript:

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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 2

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: dG m = dμ = -S m dT + V m dP At constant T, dμ = V m dP = (RT/P) dP If the gas undergoes an isothermal change from P 1 to P 2 :. μ (T, P 2 ) - μ (T, P 1 ) = RT ln (P 2 /P 1 ) Let P 1 be the standard pressure P˚ μ (T, P 2 ) – μ˚(T) = RT ln (P 2 / P˚) μ = μ˚(T) + RT ln (P/ P˚) pure ideal gas 3

4 1.2 Chemical Potential in an Ideal Gas Mixture An ideal gas mixture is a gas mixture having the following properties: 1) The equation of state PV=n tot RT obeyed for all T, P & compositions. (n tot = 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. 4 At equilibrium, P* i = P i Mole fraction of i(n i /n tot )

5 1.2 Chemical Potential in an Ideal Gas Mixture Let μ i – 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 x 1, x 2,….x i The pure gas i is at temp, T & pressure, P* i. P* i at equilibrium equals to the partial pressure of i, P i in the mixture: Phase equilibrium condition becomes: gas in the mixture pure gas (ideal gas mixture) 5 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: (for standard state, ) 6

7 2. Ideal-Gas Reaction Equilibrium All the reactants and products are ideal gases For the ideal gas reaction: the equilibrium condition: Substituting into μ A, μ B, μ C and μ D : 7

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

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

10 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) 10

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 C & 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 C. 11 1bar =750torr

12 Answer (Example 1) 12 Mole fraction: P = 762 torr, Partial pressure: Standard pressure, P 0 = 1bar =750torr.

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

14 3. Temperature Dependence of the Equilibrium Constant The ideal-gas equilibrium constant (Kp 0 ) is a function of temperature only. Differentiation with respect to T: From 14

15 3. Temperature Dependence of the Equilibrium Constant Since, This is the Vant Hoff equation. The greater the | ΔH 0 |, the faster changes with temperature. Integration: Neglect the temperature dependence of ΔH 0, 15

16 Example 2 Find at 600K for the reaction by using the approximation that ΔH 0 is independent of T; Note: 16 Substance kJ/mol NO2 (g) N2O4 (g)

17 Answer (Example 2) 17 If ΔH 0 is independent of T, then the vant Hoff equation gives From

18 3. Temperature Dependence of the Equilibrium Constant Since, the vant Hoff equation can be written as: The slope of a graph of ln K p 0 vs 1/T at a particular temperature equals –ΔH 0 /R at that temperature. If ΔH 0 is essentially constant over the temperature range, the graph of lnK p 0 vs 1/T is a straight line. The graph is useful to find ΔH 0 if Δ f H 0 of all the species are not known. 18

19 Example 3 Use the plot ln K p 0 vs 1/T for for temperature in the range of 300 to 500K Estimate the ΔH Plot of lnK p 0 vs 1/T

20 Answer (Example 3) 20 T -1 = K -1, lnK p 0 = T -1 = K -1, lnK p 0 = 0.0. The slope: From So,

21 4. Ideal-Gas Equilibrium Calculations Thermodynamics enables us to find the K p 0 for a reaction without making any measurements on an equilibrium mixture. K p 0 - obvious value in finding the maximum yield of product in a chemical reaction. If ΔG T 0 is highly positive for a reaction, this reaction will not be useful for producing the desired product. If ΔG T 0 is negative or only slightly positive, the reaction may be useful. A reaction with a negative ΔG T 0 is found to proceed extremely slow - + catalyst 21

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) n 1,0,n 2,0 ….. Of the mixture. The equilibrium composition is related to the initial composition by the equilibrium extent of reaction (ξ eq ). Our aim is to find ξ eq. 22

23 4. Ideal-Gas Equilibrium Calculations Specific steps to find the equilibrium composition of an ideal-gas reaction mixture: 1) Calculate ΔG T 0 of the reaction using and a table of Δ f G T 0 values. 2) Calculate K p 0 using [If Δ f G T 0 data at T of the reaction are unavailable, K p 0 at T can be estimated using which assume ΔH 0 is constant] 23

24 4. Ideal-Gas Equilibrium Calculations 3) Use the stoichiometry of the reaction to express the equilibrium mole numbers (n i ) in terms of the initial mole number (n i,0 ) & the equilibrium extent of reaction (ξ eq ), according to n i =n 0 +ν i ξ eq. 4) (a) If the reaction is run at fixed T & P, use (if P is known) & the expression for n i from n i =n 0 +ν i ξ eq to express each equilibrium partial pressure P i in term of ξ eq. (b) If the reaction is run at fixed T & V, use P i =n i RT/V (if V is known) to express each Pi in terms of ξ eq 24

25 Ideal-Gas Equilibrium Calculations 5) Substitute the P i s (as function of ξ eq ) into the equilibrium constant expression & solve ξ eq. 6) Calculate the equilibrium mole numbers from ξ eq and the expressions for n i in step 3. 25

26 n i =n 0 +ν i ξ eq Get and find n Example 4 Suppose that a system initially contains mol of N2O4 (g) and mol of NO2 (g) & the equilibrium is attained at 25 0 C and 2.00atm (1520 torr). Find the equilibrium composition. Note: 26 Substance kJ/mol NO2 (g)51.31 N2O4 (g)97.89

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

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

29 Answer (Example 4) 29 The reaction occurs at: P=2.00atm=1520 torr & P 0 =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 K p 0 =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 cm 3 vessel at 800K. Find the equilibrium amounts of all species n i =n 0 +ν i ξ eq. 4. P i =n i RT/V Get and find n 1 bar= torr, 1 atm = 760 torr R=82.06 cm 3 atm mol -1 K -1

31 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 P i =n i RT/V & substituting into We get: Substitute P 0 =1bar= torr, R=82.06 cm 3 atm mol -1 K -1,

32 Answer (Example 5) 32 We get, By using trial and error approach, solve the cubic equation. The requirements: n B >0 & n D >0, Hence, 0 < x <1. Guess if x=0, the left hand side = Guess if x =1, the left hand side = Guess if x=0.9, the left hand side = Therefore, 0.9 < x < 1.0. For x=0.94, the left hand side = For x=0.93, the left hand side= As a result, n A =1.14 mol, n B =0.07mol, n C =4.93mol, n D =0.93mol.


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