Reaction Equilibrium in Ideal Gas Mixture

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: (for standard state, )

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: • Then, • Standard equilibrium constant: (Standard pressure equilibrium constant) Ideal gas reaction equilibrium Ideal gas reaction equilibrium

11. 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

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 Eq 6.14 • 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:

17. Answer (Example 2) If ΔH0 is independent of T, then the van’t Hoff equation gives From 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. T-1 = 0.0022K-1, lnKp0 = 0.0. 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 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

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+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

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 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

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 = -0.324 @ -0.176 • 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 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

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=750.06 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 = -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.