Reaction Equilibrium in Ideal Gas Mixture

1. Reaction Equilibrium in Ideal Gas Mixture

2. Subtopics • Chemical Potential in an Ideal Gas Mixture. • Ideal-Gas Reaction Equilibrium • Temperature Dependence of the Equilibrium Constant • Ideal-Gas Equilibrium Calculations

3. The reaction Gibbs energy • The reaction Gibbs energy: • the slope of the graph of the Gibbs energy (G) plotted against the extent of reaction (ξ). • ΔrG – the difference between the chemical potentials of the reactants & products at the composition of the reaction mixture.

4. 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. 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. • In an ideal gas mixture, P*iat equilibrium equals to the partial pressure of i in the mixture. • For ideal gas mixture,

6. Chemical Potential in an Ideal Gas Mixture • From chemical potential of a pure ideal gas: (for standard state, ) • For a ideal gas mixture: (for standard state, )

7. Ideal-Gas Reaction Equilibrium • For the reaction & specialize to the case of all reactants and products are ideal gases: • For the ideal gas reaction: • the equilibrium condition: • Substitute ,

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

9. Ideal-Gas Reaction Equilibrium • Defining the standard equilibrium constant (K0p) for the ideal gas reaction: aA + bBcC + dD • Then,

10. Ideal-Gas Reaction Equilibrium • For the general ideal-gas reaction: • Repeat the derivation above, • Then, • Define: • Then, • Standard equilibrium constant: (Standard pressure equilibrium constant)

11. Ideal-Gas Reaction Equilibrium • A chemical system which has reached the reversible thermodynamic state shows no further net reaction. Since • This does not mean that nothing is occurring. • Actually, the chemical reactions in both directions continue, but at the same rate. • The result is no net change in concentrations, but there is still a lot going on chemically.

12. 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.

13. Temperature Dependence of the Equilibrium Constant • The ideal-gas equilibrium constant (Kp0) is a function of temperature only. • It is independent of P, V, # of the reaction species • Differentiation with respect to T: • From

14. Temperature Dependence of the Equilibrium Constant • Since , • This is the Van’t Hoff equation. • The greater the |ΔH0 |, the faster Kp0 changes with temperature. • Integration: • Neglect the temperature dependence of ΔH0,

15. Example 2 • Find Kp0 at 600K for the reaction by using the approximation that ΔH0 is independent of T; Note:

16. Temperature Dependence of the Equilibrium Constant • Since , the van’t Hoff equation can be: • The slope of a graph of lnKp0vs1/T at a particular temperature equals –ΔH0/R at that temperature. • If ΔH0 is essentially constant over the temperature range, the graph of lnKp0vs1/T is a straight line. • The graph is useful to find ΔH0 if ΔfH0 of all the species are not known.

17. Example 3 • Use the plot lnKp0vs1/T for for temperature in the range of 200 to 1000K • Estimate the ΔH0. Plot of lnKp0vs1/T

18. 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

19. 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.

20. 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]

21. 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

22. 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 ξeqand the expressions for ni in step 3.

23. 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. • Find the equilibrium composition. • Note:

24. 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.

25. Le Chatelier’s Principle • The statement of Le Chatelier’s principle is generally if a system at equilibrium is stressed, the reaction will shift to relieve that stress. • Chemical reactions can be exposed to stresses including a change in temperature, a change in pressure, a change in concentration of one or more of the participants and others. • Effect of changes in temperature: • The effect of an increase in temperature of a system at equilibrium is a shift in the direction which absorbs heat.

26. Le Chatelier’s Principle (2) Effect of changes in pressure: • When the pressure of a system at equilibrium increases, the reaction occurs in the direction that lowers the pressure by reducing the volume of gas. (3) Effect of varying the concentration: • Increasing the concentration of any component of a system at equilibrium will cause a shift resulting in using up some of the added substance. (4) Effect of catalysts: • Catalysts accelerate both forward & reverse reaction rates equally. • Catalysts can be used to shorten the amount of time it takes to reach equilibrium when the original concentrations do not match equilibrium concentration.

27. Answer (Example 1) • Mole fraction: • P = 762 torr, • Partial pressure: • Standard pressure, P0 = 1bar =750torr.

28. Answer (Example 1) • Use • At 7000C (973K),

29. Answer (Example 2) • If ΔH0 is independent of T, then the van’t Hoff equation gives • From • From

30. Answer (Example 3) • T-1 = 0.0040K-1, lnKp0 = 20.0. • T-1 = 0.0022K-1, lnKp0 = 0.0. • The slope: • From • So,

31. Answer (Example 4) • Get: • From • By the stoichiometry,

32. Answer (Example 4) • Since T & P are fixed: • Use

33. 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,

34. 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 cm3atm mol-1 K-1,

35. 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.