1 / 32

Reaction Equilibrium in Ideal Gas Mixture

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.

cheryl
Download Presentation

Reaction Equilibrium in Ideal Gas Mixture

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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 = -SmdT + VmdP • At constant T, • dμ = VmdP = (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*iat 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 + bBcC + 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)

  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 • 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 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.

  19. Example 3 • Use the plot lnKp0vs1/T for for temperature in the range of 300 to 500K • Estimate the ΔH0. Plot of lnKp0vs1/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 ξeqand 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 cm3atm 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 cm3atm 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.

More Related