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Chemical equilibrium

Chemical equilibrium. Thermodynamic equilibrium constant Equilibrium constant. Only related to temperature. The mass action law. Related to temperature and pressure. Predict the direction of a reaction. Left to right. Equilibrium. Right to left.

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Chemical equilibrium

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  1. Chemical equilibrium • Thermodynamic equilibrium constant • Equilibrium constant Only related to temperature The mass action law Related to temperature and pressure

  2. Predict the direction of a reaction Left to right Equilibrium Right to left Question: the thermodynamic equilibrium constant depend only on temperature, how about the equilibrium composition?

  3. Equation and the equilibrium constants • 2 H2 + O2→ 2 H2O 10 H2 + 5 O2 → 10 H2O H2 + ½ O2 → H2O H2O → H2 + ½ O2

  4. Procedure for solving equilibrium problems: • Write the balanced equation for the reaction • Write the equilibrium expression using the law of mass action • Calculate Q and determine the direction of the shift to equilibrium • Define the change needed to reach equilibrium • Define the equilibrium concentrations by applying the change to the initial concentrations • Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknown • Check your calculated equilibrium concentrations by making sure they give the corrected value of K

  5. 6.5 Equilibrium states of several types of reaction • (1) Reactions in solution • Ionization Equilibrium HAc  H+ + Ac- degree of association α c(1 – α) cα cα Q: Response of equilibrium to pH?

  6. AgBr(s) Ag++Br－ Complexation Cu2+ + 4NH3  Cu(NH3)42+

  7. (2) Heterogeneous Chemical Equilibrium Solid/liquid/Gas The Q only related on p (gas) Having no relationship with amounts of solid and liquid dissociation pressure of

  8. 298K fGm(NH4HS, s)= 55.17 kJmol-1 fGm(H2S,g)= 33.02 kJmol-1 fGm(NH3,g)= 16.64 kJmol-1 rGmΘ= ifGm,i Θ = –16.64 – 33.02+55.17= 5.51 kJmol-1

  9. p =( K )1/2×2p = 66.7 kPa Equilibrium pressure

  10. If 40.0kPa H2S in the container p(NH3)=18.9 kPa p(H2S)= 58.9 kPa p= p(NH3)+ p(H2S) = 77.8 kPa

  11. Dehydrate and hygroscopic • partial pressure of water vapor at which the two solids can coexist indefinitely; its value is Kp=1/5 atm When a solid is able to take up moisture from the air, it is described as hygroscopic

  12. . deliquescent

  13. 6.5 Experimental determination of K • Spectroscope, IR, UV, NMR • Refractive index 折光率 • Electrical conductivity 电导率 • Color • Density • Volume • Optical rotation 旋光度 • Pressure

  14. 1.564g N2O4(g) in a volume changeable container，298Kp，V=0.485dm3 n(N2O4)=1.564/92.0=0.017 mol 2NO2(g) N2O4(g) Equ. n(1-) 2n ni=n(1+) pV=ni RT=n(1+)RT, =0.167

  15. rGm= – RT ln K = 5.36 kJmol-1 （Theoretical value: 5.39 kJmol-1）

  16. For example

  17. Homework • Y: P126 15; P129:19 • P134: 28 • A: P249 9.11

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