Chapter 15Chemical Equilibrium.ppt

1、Chapter 15 Chemical Equilibrium,Copyright McGraw-Hill 2009,Double arrows ( ) denote an equilibrium reaction.,15.1 The Concept of Equilibrium,Most chemical reactions are reversible.,reversible reaction = a reaction that proceeds simultaneously in both directions,Examples:,Copyright McGraw-Hill 2009,E

2、quilibrium,Consider the reaction,At equilibrium,the forward reaction: N2O4(g) 2 NO2(g), andthe reverse reaction: 2 NO2(g) N2O4(g)proceed at equal rates.,Chemical equilibria are dynamic, not static the reactions do not stop.,Copyright McGraw-Hill 2009,Equilibrium,Lets use 2 experiments to study the r

3、eactioneach starting with a different reactant(s).,Copyright McGraw-Hill 2009,Equilibrium,Experiment #1,Copyright McGraw-Hill 2009,Equilibrium,Experiment #2,Copyright McGraw-Hill 2009,Equilibrium,Are the equilibrium pressures of NO2 and N2O4 related? Are they predictable?,Copyright McGraw-Hill 2009,

4、15.2 The Equilibrium Constant,At equilibrium,or,where Kc is the equilibrium constant,Copyright McGraw-Hill 2009,The Equilibrium Constant,Copyright McGraw-Hill 2009,The Equilibrium Constant,For the NO2 / N2O4 system:,Note: at 100C, K = 6.45,Copyright McGraw-Hill 2009,The Equilibrium Constant,reaction

5、 quotient = Qc = the value of the “equilibrium constant expression” under any conditions.,For,Q K reverse reaction favored Q = K equilibrium present Q K forward reaction favored,Copyright McGraw-Hill 2009,The Equilibrium Constant,The Law of Mass Action: Cato Maximilian Guldberg & Peter Waage, Forhan

6、dlinger: Videnskabs-Selskabet i Christiana 1864, 35.,For a reaction:,For gases:,For solutions:, = mol/L,P in atm,Copyright McGraw-Hill 2009,The Equilibrium Constant,Note: The equilibrium constant expression has products in the numerator, reactants in the denominator.,Reaction coefficients become exp

7、onents. Equilibrium constants are temperature dependent. Equilibrium constants do not have units. (pg. 622) If K 1, products favored (reaction goes nearly to completion). If K 1, reactants favored (reaction hardly proceeds).,Copyright McGraw-Hill 2009,15.3 Equilibrium Expressions,homogeneous equilib

8、ria = equilibria in which all reactants and products are in the same phase.,CaO and CaCO3 are solids. Pure solids and liquids are omitted from equilibrium constant expressions.,Ex:,The equilibrium constant expression is,K = CO2,heterogeneous equilibria = equilibria in which all reactants and product

9、s are not in the same phase.,Copyright McGraw-Hill 2009,Exercise: Write the expressions for Kp for the following reactions:,Solution:,Copyright McGraw-Hill 2009,Equilibrium Expressions,A. Reverse Equations,For,Conclusion:,For,Copyright McGraw-Hill 2009,Equilibrium Expressions,B. Coefficient Changes,

10、For,Conclusion:,For,Copyright McGraw-Hill 2009,Equilibrium Expressions,C. Reaction Sum (related to Hess Law),For,For,Add 1 + 4,Copyright McGraw-Hill 2009,Equilibrium Expressions,Copyright McGraw-Hill 2009,Exercise: At 500C, KP = 2.5 1010 for,Compute KP for each of the following:,(a) At 500C, which i

11、s more stable, SO2 or SO3?,(g),O,2,1,(g),SO,(g),SO,(d),2,2,3,+,(g),SO,(g),O,(g),SO,(b),3,2,2,+,2,1,(g),SO,3,(g),O,(g),SO,3,(c),3,2,2,+,2,3,Copyright McGraw-Hill 2009,15.4 Using Equilibrium Expressions to Solve Problems,Q K reverse reaction favored Q = K equilibrium present Q K forward reaction favor

12、ed,Predicting the direction of a reaction,Compare the computed value of Q to K,Copyright McGraw-Hill 2009,Exercise #1: At 448C, K = 51 for the reaction,Predict the direction the reaction will proceed, if at 448C the pressures of HI, H2, and I2 are 1.3, 2.1 and 1.7 atm, respectively.,Solution:,0.47 5

13、1 system not at equilibrium,Numerator must increase and denominator must decrease. Consequently the reaction must shift to the right.,Copyright McGraw-Hill 2009,Exercise #2: At 1130C, K = 2.59 102 for,At equilibrium, PH2S = 0.557 atm and PH2 = 0.173 atm, calculate PS2 at 1130C.,Solution:,PS2 = 0.268

14、 atm,Copyright McGraw-Hill 2009,Exercise #3: K = 82.2 at 25C for,Initially, PI2 = PCl2 = 2.00 atm and PICl = 0.00 atm. What are the equilibrium pressures of I2, Cl2, and ICl?,Solution:,Initial 2.00 atm 2.00 atm 0.00 atm Change x x +2x Equilibrium (2.00 x) (2.00 x) 2x,Copyright McGraw-Hill 2009,squar

15、e root ,2 x = 18.132 9.066 x 11.066 x = 18.132 x = 18.132 / 11.066 = 1.639,PI2 = PCl2 = 2.00 x = 2.00 1.639 = 0.36 atm PICl = 2x = (2)(1.639) = 3.28 atm,Exercise #3: (cont.),Copyright McGraw-Hill 2009,Exercise #4: At 1280C, Kc = 1.1 103 for,Initially, Br2 = 6.3 102 M and Br = 1.2 102 M. What are the

16、 equilibrium concentrations of Br2 and Br at 1280C?,Initial 6.3 102 M 1.2 102 M Change -x +2x Equilibrium (6.3 102) - x (1.2 102) + 2x,Solution:,Copyright McGraw-Hill 2009,4 x2 + 0.0491x + (7.47 10-5) = 0,quadratic equation: a x2 + b x + c = 0,solution:,x = 1.779 103 and 1.050 102,Q: Two answers? Bo

17、th negative? Whats happening?,Equilibrium Conc. x = 1.779 103 1.050 102 Br2 = (6.3 102) x = 0.0648 M 0.0735 MBr = (1.2 102) + 2x = 0.00844 M 0.00900 M,Copyright McGraw-Hill 2009,Exercise #5: A pure NO2 sample reacts at 1000 K,KP is 158. If at 1000 K the equilibrium partial pressure of O2 is 0.25 atm

18、, what are the equilibrium partial pressures of NO and NO2.,Solution:,Initial ? 0 atm 0 atm Change Equilibrium 0.25 atm,+0.25,+0.50,+0.50 atm,0.50,PNO2,Copyright McGraw-Hill 2009,Exercise #5: (cont.),= 3.956 104,Copyright McGraw-Hill 2009,Exercise #6: The total pressure of an equilibrium mixture of

19、N2O4 and NO2 at 25C is 1.30 atm. For the reaction:KP = 0.143 at 25C. Calculate the equilibrium partial pressures of N2O4 and NO2.,PNO2 + PN2O4 = 1.30 atm,Copyright McGraw-Hill 2009,PNO22 + 0.143 PNO2 0.1859 = 0,PN2O4 = 1.30 atm - PNO2,PN2O4 = 1.30 atm - PNO2 = 1.30 0.366 = 0.934 atmPN2O4 = 0.93 atm,

20、Copyright McGraw-Hill 2009,15.5 Factors That Affect Chemical Equilibrium,“If an equilibrium system variable is changed, the equilibrium will shift in the direction (right or left) that tends to reduce the change.”,Example: N2, H2, and NH3 are at equilibrium in a container at 500C.,(continued on next

21、 5 slides),Le Chteliers Principle,Copyright McGraw-Hill 2009,Case I : Change: N2 is addedShift: ?,to the right,Q: Why? Ans: N2 has increased. Which direction will decrease N2?,Copyright McGraw-Hill 2009,Case II: Change: compress the systemShift: ?,to the right,Q: Why? Ans: Total pressure has increas

22、ed. Which direction will decrease the total pressure? Recall: P n,N2 H2NH3,Copyright McGraw-Hill 2009,Case III: Change: increase the temperatureShift: ?,to the left,Q: Why? Ans: Temperature has increased. Which direction decreases the temperature?Recall, the reaction is exothermic.,Copyright McGraw-

23、Hill 2009,Case IV: Change: add helium at constant volumeShift: ?,none,Q: Why? Ans: Helium is not a reactant or product. Adding helium (at constant V) does not change PN2, PH2 or PNH3. Hence the equilibrium will not shift.,Copyright McGraw-Hill 2009,Case V: Change: add helium at constant total pressu

24、reShift: ?,to the left,Q: Why? Ans: If the total pressure is constant, PN2 + PH2 + PNH3 must decrease. Which direction increases this sum?Recall: P n,Copyright McGraw-Hill 2009,Exercise: Hydrogen (used in ammonia production) is produced by the endothermic reaction,750C,Ni,Assuming the reaction is initially at equilibrium, indicate the direction of the shift (L, R, none) if,H2O(g) is removed. The temperature is increased. The quantity of Ni catalyst is increased. An inert gas (e.g., He) is added. H2(g) is removed. The volume of the container is tripled.,Left Right None None Right Right,