2. Chapter 13 Chemical Equilibrium  The state where the concentrations of all reactants and products remain constant with time  Equilibrium is reached when rates of forward and reverse reactions are equal

3. Equilibrium Condition  Dynamic Condition- rates of forward and reverse reactions are equal  Law of Mass Action- general description of the equilibrium condition  - generic equation jA + kB  lC + mD A,B,C,D  chemical substances J,k,l,m  coefficients

4. Equilibrium Expression K=[C] l [D] m [A] j [B] k Write the equilibrium expression for: 1) PCl 5 (g)  PCl 3 (g) + Cl 2 (g) 2) Cl 2 O 7 (g) + 8H 2 (g)  2HCl(g) + 7H 2 O(g)

5. Calculating the Values of K  Calculate the equilibrium constant, K, for the following reaction at 25°C. H 2 (g) + I 2 (g)  2HI(g) If the equilibrium concentrations are [H 2 ] = 0.106 M, [I 2 ]= 0.022M, and [HI]=1.29M Using the same example, calculate the equilibrium concentration of HI if H 2 =.81M and I 2 =.035 M and K= 7.1x 10 2

6. Equilibrium Expressions involving Pressures Ideal Gas Law PV= nRT P= pressure(atm) V= volume(L) n= # of moles of gas(mol) R= universal gas constant=.08206Latm/Kmol T= temp. in K(273)

7. Rearrange the Ideal Gas Law for Pressure P=(n/v) RT or P= CRT where C= molar concentration of the gas Kp= [C] l [D] m /[A] j [B] k

8. What is the relationship between K c and K p Kp= K(RT)  n  n= sum of the gaseous products coefficients minus the sum of the gaseous reactants coefficients

9. Ex. 13.4 Calculating Values of Kp Calculate the value of Kp for the following reaction at 25°C if P NOCl = 1.2 atm, P NO = 5.0x10 -2 atm, and P Cl2 = 3.0x10 -1 atm. 2NO(g) + Cl 2 (g)  2NOCl(g)

10. Calculating K from Kp  Calculate the value of K at 25°C for the reaction 2NO(g) + Cl 2 (g)  2NOCl(g) If Kp= 1.9x10 3 Kp= K(RT)  n T = 273 +25= 298K  n= 2-(2+1)= -1

11. Heterogeneous Equilibria  Involves more than one phase CaCO 3 (s)  CaO(s) + CO 2 (g) K= [CaO] [CO 2 ] / [CaCO 3 ] K= [CO 2 ]

12. Write the expressions for K and Kp  NH 4 NO 2 (s)  N 2 (g) + 2H 2 O(g)  HCl(g) + NH 3 (g)  NH 4 Cl(s)

13. Applications of the Equilibrium Constant  Reaction Quotient(Q)- using initial concentrations in the law of mass action N 2 (g) + 3H 2 (g)  2NH 3 (g) Compare Q to K  3 Possible Cases 1) Q= K, equilibrium, no shift 2) Q  K, system shifts to left 3) Q  K, system shifts to right

14. Predicting the shift H 2 (g) + I 2 (g)  2HI(g) K= 7.1x10 2 at 25°C a. Q= 427 Q  K shift to right b. Q=1522 Q  K, shift left

15. Con. [H 2 ] 0 =.81M [I 2 ] 0 =.44M [HI] 0 =.58M Q= ? [H 2 ] 0 =.078M [I 2 ] 0 =.033M [HI] 0 =1.35M Q=?

16. Example 13.8 N 2 O 4 (g)  2NO 2 (g) Kp=.133atm P N 2 O 4 =2.71atm What is P NO 2 ?

17. Example 2,p. 627  1.00 L flask initially contained 0.298 mol PCl 3 (g) and 8.70x10 -3 mol PCl 5 (g). After reaching equilibrium 2.00x10 -3 mol of Cl 2 (g) was found in the flask. Calculate the equilibrium concentrations of all species and value of K. PCl 5 (g)  PCl 3 (g) + Cl 2 (g) K= [PCl 3 ][Cl 2 ]/[PCl 5 ]

18. Solving Equilibrium Steps 1. Write a balanced equation 2. Write the equilibrium expression 3. List the initial concentrations 4. Calculate Q and determine the direction of the shift 5. Use ICE box 6. Solve for unknown(s)

19. Ex. 13.10, p. 628 CO(g) + H 2 O(g)  CO 2 (g) + H 2 (g) K= 5.10 Calculate the equilibrium concentration of all species if 1.00 mol of each component is mixed in a 1.000 L flask.

20. Example, p. 632 H 2 (g) + F 2 (g)  2HF(g) K= 1.15x10 2 Suppose 3.000 mol of H 2 and 6.000 mol of F 2 are mixed in 3.000 L flask. Calculate the equilibrium concentration for all species.

21. Le Chatelier Principle  If a system at equilibrium is subjected to a stress(change), the equilibrium will shift in an attempt to reduce the stress 1. Effect of a change in concentration a. If a reactant/product is added, the system will shift away from the added component b. If a reactant/product is removed, the system will shift toward the removed component

22. 2. Effect of a Change in Pressure a. Add or remove a gaseous reactant/product- add, shift away -remove, shift towards b. Add an inert gas - there is no effect on equilibrium c. Change the volume of the container - when the container volume is reduced, the system will shift toward the side involving the smaller # of gaseous molecules - when the volume is increased, it will shift toward the side with the larger # of molecules

23. Example of Pressure N 2 (g) + 3H 2 (g)  2NH 3 (g) If volume is decreased, then ? Shift right, 4  2 If volume is increased, then ? Shift left, 2  4 Determine the shift if the volume is reduced: P 4 (s) + 6Cl 2 (g)  4PCl 3 (l) Shift right, P 4 (s) and PCl 3 (l), look only at Cl 2 PCl 3 (g) + 3NH 3 (g)  P(NH 2 ) 3 (g) + 3HCl(g) No effect 4 on each side

24. 3. Effect of a change in Temperature a. If  H is positive, it is an endothermic reaction  energy is viewed as a reactant b. If  H is negative, it is an exothermic reaction  energy is viewed as a product N 2 (g) + O 2 (g)  2NO(g)  H= 181kJ 181kJ + N 2 (g) + O 2 (g)  2NO(g) Shift right 2SO 2 (g) + O 2 (g)  2SO 3 (g)  H= -198kJ 2SO 2 (g) + O 2 (g)  2SO 3 (g) + -198kJ Shift left

25. Summarizing Le chatelier N 2 O 4 (g)  2NO 2 (g)  H = 58kJ 58kJ + N 2 O 4 (g)  2NO 2 (g)

26. K sp The Solubility-Product Constant  Deals with equilibria associated with solids dissolving to form aqueous solutions AgCl(s)  Ag + (aq) + Cl - (aq) K= [Ag + ] [Cl - ]/ [AgCl] so Ksp= [Ag + ] [Cl - ]

27. K sp Examples Determine the Ksp of calcium fluoride given that its molar solubility is 2.14x 10 -4 M. CaF 2 (s)  Ca 2+ (aq) + 2F - (aq) Calculate the molar solubility(mol/L) of silver chloride that has a K sp = 1.77x10 -10. AgCl(s)  Ag + (aq) + Cl - (aq)

28. Con. Calculate the molar solubility of Tin(II) hydroxide if Ksp=5.45x10 -27. Sn(OH) 2 (s)  Sn 2+ (aq) + 2OH - (aq)