1. Equilibrium Calculations

2. Law of chemical equilibrium  For an equilibrium  a A + b B c C + d D  K = [C] c [D] d  [A] a [B] b  K is the equilibrium constant for that reaction.  The [ ] mean concentration in molarity

3. Problem  2NH 3 (g) N 2 (g) + 3H 2 (g)  Calculate the equilibrium constant for the above reaction if it comes to equilibrium with the following concentrations: N 2 =.59 M, H 2 =3.1 M, and NH 3 = 1.03 M

4. Answer  K = [N 2 ][H 2 ] 3  [NH 3 ] 2  K = [.59][3.1] 3  [1.03] 2  K = 16.6

5. Equilibrium by phase  Equilibrium depends on the concentration of the reactants.  We can calculate the concentration of a gas or of anything dissolved (aqueous).  Insoluble solids or liquids won’t have a concentration.  They in essence are removed from the equilibrium.

6. So using that  What would the equilibrium expression look like for the following reaction?  2 H 2 O 2(l) 2 H 2 O (l) + O 2(g)  We ignore the liquids (and solids).  K = [O 2 ]

7. Water  2 H 2 O (l) ⇌ 2 H 2 O + (l) + OH - (g)  K w = [OH - ] [H 3 O + ]  K w is the equilibrium constant for water, it equals 1 x 10 -14 M  We have already used the equation

8. Another problem  2 SO 2 (g) +O 2 (g) 2 SO 3 (g)  K = 4.34, for the above reaction. Calculate the concentration of SO 3 if the SO 2 =.28 M and O 2 =.43 M at equilibrium.

9. Answer  K = [SO 3 ] 2  [SO 2 ] 2 [O 2 ]  4.34 = [SO 3 ] 2  [.28] 2 [.43]  [SO 3 ] =.38 M

10. Reaction Quotient, Q  The reaction quotient, Q, gives us a value of current concentrations of all chemicals before they complete reacting towards equilibrium.  It is calculated the same way as K except you use current concentrations.  Comparing the Q value to the K value allows you to determine which way the reaction will shift.

11. K and Q values  K and Q are calculated by products over reactants.  Larger values mean higher products, smaller mean higher reactants.  If K is larger than Q, it means that Q has a higher concentration of reactants so the reaction will shift to the right.  If K is smaller than Q, it means that Q has a higher concentration of products so the reaction will shift to the left.  If K equals Q you are at equilibrium, no shift

12. Easy way to remember  Greater than, less than signs are an arrow!!!  Put K first, then Q and determine which is larger.  K > Q, shift right  K < Q, shift left  K = Q, No shift

13. Q problem  The K is.060 for the reaction:  N 2 + 3 H 2 ⇌2 NH 3  Calculate Q and determine which way the reaction will shift [N 2 ] (M)[H 2 ] (M)[NH 3 ](M) Case 11.0x10 -5 2.0x10 -3 1.0x10 -3 Case 21.5x10 -5 3.54x10 -1 2.0x10 -4 Case 35.01.0x10 -2 1.0x10 -4

14. Answer  Q = [NH 3 ] 2  [N 2 ][H 2 ] 3  Case 1, Q = 1.25 x10 7  K < Q, reaction shifts left  Case 2, Q =.060  K = Q, equilibrium  Case 3, Q =.002  K> Q, reaction shifts right

15. Solution Equilibrium  All dissociations we have done are equilibriums.  Before we simply stated something was soluble or insoluble.  Actually everything dissolves to some extent, and some dissolved substance fall out of solution.  Higher concentrations force more solute to fall out of solution.  So there is a maximum concentration of solute a solution can hold (saturation)

16. Solution equilibrium  For Example:  Lead (II) Bromide  PbBr 2 (s) Pb 2+ (aq) + 2 Br - (aq)  What would the equilibrium expression look like?  K sp = [Pb 2+ ][Br - ] 2  Equilibrium constants for dissociations are called solubility products, and are denoted by K sp.

17. Cont.  The K sp value for PbBr 2 is 5.0 x 10 -6.  PbBr 2 (s) Pb 2+ (aq) + 2 Br - (aq)  I will assume 1 L, and that I start with x moles of PbBr 2.  Using the balanced equation I have x mole/L of Pb 2+ and 2x mole/L of Br -  So  5.0 x 10 -6 = x (2x) 2  5.0 x 10 -6 = 4x 3  x=[Pb 2+ ] =.011 M 2x = [Br - ]=.022M

18. This means  A solution of PbBr 2 would be saturated with [Pb 2+ ] =.011 M and [Br - ]=.021M  These are both very low concentrations so we say the is compound is insoluble.

19. Problems  Calculate the saturation concentrations of solutes in aluminum hydroxide K sp = 5.0 x10 -33, and Barium sulfate K sp = 1.4 x10 -14

20. Answer  Al(OH) 3 Al 3+ (aq) + 3 OH - (aq)  K sp = [Al 3+ ][OH - ] 3  5.0x10 -33 = x (3x) 3  x = [Al 3+ ]= 3.7x10 -9 M  3x = [OH - ] = 1.1 x10 -8 M

21. Answer  BaSO 4 Ba 2+ (aq) + SO 4 2- (aq)  K sp = [Ba 2+ ][SO 4 2- ]  1.4x10 -14 = x (x)  x = [Ba 2+ ]= [SO 4 2- ] = 1.2 x10 -7 M