1. Equilibrium Just the Beginning

2. Reactions are reversible  A + B C + D ( forward)  C + D A + B (reverse)  Initially there is only A and B so only the forward reaction is possible  As C and D build up, the reverse reaction speeds up while the forward reaction slows down.  Eventually the rates are equal

3. Reaction Rate Time Forward Reaction Reverse reaction Equilibrium

4. What is equal at Equilibrium?  Rates are equal  Concentrations are not.  Rates are determined by concentrations and activation energy.  The concentrations do not change at equilibrium.  or if the reaction is verrrry slooooow.

5. Law of Mass Action  For any reaction  jA + kB lC + mD  K = [C] l [D] m PRODUCTS power [A] j [B] k REACTANTS power  K is called the equilibrium constant. It has no units.  is how we indicate a reversible reaction

6. Try one together 2SO 2(g) + O 2(g) 2SO 3(g) where the equilibrium concentrations are [SO 2 ] = 1.50M; [O 2 ] = 1.25M and [SO 3 ] = 3.5M K = [SO 3 ] 2 / [SO 2 ] 2 [O 2 ] K = (3.50) 2 /(1.50) 2 (1.25) K=4.36

7. Try another one together 2SO 2(g) + O 2(g) 2SO 3(g) where the equilibrium concentrations are [SO 2 ] = 1.50M; [O 2 ] = 1.25M and [SO 3 ] = 3.5M Now find the k for the reverse reaction K = [SO 2 ] 2 [O 2 ]/ [SO 3 ] 2 K = (1.50) 2 (1.25)/(3.50) 2 K=0.230

8. Playing with K  If we write the reaction in reverse. Then the new equilibrium constant is  K ’ = 1/K

9. Try one together SO 2(g) + 1/2 O 2(g) SO 3(g) where the equilibrium concentrations are [SO 2 ] = 1.50M; [O 2 ] = 1.25M and [SO 3 ] = 3.5M K = [SO 3 ] / [SO 2 ][O 2 ] 1/2 K = (3.50)/(1.50)(1.25) 1/2 K= 2.09

10. Playing with K  If we multiply the equation by a constant, n  Then the equilibrium constant is  K’ = K n

11. K is CONSTANT  At any temperature.  Temperature affects rate.  The equilibrium concentrations don’t have to be the same, only K.  Equilibrium position is a set of concentrations at equilibrium.  There are an unlimited number.

12. Equilibrium Constant One for each Temperature

13. Calculate K  N 2 + 3H 2 2NH 3 Initial At Equilibrium  [N 2 ] 0 =1.000 M [N 2 ] = 0.921M  [H 2 ] 0 =1.000 M [H 2 ] = 0.763M  [NH 3 ] 0 =0 M [NH 3 ] = 0.157M  K = 6.03 x 10 -2

14. Calculate K  N 2 + 3H 2 2NH 3 Initial At Equilibrium  [N 2 ] 0 = 0 M [N 2 ] = 0.399 M  [H 2 ] 0 = 0 M [H 2 ] = 1.197 M  [NH 3 ] 0 = 1.000 M [NH 3 ] = 0.203M  K = 6.02 x 10 -2

15. Equilibrium and Pressure  Some reactions are gaseous  PV = nRT  P = (n/V)RT  P = CRT  C is a concentration in moles/Liter  C = P/RT

16. Equilibrium and Pressure 2SO 2 (g) + O 2 (g) 2SO 3 (g) Kp = (P SO3 ) 2 (P SO2 ) 2 (P O2 ) K = [SO 3 ] 2 [SO 2 ] 2 [O 2 ]

17. Equilibrium and Pressure  K = (P SO3 /RT) 2 (P SO2 /RT) 2 (P O2 /RT)  K = (P SO3 ) 2 (1/RT) 2 (P SO2 ) 2 (P O2 ) (1/RT) 3  K = Kp (1/RT) 2 = Kp RT (1/RT) 3

18. General Equation  jA + kB lC + mD  K p = (P C ) l (P D ) m = (C C xRT ) l (C D xRT ) m (P A ) j (P B ) k (C A xRT ) j (C B xRT ) k  K p = (C C ) l (C D ) m x(RT) l+m (C A ) j (C B ) k x(RT) j+k  K p = K (RT) (l+m)-(j+k) = K (RT)  n   n= (l+m)-(j+k) = Change in moles of gas

19. Homogeneous Equilibria  So far every example dealt with reactants and products where all were in the same phase.  We can use K in terms of either concentration or pressure.  Units depend on reaction.

20. Heterogeneous Equilibria  If the reaction involves pure solids or pure liquids the concentration of the solid or the liquid doesn’t change.  As long as they are not used up we can leave them out of the equilibrium expression.  For example

21. For Example  H 2 (g) + I 2 (s) 2HI(g)  K = [HI] 2 [ H 2 ][ I 2 ]  But the concentration of I 2 does not change.  K= [HI] 2 [ H 2 ]

22. Write the equilibrium constant for the heterogeneous reaction

23. The Reaction Quotient, Q  Tells you the direction the reaction will go to reach equilibrium  Calculated the same as the equilibrium constant, but for a system not at equilibrium  Q = [Products] coefficient [Reactants] coefficient  Compare value to equilibrium constant

24. What Q tells us  If Q<K  Not enough products  Shift to right  If Q>K  Too many products  Shift to left  If Q=K system is at equilibrium

25. Example  for the reaction  2NOCl(g) 2NO(g) + Cl 2 (g)  K = 1.55 x 10 -5 M at 35ºC  In an experiment 0.10 mol NOCl, 0.0010 mol NO(g) and 0.00010 mol Cl 2 are mixed in 2.0 L flask.  Which direction will the reaction proceed to reach equilibrium?

26. Solution  K = 1.55 x 10 -5 M  Q = [NO] 2 [Cl] / [NOCl] 2  Q = (.001mol/2.0L) 2 (.0001mol/2.0L) / (.1mol/2.0L) 2  Q = (5 x 10 -4 ) 2 (5 x 10 -5 ) / (5 x 10 -2 ) 2  Q = 5 x 10 -9  Q<K, therefore more product will need to be formed to reach equilibrium

27. Solving Equilibrium Problems  Given the starting concentrations and one equilibrium concentration.  Use stoichiometry to figure out other concentrations and K.  Learn to create a table of initial and final conditions.

28. Consider the following reaction at 600ºC  2SO 2 (g) + O 2 (g) 2SO 3 (g)  In a certain experiment 2.00 mol of SO 2, 1.50 mol of O 2 and 3.00 mol of SO 3 were placed in a 1.00 L flask. At equilibrium 3.50 mol of SO 3 were found to be present. Calculate  The equilibrium concentrations of O 2 and SO 2, K and K P

29. Solution K = [SO 3 ] 2 / [SO 2 ] 2 [O 2 ] K = (3/3.5) 2 /(2/3.5) 2 (1.5/3.5) K = (.857) 2 / (.571) 2 (.429) K = 5.25 K p = k(RT)  n K p = 5.25[.0821*873] -1 K p =.073