1. Predicting Equilibrium QCQC

2. Determine the reaction quotient for a system. Determine if a system is at equilibrium and, if not, which reaction is favoured.

3. The reaction quotient, Q enables us to determine if a reaction has reached equilibrium - trial K C Q is compared to the know K c for the reaction. It also indicates which reaction, forward or reverse, must continue (favoured) in order for equilibrium to be achieved. Q is calculated exactly like K c - using the concentrations at that moment.

4. Q = K c The system is at equilibrium. Forward rate = Reverse rate Concentrations remain constant Q = 5.0 K c = 5.0 Q Reactants Products

5. Q > K C The system is NOT at equilibrium. Reverse reaction is favoured. Must increase reactant concentration. Q = 10 K c = 1.0 Q Reactants Products

6. Q < K C The system is NOT at equilibrium. The forward reaction is favoured. Must increase product concentrations. Q = 0.1 K c = 1.0 Q Reactants Products

7. K c = 3.9 x 10 -2 Is the system at equilibrium with: [HI] = 0.75 M[H 2 ] = 0.14 M[I 2 ] = 0.10 M If not, which reaction must continue for equilibrium to be established? 2 HI (g) ↔ H 2 (g) + I 2 (g) Q c = [H 2 ][I 2 ] [HI] 2

8. Q C = 2.5 x 10 -2 K C = 3.9 x 10 -2 Q < K This means the system is NOT at equilibrium. The forward reaction is favoured – must continue [HI] is decreasing, [H 2 ] and [I 2 ] are increasing. Q c = [0.14][0.10]= 2.5 x 10 -2 [0.75] 2 [H 2 ][I 2 ] [HI] 2 QcQc =

9. After 2 hrs the reaction was checked - 8.50 moles N 2, 11.0 moles O 2 and 2.20 moles NO, in a 5.00 L container. If K c = 0.035, is the system at equilibrium? Which reaction is favoured? Which concentrations are increasing/decreasing? N 2(g) + O 2(g) 2 NO (g)

11. Q = 0.0518 K = 0.035 Q >K This means the system is NOT at equilibrium. The reverse reaction is favoured. [NO] is decreasing, [O 2 ] and [N 2 ] are increasing.