Predicting Equilibrium

OUTCOME QUESTION(S): C12-4-11 PREDICTING EQUILIBRIUM Use the value of the reaction quotient, Q to explain how far a system at equilibrium has gone towards completion. Write solubility product (Ksp) expressions from balanced chemical equations for salts with low solubility and solve related problems. Vocabulary & Concepts

Q Reaction quotient (Q) A trial KC that allows for the prediction of how a reaction will reach equilibrium: Q is calculated exactly like Keq Using the concentrations at that moment Q is compared to the known constant, Keq Q indicates which reaction, forward or reverse, must increase – favoured – to reach equilibrium Similar to: “checking” to see if pasta is cooked – is it ready, undercooked or overcooked? Q

There are three possibilities: Reactants Products Kc = 5.0 1. Q = Keq The system is AT equilibrium. Qc = [products] [reactants] Qc = 5.0 Prediction: Forward rate = Reverse rate Concentrations remain constant Time Concentration teqbm Reactants Products “Checking” – it’s done!

“Checking” – it’s undercooked! There are three possibilities: Reactants Products Kc = 5.0 2. Q < Keq The system is NOT at equilibrium. Qc = [products] [reactants] Qc = 1.0 Prediction: Forward rate favoured (must ↑ ) [Products] must increase Concentration “Checking” – it’s undercooked! Reactants Products Time

“Checking” – it’s overcooked! There are three possibilities: Reactants Products Kc = 5.0 2. Q > Keq The system is NOT at equilibrium. Qc = [reactants] [products] Qc = 10.0 Prediction: Reverse rate favoured (must ↑ ) [Reactants] must increase Concentration “Checking” – it’s overcooked! Reactants Products Time

2 HI(g) ↔ H2 (g) + I2 (g) Kc = 3.9 x 10-2 Show the system is not at equilibrium with the following concentrations, and establish what must happen to reach equilibrium. 2 HI(g) ↔ H2 (g) + I2 (g) Kc = 3.9 x 10-2 0.75 M 0.14 M 0.10 M [H2][I2] [HI]2 Qc = Q < K System is NOT at equilibrium. Prediction: Forward rate favoured (must ↑ ) [Products] must increase [0.14][0.10] [0.75]2 Qc = Qc = 2.5 x 10-2 Undercooked!

Initially 6. 0 mol of SO2, 3. 0 mol of O2 are mixed in a 1 Initially 6.0 mol of SO2, 3.0 mol of O2 are mixed in a 1.0 L reaction container. At equilibrium, 0.50 mol of O2 are found to remain. Kc= 5.6 x 10-3 Is this system at equilbrium? 2 SO3 (g) 2 SO2 (g) + O2 (g) 1 [I] 6.0 3.0 [C] + 5.0 - 5.0 - 2.5 [E] 5.0 1.0 0.50 Kc = [SO2]2[O2] [SO3]2

2 SO3 (g) 2 SO2 (g) + O2 (g) [E] 5.0 1.0 0.50 [SO2]2[O2] [SO3]2 Qc = Kc= 5.6 x 10-3 [SO2]2[O2] [SO3]2 Qc = Q > K System is NOT at equilibrium. Prediction: Reverse rate favoured (must ↑ ) [Reactants] must increase [1.0]2[0.50] [5.0]2 Qc = Qc = 0.02 Overcooked!

CAN YOU / HAVE YOU? C12-4-11 PREDICTING EQUILIBRIUM Use the value of the reaction quotient, Q to explain how far a system at equilibrium has gone towards completion. Write solubility product (Ksp) expressions from balanced chemical equations for salts with low solubility and solve related problems. Vocabulary & Concepts