Predicting Directions of a Reaction

1. Predicting Directions of a Reaction SCH4U/ AP

2. Predicting the Direction of a Reaction • So far, you have worked with reactions that have reached equilibrium. What if a reaction has not yet reached equilibrium? How can we predict the direction in which the reaction must proceed to reach equilibrium?

3. The Reaction Quotient (Q) • To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression. • Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.

4. The Reaction Quotient (Q) The reaction quotient (Qc)is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression. IF • Qc > Kcsystem proceeds from right to left to reach equilibrium (reverse reaction favoured) • Qc = Kcthe system is at equilibrium • Qc < Kcsystem proceeds from left to right to reach equilibrium (forward reaction favoured)

5. If Q = K, the system is at equilibrium.

6. If Q > K, there is too much product and the equilibrium shifts to the left.

7. If Q < K, there is too much reactant, and the equilibrium shifts to the right.

8. Sample Problem Ammonia is one of the worlds most important chemicals, in terms of the quantity manufactured. It is produced industrially via the Haber Process: N2(g) + 3 H2(g)  2 NH3 (g) At 500°C, the value of Kc for the reaction is 0.40. The following gases are present in a container at this temperature. [N2] = 0.10 mol/L, [H2] = 0.30 mol/L. [NH3] = 0.20 mol/L Is this mixture of gases at equilibrium? If not which direction will the reaction go to reach equilibrium?

9. Solution • Qc= [NH3]2 [N2][H2]3 = (0.20)2 (0.10)(0.30)3 = 14.8 Therefore, Qc > 0.40 The system is not at equilibrium. The reaction will proceed by moving left (i.e., the reverse reaction must take place)

10. Practice Question #1

11. Practice Question #1 Solution

12. Practice Question #2

13. Practice Question #2 Solution

14. Practice Question #3

15. Practice Question #3 Solution

16. Le Chatalier’s Principle • If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. • In other words, When a chemical system at equilibrium is disturbed by a change in a property, the system adjusts in a way that opposes the change.

17. Le Chatalier’s Princple Le Chatelier’s Principle: if you disturb an equilibrium, it will shift to undo the disturbance. Equilibrium shift = movement of a system at equilibrium, resulting in a change in the concentrations of reactants and products • https://www.youtube.com/watch?v=dIDgPFEucFM

18. Le Chataliers Principle • System starts at equilibrium. • A change/stress is then made to system at equilibrium. • Change in concentration • Change in temperature • Change in volume/pressure • System responds by shifting to reactant or product side to restore equilibrium.

19. Le Chataliers Principle Change in Reactant or Product Concentrations • Adding a reactant or product shifts the equilibrium away from the increase. • Removing a reactant or product shifts the equilibrium towards the decrease • To optimize the amount of product at equilibrium, we need to flood the reaction vessel with reactant and continuously remove product.

20. Le Châtelier’s Principle N2(g)+ 3H2(g)↔ 2NH3(g) Change in Reactant or Product Concentrations • If H2 is added while the system is at equilibrium, the system must respond to counteract the added H2 • That is, the system must consume the H2 and produce products until a new equilibrium is established. • Equilibrium shifts to the right. • Therefore, [H2] and [N2] will decrease and [NH3] increases.

21. N2(g) + 3H2(g) 2NH3(g) Equilibrium shifts left to offset stress Add NH3 Change in Reactant or Product Concentrations

22. aA + bB cC + dD Change in Reactant or Product Concentrations Change Shifts the Equilibrium Increase concentration of product(s) left Decrease concentration of product(s) right Increase concentration of reactant(s) right Decrease concentration of reactant(s) left

23. Le Châtelier’s Principle Effect of Temperature Changes • The equilibrium constant is temperature dependent. • For an endothermic reaction, ΔH > 0 and heat can be considered as a reactant. • For an exothermic reaction, ΔH < 0 and heat can be considered as a product.

24. Effect of Temperature Changes

25. Effect of Temperature Changes Adding heat (i.e. heating the vessel) favors away from the increase: • if ΔH = + (Endothermic), adding heat favors the forward reaction, • if ΔH = - (Exothermic), adding heat favors the reverse reaction. Removing heat (i.e. cooling the vessel), favors towards the decrease: • if ΔH = + (Endothermic), cooling favors the reverse reaction, • if ΔH = - , (Exothermic), cooling favors the forward reaction.

26. Le Châtelier’s Principle Effects of Volume and Pressure • As volume is decreased pressure increases. • The system shifts to decrease pressure. • An increase in pressure favors the direction that has fewer moles of gas. • Decreasing the number of molecules in a container reduces the pressure. • In a reaction with the same number of product and reactant moles of gas, pressure has no effect.

28. A (g) + B (g) C (g) Effects of Volume and Pressure Change Shifts the Equilibrium Increase pressure Side with fewest moles of gas Decrease pressure Side with most moles of gas Increase volume Side with most moles of gas Decrease volume Side with fewest moles of gas

29. Le Châtelier’s Principle Adding a Catalyst • does not shift the position of an equilibrium system • system will reach equilibrium sooner

30. Le Châtelier’s Principle Adding Inert Gases • pressure of a gaseous system at equilibrium can be changed by adding a gas while keeping the volume constant • If the gas is inert in the system, for example, if it is a noble gas or if it cannot react with the entities in the system, the equilibrium position of the system will not change

31. Le Châtelier’s Principle Summary

32. As4O6(s) + 6C(s) ⇄ As4(g) + 6CO(g) • add CO • to left • add C • no shift • remove C • no shift • add As4O6 • no shift • remove As4O6 • no shift • remove As4 • to right • decrease volume • to left • add Ne gas • no shift

33. P4(s) + 6Cl2(g) ⇄ 4PCl3(l) • decrease volume • to right • increase volume • to left • add P4 • no shift • remove Cl2 • to left • add Kr gas • no shift • add PCl3 • no shift

34. energy + N2(g) + O2(g) ⇄ 2NO(g) • endo or exo? • endothermic • increase temp • to right • increase volume • no shift • decrease temp • to left

35. Homework • SCH4U • Read 350- 369 • pg 352- # 21-25 • pg 353 # 1-5 • Pg 356 # 26-28 • Pg 366 # 28-33 • Pg 370 # 1-5 • SCH4U AP • Read 15.6 – 15.7 • Q 51 – 72 • Virtual Labhttp://dept.harpercollege.edu/chemistry/chm/100/dgodambe/thedisk/equil/equil.htm