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 > Kc system proceeds from right to left to reach equilibrium (reverse reaction favoured)
Qc = Kc the system is at equilibrium
Qc < Kc system proceeds from left to right to reach equilibrium (forward reaction favoured)

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

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

7. there is too much reactant, and the equilibrium shifts to the right.
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 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
= (0.20) (0.10)(0.30)3
= 14.8
Therefore, Qc > 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

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
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.
N2 (g) H2 (g) ↔ NH3 (g)

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

22. Change in Reactant or Product Concentrations
aA + bB cC + dD
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. Effects of Volume and Pressure
A (g) + B (g) C (g)
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
add As4O6
remove As4O6
no shift
remove As4
to right
decrease volume
to left
add Ne gas

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

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
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