1. Le Châtelier’s Principle

2. Le Chatelier’s Principle
The French chemist Henri Le Chatelier ( ) studied how the equilibrium position shifts as a result of changing conditions
If a closed system at equilibrium is subjected to a stress, a shift will occur to counteract the stress and establish a new equilibrium.

3. Cont.
Equilibrium systems may be subjected to a variety of different external stresses. No closed equilibrium system can “stress itself.”
The most common external stresses are changes in concentration, temperature, partial pressure of gases, or volume.
Addition of a catalyst or an inert gas to an equilibrium system might be considered to be a stress, but these changes do not result in equilibrium shifts (in ideal systems).

4. Stresses to Equilibrium Systems
The response of equilibrium system to these stresses is explained by Le Chatelier’s principle:
We have:
1) Equilibrium
2) Disturbance of equilibrium (stress)
3) Shift to restore equilibrium
Le Chatelier’s principle predicts how an equilibrium will shift (but does not explain why)

5. Le Châtelier and the equilibrium law
The response to changes in an equilibrium can be explained via the equilibrium law
Consider C2H4(g) + H2O(g)  C2H5OH(g)
Sample values
[C2H5OH]
[0.150]
, 300 =
[0.0222] [0.0225]
Keq =
[C2H4] [H2O]
What happens if 1.0M C2H5OH is added?
Now the expression = 2300
Recall Keq does not change (for a given temp)
To reestablish equilibrium we must reduce 2300 to 300 ( top,  bottom = shift left)
The equilibrium law explains Le Chatelier’s principle (compensating for stresses)

6. Temperature Changes
Endothermic: ΔH = positive (+) OR Heat + H2O (l) ↔ H2O (g) Exothermic: ΔH = negative (-) OR CH4 (g) + 2 O2 (g) —> CO2 (g) + H2O (g) + Heat If you add or remove heat the effect will depend upon whether the reaction is endothermic or exothermic. We will treat HEAT as either a reactant or product as before.

7. Pressure and equilibrium
Gas pressure will change if volume changes since V 1/P
C2H4(g) + H2O(g)  C2H5OH(g)
[C2H5OH]
[0.150]
, 300 =
[0.0222] [0.0225]
Keq =
[C2H4] [H2O]
If volume is reduced by half (P increases), we will have [0.300]/[0.0444][0.0450] = Q<K
To get back to 300, we must have a shift to the right.
When pressure increases, the system will always shift towards the side of the reaction with fewer gas particles.
If pressure is increased by adding an inert gas [ ]’s do not change

8. Catalysts, Le Châtelier questions
The last factor to consider is the addition of a catalyst: this does not affect an equilibrium
A catalyst speeds both forward and reverse reactions (by lowering the activation energy)
It allows us to get to equilibrium faster, but it does not alter equilibrium concentrations
Q- predict the colour of the “NO2 tubes” if they are heated and/or cooled (the reaction is endothermic when written as):
N2O4 (colourless)  2NO2 (brown)

9. Answer N2O4 (colourless) + heat  2NO2 (brown)
Increasing temp causes shift to right (brown)
Cooling causes shift to left (colourless)
Homework:
Read Section 7.4
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