Presentation on theme: "UNIT 3: EQUILIBRIUM 3.3.1- 3.3.5 Le Châtelier's Principle."— Presentation transcript:
UNIT 3: EQUILIBRIUM Le Châtelier's Principle
Le Châtelier's Principle If a system at equilibrium is subjected to an external stress, the equilibrium will shift to minimize the effects of that stress. Copper sulfate demo Cu(H 2 O) 4 + 4NH 3 Cu(NH 3 ) H 2 0
3.3.2 Changes in Concentration Consider the following equilibrium system: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) (colourless) (red) If more Fe 3+ is added to the reaction, what will happen? According to Le Châtelier's Principle, the system will react to minimize the stress. Since Fe 3+ is on the reactant side of this reaction, the rate of the forward reaction will increase in order to "use up" the additional reactant. This will cause the equilibrium to shift to the right, producing more FeSCN 2+. For this particular reaction we will be able to see that this as happened, as the solution will become a darker red colour.
There are a few different ways we can say what happens here when we add more Fe 3+ ; these all mean the same thing: 1. equilibrium shifts to the right 2. equilibrium shifts to the product side 3. the forward reaction is favoured
What happens? How does this cause the concentrations of the reaction participants to change? Fe 3+ -since this is what was added to cause the stress, the concentration of Fe 3+ will increase. (a shorthand way to indicate this: [Fe 3+ ] ↑ ) SCN - -equilibrium will shift to the right, which will uses up the reactants. The concentration of SCN - decreases (or [SCN - ] ↓ ) FeSCN 2+ -as the forward reaction rate increases, more products are produced, so the concentration of FeSCN 2+ will increase. ([FeSCN 2+ ] ↑ )
How about the value of K eq ? The concentration of some reaction participants have increased, while others have decreased. Once equilibrium has re- established itself, the value of K eq will be unchanged. * The value of K eq does not change when changes in concentration cause a shift in equilibrium. *
What if we add more FeSCN 2+ ? Again, equilibrium will shift to use up the added substance. In this case, equilibrium will shift to favour the reverse reaction, to use up the additional FeSCN 2+. equilibrium shifts to the left equilibrium shifts to the reactant side the reverse reaction is favoured How do the concentrations of reaction participants change? [Fe 3+ ] ↑ as the reverse reaction is favoured. [SCN - ] ↑ as the reverse reaction is favoured. [FeSCN 2+ ] ↑ because this is the substance that was added
Concentration can also be changed by removing a substance from the reaction. This is often accomplished by adding another substance that reacts with something already in the reaction.
Let's remove SCN - from the system. What will happen now? Equilibrium will shift to replace SCN -. The reverse reaction will be favoured because that is the direction that produces more SCN -. How do the concentrations of reaction participants change? [Fe 3+ ] ↑ as the reverse reaction is favoured.[SCN - ] ↑ as the reverse reaction is favoured. [FeSCN 2+ ] ↓ because it is being used up to produce more [SCN - ].
Practice problems 3.3.1
3.3.3 Changes in Volume & Pressure Changing the pressure or volume of an equilibrium system will only affect the reaction if gases are present. Equal volumes of gases contain an equal number of particles and, one mole of gas occupies a volume of 22.4 L. So, two moles of any gas will occupy a volume of 44.8 L and one-half mole would occupy 11.2 L.
How does changing pressure and volume affect equilibrium systems? If you increase the pressure of a system at equilibrium (typically by reducing the volume of the container), the stress will best be reduced by favouring the side with the fewest moles of gas, since fewer moles will occupy less volume. If you decrease the pressure (by increasing the volume), equilibrium will shift to favour the side with the most moles of gas, since more moles will occupy a greater volume. If both sides of the equation have the same number of moles of gas, then there will be no change in the position of equilibrium.
When considering the effect of changing volume or pressure on equilibrium systems, be sure to only count the number of moles of GASES on each side of the equation. Solids, liquids, and aqueous solutions are not affected by changing pressure and volume.
Example: Predict the effect on equilibrium when the pressure is increased for the following reaction: N 2 O 4 (g ) 2 NO 2 (g) The reactant side has 1 mole of a gas; the product side has 2 moles. Increasing the pressure favours the side with the fewest moles of gas, so the equilibrium will shift to the left (or reverse reaction is favoured).
Practice Problems 3.3.3
3.3.4 Changes in Temperature When temperature affects a system at equilibrium, there are two important consequences: 1. an increase in temperature will favour that reaction direction that absorbs heat (i.e. the endothermic reaction) 2. the value of K eq will change
Consider the following equilibrium system N 2 O 4 (g ) 2 NO 2 (g) Δ H° = kJ We see by the sign of Δ H° that the forward reaction is endothermic. Heat is absorbed when the forward reaction proceeds By adding more heat, equilibrium will shift to use up the additional heat, thus favouring this forward direction.
Why will K eq change, When temperature changes cause an equilibrium shift, one entire side of the reaction equation is favoured over the other side. Mathematically, this will alter the value of K eq as follows: K eq = [products] [reactants] if the forward reaction is favoured: 1. more products are produced; fewer reactants 2. K eq will increase if the reverse reaction is favoured: 1. fewer products; more reactants 2. K eq will decrease
Removing heat (making the system colder) will favour the exothermic reaction - the exothermic reaction releases heat to the surroundings, thus "replacing" the heat that has been removed. Adding heat (making system warmer) will favor the endothermic reaction
3.3.5 Addition of a Catalyst How will adding a catalyst affect the following: N 2 (g) + O 2 (g) 2 NO (g) Adding a catalyst will not affect the position of an equilibrium. A catalyst speeds up both the forward and the reverse reactions. Generally, a catalyst will help a reaction to reach the point of equilibrium sooner, but it will not affect the equilibrium otherwise.
Practice Problems Assignment Le Chatelier’s principle Lab