Mr Chiasson Advanced Chemistry 12 1 Unit C: Chemical Kinetics and Chemical Equilibrium

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1 Mr Chiasson Advanced Chemistry 12 1 Unit C: Chemical Kinetics and Chemical Equilibrium Le Châtelier's Principle Think back to our escalator example, with you walking up a downward moving escalator. With the rate of the moving stairs and your walking evenly matched, you appear to be at a standstill. But what happens if the escalator begins moving just a little faster? If you want to maintain the same position you had, at some specific point between the bottom and the top of the stairs, you'll also need to make some adjustments. Chemical systems at equilibrium tend to make these adjustments as well. A French chemist, Henri Louis Le Châtelier, was the first to describe what we now call: Le Châtelier's Principle If an external stress is applied to a system at equilibrium, the system will adjust itself in such a way as to partially offset the stress as the system reaches a new equilibrium position. (Chang, 2007) Equilibrium is all about rates - the rate of the forward reaction is equal to the rate of the reverse reaction. External stresses are factors that will cause the rate of either the forward or reverse reaction to change, throwing the system out of balance. Le Châtelier's Principle allows us to predict how this will affect our system. In our unit on Kinetics we examined factors that influenced reaction rates. Recall these factors: 1. concentration 2. pressure and volume 3. temperature, and 4. catalysts We'll see how changing these factors affects a system at equilibrium. 3.2 Changes in Concentration Consider the following equilibrium system: (aq) + (aq) (aq) (colourless) (red)

2 Mr Chiasson Advanced Chemistry 12 2 If more is added to the reaction, what will happen? According to Le Châtelier's Principle, the system will react to minimize the stress. Since 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. 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 ; these all mean the same thing: equilibrium shifts to the right equilibrium shifts to the product side the forward reaction is favoured How does this cause the concentrations of the reaction participants to change? Since this is what was added to cause the stress, the concentration of will increase. (a shorthand way to indicate this: [ ] ) (Reminder: the square brackets represent "concentration") Equilibrium will shift to the right, which will use up the reactants. The concentration of will decrease (or [ ] ) as the rate of the forward reaction increases. With the forward reaction rate increases, more products are produced, and the concentration of will increase. ([ ] ) How about the value of Keq? Notice that the concentration of some reaction participants have increased, while others have decreased. Once equilibrium has re-established itself, the value of Keq will be unchanged. The value of Keq does not change when changes in concentration cause a shift in equilibrium. What if we add more? Again, equilibrium will shift to use up the added substance. In this case, equilibrium will shift to favour the reverse reaction, since the reverse reaction will use up the additional. equilibrium shifts to the left equilibrium shifts to the reactant side the reverse reaction is favoured

3 Mr Chiasson Advanced Chemistry 12 3 How do the concentrations of reaction participants change? [ ] as the reverse reaction is favoured [ ] as the reverse reaction is favoured [ ] 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 (in a side reaction) with something already in the reaction. Let's remove from the system (perhaps by adding some Pb 2+ ions - the lead(ii) ions will form a precipitate with, removing them from the solution). What will happen now? Equilibrium will shift to replace - the reverse reaction will be favoured because that is the direction that produces more. equilibrium shifts to the left equilibrium shifts to the reactant side the reverse reaction is favoured How do the concentrations of reaction participants change? [ ] as the reverse reaction is favoured [ ] as the reverse reaction is favoured (but it also because it was removed) [ ] because this is the substance that was added 3.3 Changes in Volume & Pressure Changing the pressure or volume of a container enclosing an equilibrium system will only affect the reaction if gases are present. You may remember from earlier chemistry classes that equal volumes of gases contain an equal number of particles and, under standard conditions of temperature and pressure (STP), one mole of gas occupies a volume of 22.4 L. This is known as the molar volume of gases. 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 reaction favouring the side with the fewest moles of gas, since fewer moles will occupy the smallest volume.

4 Mr Chiasson Advanced Chemistry 12 4 Conversely, if you decrease the pressure (by increasing the volume of the container), 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 will not be affected by changing pressure and volume. Here's an example. Predict the effect on equilibrium when the pressure is increased for the following reaction: N 2 O 4 (g ) 2NO 2 (g) The reactant side of the equation 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 (the reverse reaction will be favoured). 3.4 Changes in Temperature When temperature is the stress that affects a system at equilibrium, there are two important consequences: an increase in temperature will favour that reaction direction that absorbs heat (i.e. the endothermic reaction) the value of Keq will change Consider the following equilibrium system N 2 O 4 (g ) 2NO 2 (g) ΔH = kj We see by the sign of ΔH that the forward reaction is endothermic. Heat is absorbed (required as a reactant) when the reaction proceeds as N 2 O 4 (g ) 2NO 2 (g) By adding more heat, equilibrium will shift to use up the additional heat, thus favouring this forward direction. Why will Keq change, when it did not change when concentration, pressure, and volume were the applied stresses? When temperature changes cause an equilibrium to shift, one entire side of the reaction equation is favoured over the other side. Mathematically, this will alter the value of Keq as follows:

5 Mr Chiasson Advanced Chemistry 12 5 Keq = [products] [reactants] if the forward reaction is favoured if the reverse reaction is favoured more products are produced; fewer reactants fewer products; more reactants Keq will increase Keq will decrease So in our example given above, increasing the temperature will favour the forward direction. The value of Keq will increase. 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. 3.5 Addition of a Catalyst The addition of a catalyst to an equilibrium system is our final stress factor. How will adding a catalyst affect the following: N 2 (g) + O 2 (g) 2NO (g) Adding a catalyst to this, or any other equilibrium system, will not affect the position of an equilibrium. A catalyst speeds up both the forward and the reverse reactions, so there is no uneven change in reaction rates. Generally, a catalyst will help a reaction to reach the point of equilibrium sooner, but it will not affect the equilibrium otherwise.