Equilibrium III

Le Chatelier's Principle

Remember this equilibrium? (How could you forget)

N₂O₄ = 2 NO₂

We said that both the forward and backward reactions are elementary steps and so we can write the rate laws for these reactions

1) N₂O₄ ® 2 NO₂ R_f = k_f [N₂O₄]

2) 2 NO₂ ® N₂O₄ R_b = k_b [NO₂]²

If at equilibrium at 100ºC, [N₂O₄] = 4.27 X 10^-3 M and [NO₂] = .0300 M, then

K_c = (.0300)²/ 4.27 X 10^-3 = .211

At equilibrium the forward and backward reaction rates are equal. If I increase the concentration of nitrogen dioxide, say to .060 M, I will increase the rate of the backward reaction. The forward and backward rates are no longer equal. I have disturbed the equilibrium. Now

Q = (.0600)²/ 4.27 X 10^-3 = .843> K_c

The backward rate is now faster than the forward rate so the reaction is said to shift to the left to reactants. The concentration (or pressure) of products will decrease and the concentration of reactants will increase. Disturbing the equilibrium causes the reaction to shift to undo the disturbance.

Le Chatelier's principle states if a system at equilibrium is disturbed by a change in temperature, pressure, or concentration, the system undergoes a change in a direction that counteracts the disturbing influence, and if possible, returns the system to equilibrium.

So

1. Adding or removing a reactant or product. A reaction will proceed in a direction that consumes a substance that is added, or replaces a substance that is removed.

2. Volume or pressure changes - A decrease in volume will cause an increase in pressure (Boyle's Law). Since molar concentration = moles/volume, a decrease in volume will change concentrations. The system will respond to an increase in pressure by trying to lower the pressure. One way to lower the pressure is by reducing the number of moles of gas (P µ n). Consider

N₂O₄ = 2 NO₂

An increase in pressure can be compensated by a shift to the left, reducing the number of moles of gas and thus reducing the pressure. So an increase in pressure will shift the reaction to reduce the number of moles of gas.

3. Temperature - We can consider heat to be a product (of an exothermic reaction) or a reactant (of an endothermic reaction). Again consider

N₂O₄ + 13.9 kJ = 2 NO₂ (This energy is to convert one mole of reactant to 2 moles of product by breaking the N-N bond in N₂O₄).

If we increase the temperature, more heat energy is available to the system. The system will shift to the right to consume this disturbance. At equilibrium we will have more product, so the equilibrium constant will increase when the temperature increases in an endothermic reaction. Temperature is the only condition that will change an equilibrium constant.

Now consider

N₂ + 3 H₂ = 2 NH₃ + 92.38 kJ

This system producing ammonia will shift to the left to consume additional energy. So an increase in T will shift this equilibrium to the left to reactants. This decreases the equilibrium constant. Temperature's effects are further discussed below.

4. Addition of a catalyst - A catalyst increases the rate of both the forward and backward reaction by the same amount, (demonstrated below). So a catalyst will get the reaction to an equilibrium state quicker, but it will not change the equilibrium concentrations compared to the non-catalyzed reaction.

T and the addition of a catalysts

The following table was used to show how increasing T shift a reaction in the endothermic direction. The numbers calculated are using the T and E_a to determine the fraction of particles that have enough energy to react (from exp[ - E_a/RT ] )

T \ Ea	Ea, forward= 20 kJ/mol	Ea,backward= 30 kJ/mol
300 K	3.3 X 10^-4	6.0 X 10^-6
400 K	2.4 X 10^-3	1.2 X 10^-4

Note the ratio (2.4 X 10^-3/3.3 X 10^-4) for increasing the T for 20 kJ/mol is ~7X.

The ratio for increasing the T for 30 kJ/mol is ~20X. So increasing the T will increase the rate going both directions, but it will have a larger effect in the endothermic direction (backwards in this example). So increasing the T will result in a shift of this reaction backwards

Adding a catalyst will reduce the E_a for the forward and backward reaction, and will thus increase the rate going both directions.

Above we can see that the ratio of the forward to backward rate at 300K is about 55. Consider the effect of adding a catalyst on the forward and backward rates. For example, consider a catalyst that lowered the E_a for the above reaction by 10 kJ/mol.

T \ Ea	Ea, forward= 10 kJ/mol	Ea,backward= 20 kJ/mol
300 K	1.8 X 10^-2	3.3 X 10^-4

The ratio is still ~55. So we get to equilibrium faster, but the position of equilibrium will not change with the addition of a catalyst.

Summary of Le Chatelier’s Principle

Increase in [ ], Reaction shifts to undo the change in concentration. An increase in concentration will cause the reaction to shift to reduce the [ ].

Increase in partial P of a reactant or product (remember an increase in P can be caused by a decrease in V) – The rxn will shift to the side with less moles of gas to reduce the P.

Increase in T – The rxn will shift in the endothermic direction to consume the added energy. A change in T is the only thing that changes the equilibrium constant.

Addition of a Catalyst – This will cause the rxn to reach equilibrium faster, but it will not change the equilibrium constant or [ ]s.

Nice illustration of Le Chatelier’s principle is Hemoglobin uptake of O₂

1. H Hb + O₂ = H⁺ + Hb O₂^- (all aqueous and note this really involves the uptake of 4 molecules of oxygen and thus 4 equilibrium reactions.

Other important blood equilibria are

2. O_{2 (g)} = O_{2 (aq)}

3. HCO₃^- + H⁺ = H₂O + CO₂

High altitude (above 9,000 feet) shifts equilibrium 1 and 2 to the left due to the reduced partial pressure of oxygen. Breathing at a higher rate, trying to get enough oxygen at high altitude, releases CO₂ and shifts the 3^rd rxn to the right. How does this last shift effect rxn 1?

Applications - Solving for an equilibrium concentration

1^st example involves the water gas shift reaction (K_c = 4.06 @ 500°C)

All [ ]₀ were 0.10 M and what will be the equilibrium [ ].

CO + H₂O = CO₂ + H₂

Initial	.10	.10	.10	.10
Change	-x	-x	+x	+x
Equilibrium	.10 - x	.10 - x	.10 + x	.10 + x

Plug equilibrium amounts into equilibrium expression.

K_c = 4.06 = (.10 + x)² / (.10 – x)²

Take square root of both sides and solve for x. x = .0337 and the equilibrium concentrations are

Equilibrium

.10 – x = .0663

.0663

.10 + x=.134

.134 M

You can check your work by substituting these values into the equilibrium expression and getting K_c.

2^nd Example – Can the decomposition of water at 1000°C be used as a source for hydrogen gas. Lets use an initial concentration of gaseous water 0.10 M and the equilibrium constant is 7.3 X 10^-18. Determine the equilibrium hydrogen gas concentration.

2 H₂O = 2 H₂ + O₂

Initial	0.10	0	0
Change	-2x	+2x	+x
Equilibrium	0.10 - 2x	2x	x

When you write the algebraic expressions for the equilibrium concentrations, and substitute into the equilibrium expression, you get an ugly looking equation that will be difficult to solve for x. Since the equilibrium constant is small, lets see if we can neglect the x in .010 - 2x. Solving for x, I got x = 2.6 X 10^-7, which is mighty small compared to 0.10. So it appears our approximation is justified. This will give us an equilibrium [H₂] = 5.2 X 10^-7 M at equilibrium, which doesn't sound very promising.