• Reversible Reactions and Equilibrium
• Reversible Reactions
• Chapter 18
• Reversible Reactions and Equilibrium
• Reversible reactions are those reactions that occur simultaneously in both directions.
• Ultimately,a chemical equilibrium will be reached. This occurs when the forward and reverse reactions take place at the same rate.
– There is no net change in the actual amounts of the components of the system?
– Reversible Reactions and Equilibrium
• 2SO
2
(g) + O
2
(g) ---> 2SO
3
(g)
• 2SO
2
(g) + O
2
(g) <--- 2SO
3
(g)
• 2SO
2
(g) + O
2
(g) <---> 2SO
3
(g)
• 2SO
2
(g) + O
2
(g) <--->> 2SO
3
(g)
• Note: The rate of change reaches an equilibrium. The concentration of reactants and products is not equal.
• Reversible Reactions and Equilibrium
• Equilibrium position of a reaction is given as a relative concentration of the components of the reaction at equilibrium. Huh?
– If, at equilibrium, the reaction favors component B by 99% then A is at
1%
A -----> B
1%<----- 99%
• Reversible Reactions and Equilibrium
• If all of your reactants have been changed to products, then the reaction is said to be irreversible.
• Catalysts reduce the amount of time needed to establish equilibrium, both forward and back.
• Processes That Reach Equilibrium
• Vapor-Liquid Equilibrium
• The rate of change from liquid to vapor is equal. A closed system.
• H
2
O(l)   H
2
O(g)
• Processes That Reach Equilibrium
• Solutions of solids
• This occurs when solutes are dissolved in a solvent.
• O
2
(g)   O
2
(aq)
• Gas phase equilibrium
N
2
(g) + 3H
2
(g)   2NH
3
(g)
• Processes That Reach Equilibrium
• Acids and bases
• CH
3
COOH(aq) + H
2
O   CH
3
COO + H
3
O +
• Practice #2, 3, 4
• Reversible Reactions and Equilibrium
• Le Chatelier’s Principle
– Nature strives for stability.

– A system that experiences a change from equilibrium will make adjustments to restore the equilibrium.
• This new equilibrium may not be the same as the original equilibrium.
• This is called a shift in the position of equilibrium.
• Reversible Reactions and Equilibrium
• His principle: If a stress is applied to a system in dynamic equilibrium, the system changes to relieve the stress.
– Stressors include:
• changes in concentration of reactants or products
• changes in temperature
• changes in pressure.
• Reversible Reactions and Equilibrium
• Concentration: changing the concentration of reactants or products at equilibrium disturbs the equilibrium.
– The system will adjust to minimize the change.
• Watch this:
• H
2
CO
3
(aq) <------>> CO
2
(aq) + H
2
O(l)
• <1% >99%
– Reversible Reactions and Equilibrium
• H
2
CO
3
(aq) <------>> CO
2
(aq) + H
2
O(l)
• <1% >99%
• At equilibrium H
2
CO
3 is less than 1%
• If you add CO
2 at equilibrium this happens,
– The ratio of carbonic acid to carbon dioxide changes from 99:1 to
99.5:0.5
– This results in a new ratio of 199:1 wow!
– So now what?
• Reversible Reactions and Equilibrium
• Well, nature strives for stability so, the reaction will shift in order to regain the 99:1 ratio thus forming more carbonic acid.
– How does your body maintain the correct balance of CO
2
and H
2
CO
3
?
• H
2
CO
3
(aq) <------>> CO
2
(aq) + H
2
O(l)
• Moral of the story, if you add more product you will shift the reaction to the left(reverse.)
• Reversible Reactions and Equilibrium
• What happens if you remove some product?
– The shift is in the direction of the products.
– Stability must be maintained.
• What is the practical application of this process?
– Egg laying hens
– Carbonic acid in your blood.
• Reversible Reactions and Equilibrium
• Temperature
– Increasing the temperature shifts the reaction in the direction that absorbs heat.

– Why?
• Because the heat absorption tries to reduce the amount of heat that was introduced. Heat is the stressor.
• Reversible Reactions and Equilibrium
• Pressure
• Applicable only to gases with unequal quantities of moles of reactants and products.
• Let’s make ammonia.
• Reversible Reactions and Equilibrium
• Making ammonia
• N
2
(g) + 3H
2
(g)<-----> 2NH
3
(g)
• What conditions of heat and pressure would produce the greatest amount of ammonia?
• Reversible Reactions and Equilibrium
• Practice time
– 13 page 625
• Reversible Reactions and Equilibrium
• Equilibrium Constants
– is the ratio of the product concentrations to the reactant concentrations at equilibrium, with each concentration raised to a power equal to the number of moles of that substance in the balanced chemical equation. Huh?
– Reversible Reactions and Equilibrium
• Equilibrium Constants
– the reaction: N
2
O
4
(g) <----> 2NO
2
(g)
– Which is the reactant? Which is the product?
– How many moles of each are there?
– K eq
= [ NO
2
] 2
[ N
2
O
4
]
[ XX ] indicate moles/liter
• Reversible Reactions and Equilibrium
• Equilibrium Constants
• Provide information about whether the reaction favors reactants or products.
– K eq
is a ratio of products / reactants.
– If K eq
is > 1, products are favored.
– If K eq is < 1, reactants are favored.
– Let’s return to our original reaction.
• Reversible Reactions and Equilibrium
N
2
O
4
(g) <----> 2NO
2
(g)
If a liter of gas contains 0.0045 mol of N
2
O
4 and 0.030 mol of NO
2 what is the K eq
?
– K eq
= [ NO
2
] 2
[ N
2
O
4
]
– K eq
= [0.030 mol/L NO
2
] 2
[0.0045 mol/L N
2
O
4
]
= .20 mol/L

• Reversible Reactions and Equilibrium
– N
2
O
4
(g) <----> 2NO
2
(g) at 10 o C
– K eq
= [0.030 mol/L NO
2
] 2
[0.0045 mol/L N
2
O
4
]
= .20 mol/L
• Reversible Reactions and Equilibrium
• What you need to know:
– Equilibrium is a dynamic process. Huh?
– The only thing “equal” is the rate of change from products to reactants to products.
– Catalysts lower the activation energy of a reaction. This increases the rate of reactivity.
– Adding a reactant or removing a product can increase the yield of a reaction.
– An increase in temperature shifts the reaction towards the direction that absorbs heat.
– For reactions involving gases with unequal mole ratio amounts, an increase in pressure shifts the equilibrium toward the fewest molecules.
• Reversible Reactions and Equilibrium
• From start to finish. Ozone(O
3
) screens out about 95% of the ultraviolet rays from the sun. It is formed from the action of the UV rays on O
2 and readily decays to O
2
– Write a balanced equation for the reaction.
3O
2
(g) <---> 2O
3
(g)
Identify the reactant, product, and corresponding coefficients.
3O
2
(g) = 3 mol of reactant
2O
3
(g) = 2 mol of product
• Reversible Reactions and Equilibrium
• Write an equillibrium constant for the reaction.
– K eq
= [ O
3
] 2
[O
2
] 3
– Finally, when a reaction reaches equilibrium,the reactants and products have a fixed numerical relationship which is represented by the equilibrium constant. If the constant is greater than 1, products are favored. If the constant is less than 1, reactants are favored.
• Review??
• Page 607 #’s 1-4
• Page 612 #’s 5-8
• Page 625 # 13