Chapter 14

Equilibrium occurs when there is a constant ratio between the concentration of the reactants and the products. Different reactions have different equilibriums. Some may appear to be completely products, however, all reactions have some reactants present. A reaction may look "finished" when equilibrium is reached, but actually the forward and reverse reactions continue to happen at the same rate. A reverse reaction is when the written reaction goes from right to left instead of the forward reaction which proceeds from left to right. This is why equilibrium is also referred to as "steady state".

The following factors which are effective in changing the equilibrium:

a. temperature b. concentration c. pressure d. catalyst Except for the last one, each of these factors

affects the state of equilibrium in a system. Catalysts have no effect on the final equilibrium state since they affect the rates of both the forward and reverse reactions by lowering the activation energy requirement for both processes.

When a system reaches equilibrium there will be no changes in the reactant or the product until something alters the system. The factors that can alter an equilibrium are called as stresses. A decrease in volume increases the pressure and the conc. of both NO₂ and N₂O₄, disturbing the equilibrium and temporarily making the mixture darker. The color fades as more N₂O₄ is formed to reduce the pressure and restore equilibrium.

Temperature affects equilibrium:At 25⁰C the above system at equilibrium is a medium shade of brown. When cooled to 0⁰C the color of the mixture of gases in the flask is lighter. At 100⁰C the contents of the flask are darker. If a system at equilibrium is disturbed by applying stress the system attains a new equilibrium position to accommodate the change and tends to release the stress. This is Le Chatelier’s principle.

Le Chatelier defined stress as a change in the temperature or pressure of the system or a change in the concentration of a component.

Suppose you have an equilibrium established between four substances A, B, C and D.

A+2B↔C+DWhat would happen if you changed the

conditions by increasing the concentration of A?

According to Le Chatelier, the position of equilibrium will move in such a way as to counteract the change. That means that the position of equilibrium will move so that the concentration of A decreases again by reacting it with B and turning it into C + D. The position of equilibrium moves to the right.

A+2B↔C+D

This is a useful way of converting the maximum possible amount of B into C and D

What would happen if you changed the conditions by decreasing the concentration of A?

According to Le Chatelier, the position of equilibrium will move so that the concentration of A increases again. That means that more C and D will react to replace the A that has been removed. The position of equilibrium moves to the left.

A constant, characteristic for each chemical reaction; relates the specific concentrations of all reactants and products at equilibrium at a given temperature and pressure.

Suppose we had A + B  <----->    C + D            [product] Keq = [reactant]    [C] [D]

Keq =   [A] [B]  Suppose we had H2 + I2    <------>  2 HI     [HI]

Keq = [H2 ] [I2 ] 

we've got 2 HI   [HI] [HI]                         [HI]2

Keq =   [H2 ] [I2 ]         =             [H2 ] [I2 ] 

At a temperature of 25⁰C, the following concentrations (in mol/L) of the reactants and products for the reaction involving carbonic acid and water are present.

[H₂CO₃]=3.3x10-2 M, [H₃O+]=1.19x10-4 M, and [HCO₃-]=1.19x10-4 M

What is the Keq value for the following reaction at equilibrium in a dilute aqueous solution?

H₂CO₃(aq)+H₂O(l)↔H₃O+(aq) +HCO₃-(aq)

Page 527Practice problem 1,2,3.

The equilibrium constant for a slightly soluble ionic solid in equilibrium with its ions in a saturated solution.

The solubility product constant is equal to the product of the concentrations of the ions involved in the equilibrium, each raised to the power of its coefficient in the equilibrium equation.

Ionic compounds normally dissociate into their constituent ions when they dissolve in water. For example, for calcium sulfate:

CaSO₄(s)↔Ca2+(aq) +SO₄2-(aq) As for the previous example, the

equilibrium expression is: K= [Ca2+ ][SO₄2- ] /[CaSO₄(s)]

where K is called the equilibrium (or solubility) constant and curly brackets indicate activity.

Consider the dissociation of the salt CaF₂ in water

CaF₂(s)↔Ca2+(aq) + 2F- (aq) Most parts of the ocean are nearly saturated with CaF₂. In a region where

evaporation raises the concentration of dissolved materials the mineral fluorite, CaF₂ may precipitate. A saturated solution of CaF₂ at 25⁰C is 3.4x10-4 M. Calculate the solubility product constant for CaF₂.

An ion that is present in two or more substances involved in an ionic chemical equilibrium is called as a common ion.

The common-ion effect is a term used to describe the effect on a solution of two dissolved solutes that contain the same ion.

The presence of a common ion suppresses the ionization of a weak acid or a weak base.

If both sodium acetate and acetic acid are dissolved in the same solution they both dissociate and ionize to produce acetate ions.

Sodium acetate is a strong electrolyte so it dissociates completely in solution. Acetic acid is a weak acid so it only ionizes slightly. According to Le Chatelier's principle, the addition of acetate ions from sodium acetate will suppress the ionization of acetic acid and shift its equilibrium to the left. Thus the percent dissociation of the acetic acid will decrease and the pH of the solution will increase.

NaCH3 COO(s) → Na+(aq) + CH₃COO-

(aq) CH₃COOH ↔ H+

(aq) + CH₃COO-(aq)

This will decrease the hydrogen ion concentration and thus the common-ion solution will be less acidic than a solution containing only acetic acid.

Page 533 practice problem 6,7,8

Page 539 Test prep all