SAT Chemistry Chemical Equilibrium - Effects Of Changing Conditions

SAT Chemistry Chemical Equilibrium - Effects Of Changing Conditions

Effect of Changing the Concentrations
When a system at equilibrium is disturbed by adding or removing one of the substances (thus changing its concentration), all the concentrations will change until a new equilibrium point is reached with the same value of K_eq.
If the concentration of a reactant in the forward action is increased, the equilibrium is displaced to the right, favoring the forward reaction. If the concentration of a reactant in the reverse reaction is increased, the equilibrium is displaced to the left. Decreases in concentra­tion will produce effects opposite to those produced by increases.

Effect of Temperature on Equilibrium
If the temperature of a given equilibrium reaction is changed, the reaction will shift to a new equilibrium point. If the temperature of a system in equilibrium is raised, the equilibrium is shifted in the direction that absorbs heat. Note that the shift in equilibrium as a result of temperature change is actually a change in the value of the equilibrium constant. This is different from the effect of changing the concentration of a reactant; when concentrations are changed, the equilibrium shifts to a condition that maintains the same equilibrium constant.

Effect of Pressure on Equilibrium
A change in pressure affects only equilibria in which a gas or gases are reactants or products. Le Chatelier’s Law can be used to predict the direction of displacement. If it is assumed that the total space in which the reaction occurs is constant, the pressure will depend on the total number of molecules in that space. An increase in the number of molecules will increase pressure; a decrease in the number of molecules will decrease pressure. If the pressure is increased, the reaction that will be favored is the one that will lower the pressure, that is, decrease the number of molecules.
An example of the application of these principles is the Haber process of making ammonia. The reaction is
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + heat (at equilibrium)
If the concentrations of the nitrogen and hydrogen are increased, the forward reaction is increased. At the same time, if the ammonia produced is removed by dissolving it into water, the forward reaction is again
favored.

Because the reaction is exothermic, the addition of heat must be considered with care. Increasing the temperature causes an increase in molecular motion and collisions, thus allowing the product to form more readily. At the same time, the equilibrium equation shows that the reverse reaction is favored by the increased temperature, so a compromise tempera­ture of about 500°C is used to get the best yield.
An increase in pressure will cause the forward reaction to be favored since the equation shows that four molecules of reactants are forming two molecules of products. This effect tends to reduce the increase in pressure by the formation of more ammonia.

EQUILIBRIA IN HETEROGENEOUS SYSTEMS
The examples so far have involved systems made up of only gaseous substances. Expression of the lvalues of systems changes when other phases are present.
Equilibrium Constant for Systems Involving Solids
If the experimental data for this reaction are studied:
CaCO₃(s) ⇌ CaO(s) + CO₂(g)

it is found that at a given temperature an equilibrium is established in which the concentra­tion of CO₂ is constant. It is also true that the concentrations of the solids have no effect on the CO₂ concentration as long as both solids are present. Therefore, the K_eq, which would conventionally be written like this:

can be modified by incorporating the concentrations of the two solids. This can be done since the concentration of solids is fixed. It becomes a new constant K, known as:
K = [CO₂]

Any heterogeneous reaction involving gases does not include the concentrations of pure solids. As another example, K for the reaction
NH₄Cl(s) ⇌ NH₃(g) + HCl(g)
is K = [NH₃] [HCl]

Acid Ionization Constants
When a weak acid does not ionize completely in a solution, an equilibrium is reached between the acid molecule and its ions. The mass action expression can be used to derive an equilib­rium constant, called the acid dissociation constant, for this condition. For example, an acetic acid solution ionizing is shown as
HC₂H₃O₂(aq) + H₂O(1) ⇌ H₃O⁺(aq) + C₂H₃O₂^-(aq)

The concentration of water in moles/liter is found by dividing the mass of 1 liter of water (which is 1,000 g at 4°C) by its gram-molecular mass, 18 grams, giving H₂O a value of 55.6 moles/liter. Because this number is so large compared with the other numbers involved in the equilibrium constant, it is practically constant and is incorporated into a new equilibrium constant, designated as K_a. The new expression is


Ionization constants have been found experimentally for many substances and are listed in chemical tables. The ionization constants of ammonia and acetic acid are about 1.8 x 10^-5. For boric acid K_a = 5.8 x 10^-10, and for carbonic acid K_a = 4.3 x 10^-7.
If the concentrations of the ions present in the solution of a weak electrolyte are known, the value of the ionization constant can be calculated. Also, if the value of K_a is known, the concentrations of the ions can be calculated.

A small value for K_a means that the concentration of the unionized molecule must be relatively large compared with the ion concentrations. Conversely, a large value for K_a means that the concentrations of ions are relatively high. Therefore, the smaller the ionization con­stant of an acid, the weaker the acid. Thus, of the three acids referred to above, the ionization constants show that the weakest is boric acid, and the strongest, acetic acid. It should be remembered that, in all cases where ionization constants are used, the electrolytes must be weak in order to be involved in ionic equilibria.

Ionization Constant of Water
Because water is a very weak electrolyte, its ionization constant can be expressed as follows:


From this expression, we see that for distilled water [H₃O⁺] = [OH^-] = 1 x 10^-7. Therefore, the pH, which is -log[H₃O⁺], is
pH = -log[1 x 10^-7)
pH = - [-7] = 7 for a neutral solution
The pH range of 1 to 6 is acid, and the pH range of 8 to 14 is basic. See the chart below.



Sample Problem _____________________________________

(This sample incorporates the entire discussion of dissociation constants, including finding the pH.)
Calculate (a) the [H₃O⁺], (b) the pH, and (c) the percentage dissociation for 0.10M acetic acid at 25°C. The symbol K, is used for the acid dissociation constant. K_a for HC₂H₃O₂ is 1.8 x 10^-5.


Because a weak acid, such as acetic, at concentrations of 0.01M or greater dissociates very little, the equilibrium concentration of the acid is very nearly equal to the original concentra­tion, that is,
0.10 -x ≅ 0.10
Therefore, the expression can be changed to