CK-12-Chemistry Intermediate

http://www.ck12.org Chapter 19. Equilibrium

This is because the overall number of gas molecules would increase and so would the pressure. When the pressure
of a system at equilibrium is decreased by providing more total volume, the reaction that produces more total moles
of gas becomes favored. This is summarized in theTable19.3.

TABLE19.3:Stresses and Responses to Pressure Changes

Stress Response

pressure increase reaction produces fewer gas molecules

pressure decrease reaction produces more gas molecules

Like changes in concentration, the Keqvalue for a given reaction is unchanged by a change in pressure.

It is important to remember when analyzing the effect of a pressure change on equilibrium that only gases are
affected. If a certain reaction involves liquids or solids, they should be ignored. Calcium carbonate decomposes
according to the equilibrium reaction:

CaCO 3 (s)⇀↽CaO(s)+O 2 (g)

Oxygen is the only gas in the system. An increase in the pressure of the system has no effect on the rate of
decomposition of CaCO 3 , but it speeds the reverse reaction by forcing the oxygen molecules closer together, causing
a net shift to the left. When a system contains equal moles of gas on both sides of the equation, pressure has no
effect on the equilibrium position, as in the formation of HCl from H 2 and Cl 2.

H 2 (g)+Cl 2 (g)⇀↽2HCl(g)

Use of a Catalyst

Since a catalyst speeds up the rate of a reaction, you might think that it would have an effect on the equilibrium
position. However, catalysts have equal effects on the forward and reverse rates, so for a system at equilibrium,
these two rates remain equal. A system will reach equilibrium more quickly in the presence of a catalyst, but the
equilibrium position itself is unaffected.

Going to Completion

When one of the products of a reaction is removed from the chemical equilibrium system as soon as it is produced,
the reverse reaction cannot establish itself, and equilibrium is never reached. Reactions such as these are said to
go to completion. Reactions that go to completion tend to produce one of three types of products: (1) an insoluble
precipitate, (2) a gas, or (3) a molecular compound such as water. Examples of these reactions are shown below.

1. Formation of a precipitate: AgNO 3 (aq)+NaCl(aq)→NaNO 3 (aq)+AgCl(s)


2. Formation of a gas: Mg(s)+2HCl(aq)→MgCl 2 (aq)+H 2 (g)


3. Formation of water: HCl(aq)+NaOH(aq)→NaCl(aq)+H 2 O(l)

Response of K

Occasionally, when students apply Le Chatelier’s principle to an equilibrium problem involving a change in con-
centration, they assume that Keqmust change. This seems logical, since we talk about “shifting” the equilibrium in
one direction or the other. However, Keqis a constant for a given equilibrium at a given temperature, so it must not
change. Here is an example of how this works. Consider the simplified equilibrium below: