Chemical Equilibrium Lab. (short version)
Many of the reactions you have studied in the lab appeared to use up all of at least one of the reactants, to have "gone to
completion." In these reactions a gas escaped or a precipitate formed. But for most chemical changes we believe that an
equilibrium state is reached, with some amounts of all reactants, and products coexisting. In a closed system this occurs
when forward and backward opposite reactions occur, with equal rates. With such equal rates in opposite directions, each
material is consumed and produced at the same rate, so there is no net reaction. Each material remains at a constant concentration,
and all macroscopic properties remain constant.
The equilibrium state can be expressed by a relationship between the concentrations of reactants and products known as
the mass action expression or equilibrium constant expression. This expression has a fixed value, the equilibrium
constant, for any reaction at equilibrium at a particular temperature. For a generalized reaction: a A + b B = c C + d D,
the mass action expression is
d
[C]Ca [D]b
[A] [B]
=
K
where K is the equilibrium constant, and [ ] means the concentration of each material. A material whose
concentration is not a variable, that can't be changed at a fixed temperature, is omitted from the expression--that is,
solids, pure liquids, and water as the solvent in dilute solutions, are left out.
Besides the quantitative expression above, qualitative predictions can be made using LeChatelier's Principle.
“if a system at equilibrium is subjected to a stress, the system shifts or reacts to relieve the stress." Form instance , if
some “A” is added to the system above when it is at equilibrium, there will be a net reaction to the right (“shifts right”),
consuming some of the A and thus relieving the "stress" produced by the added A. In this lab you will observe and
explain Le Chatelier shifts in four equilibrium systems.
NaCl (s) = Na+ (aq) + Cl- (aq)
1. Solubility equilibrium for NaCl.
In a clean dry test tube place about 2 cm3 of saturated NaCl (aq). Add a cm3 of concentrated (12M) hydrochloric acid.
(CAUTION). Observe, record your observations. Discard the solution.
2. Chromate - Dichromate Equilibrium. 2 CrO4 2- (aq) + 2 H+ (aq) = Cr2O7 2- (aq) + H20
yellow
orange
In a clean test tube place about 2cm3of O.1OM K2CrO4 , a clear yellow solution. Add drops of 3M H2SO4 (CAUTION)
until the orange dichromate ion has been formed. Then add drops of 6M NaOH (CAUTION) until the yellow chromate reforms.
Shake gently or stir well during the reactions to keep the solutions well mixed. Repeat the addition of the H2SO4. Record all
observations.
In a clean test tube place another few cm3 of chromate. Add a few drops of BaCl2 (aq). Mix. Add drops of HCI (aq) until
the color changes and the precipitate dissolves. Add drops of NaOH (aq) until the color and precipitate change back. Repeat the
HCI addition. Record observations. Discard solutions.
Questions and Write-up.
Part 1. a. Write the mass-action expression ("mae") for the reaction.
b. Explain the results, using Le Chatelier's Principle.
Part 2. a. Write the mass action expression for the chromate-dichromate equilibrium reaction.
b. Look- up or otherwise find out whether or not each of the following is appreciably soluble in water, or would
BaCr2CO7
Ba(OH)2
BaSO4
precipitate when formed:
BaCrO4
--
+
c. OH , as from NaOH, reacts with H , as from acid, to form HOH, water; this is a reversible reaction that reaches an
equilibrium. With this in mind, explain the color changes in the system with NaOH and H2SO4, using Le Chatelier's Principle.
d. Explain the formation and dissolving of the precipitate with the BaCl2 solutions.