a static one, although in appearance the reaction seems to have stopped. An
example of an equilibrium is a crystal of copper sulfate in a saturated solution of
copper sulfate. Although to the observer the crystal seems to remain unchanged,
there is actually an equal exchange of crystal material with the copper sulfate in
solution. As some solute comes out of solution, an equal amount is going into
solution.
To express the rate of reaction in numerical terms, we can use the Law of
Mass Action, discussed in Chapter 9, which states: The rate of a chemical
reaction is proportional to the product of the concentrations of the reacting
substances. The concentrations are expressed in moles of gas per liter of volume
or moles of solute per liter of solution. Suppose, for example, that 1 mole/liter of
gas A 2 (diatomic molecule) is allowed to react with 1 mole/liter of another
diatomic gas, B 2 , and they form gas AB; let R be the rate for the forward reaction
forming AB. The bracketed symbols [A 2 ] and [B 2 ] represent the concentrations in
moles per liter for these diatomic molecules. Then A 2 + B 2 → 2AB has the rate
expression
R ∝ [A 2 ] × [B 2 ]where ∝ is the symbol for “proportional to.” When [A 2 ] and [B 2 ] are both 1
mole/liter, the reaction rate is a certain constant value (k 1 ) at a fixed temperature.
R = k 1 (k 1 is called the rate constant)For any concentrations of A and B, the reaction rate is
R = k 1 × [A 2 ] × [B 2 ]If [A 2 ] is 3 moles/liter and [B 2 ] is 2 moles/liter, the equation becomes
R = k 1 × 3 × 2 = 6k 1The reaction rate is six times the value for a 1 mole/liter concentration of each
reactant.
At the fixed temperature of the forward reaction, AB molecules are also
decomposing. If we designate this reverse reaction as R′, then, since
2AB (or AB + AB) → A 2 + B 2two molecules of AB must decompose to form one molecule of A 2 and one of B 2.
Thus the reverse reaction in this equation is proportional to the square of the