Barrons SAT Subject Test Chemistry, 13th Edition

molecular concentration of AB:

where k 2 represents the rate of decomposition of AB at the fixed temperature.

Both reactions can be shown in this manner:

A 2 + B 2 2AB (note double arrow)

When the first reaction begins to produce AB, some AB is available for the
reverse reaction. If the initial condition is only the presence of A 2 and B 2 gases,

then the forward reaction will occur rapidly to produce AB. As the concentration
of AB increases, the reverse reaction will increase. At the same time, the
concentrations of A 2 and B 2 will be decreasing and consequently the forward

reaction rate will decrease. Eventually the two rates will be equal, that is, R = R′.
At this point, equilibrium has been established, and

k 1 [A 2 ] × [B 2 ] = k 2 [AB]^2

or

The convention is that k 1 (forward reaction) is placed over k 2 (reverse

reaction) to get this expression. Then k 1 /k 2 can be replaced by Keq, which is

called the equilibrium constant for this reaction under the particular conditions.
In another general example:

aA + bB cC + dD

the reaction rates can be expressed as

R = k 1 [A]a × [B]b

R′ = k 2 [C]c × [D]d

Note that the values of k 1 and k 2 are different, but that each is a constant for the

conditions of the reaction. At the start of the reaction, [A] and [B] will be at their
greatest values, and R will be large; [C], [D], and R′ will be zero. Gradually R
will decrease and R′ will become equal. At this point the reverse reaction is