494 11 The Rates of Chemical Reactions
Exercise 11.8
a.Verify Eq. (11.2-18).
b.Find the units of the rate constantkffor annth-order reaction.
c.For a third-order reaction with a single reactant and negligible reverse reaction, find an
expression for the time required for 80% of the reactant to react.
d.In terms oft 1 / 2 , how long will it take for 7/8 (87.5%) of the reactant of part c to react?
Zero-Order Reactions
In the rare case that a reaction of a single reactant is zero order (the rate is independent
of the concentration of the reactant), the rate law for the forward reaction is
r−
d[A]
dt
kf[A]^0 kf (11.2-19)
If there is no reverse reaction and if the temperature is constant, the solution of this
equation is
[A]t
{
[A] 0 −kft if 0<t<[A] 0 /kf
0if[A] 0 /kf<t
(11.2-20)
where the first line of the solution is obtained from Eq. (11.2-19). The second line
of the solution is obtained from the fact that the reaction stops when no reactant
remains.
EXAMPLE11.5
A hypothetical zero-order reaction has a rate constant equal to 0.0150 mol L−^1 s−^1 at a certain
temperature. If the initial concentration of the single reactant is 1.000 mol L−^1 , find the
concentration after a reaction time of 5.00 s at this temperature.
Solution
[A]t[A] 0 −kft 1 .000 mol L−^1 −(0.0150 mol L−^1 s−^1 )(5.00 s) 0 .925 mol L−^1
Exercise 11.9
a.Find the time required for all of the reactant of the previous example to react at this temper-
ature.
b.Find an expression for the half-life of a zero-order reaction and the value of the half-life of
the reaction of the previous example.
Determination of Reaction Order Using Integrated
Rate Laws
For a reaction of a single substance with a negligible reverse reaction, we can com-
pare the integrated rate law with experimental data on the concentration of the reac-
tant. Since graphs of linear functions are easy to recognize, for each order one can
plot the appropriate function of the reactant’s concentration that will give a linear