Solving Kinetics Problems
1) Rate of Reaction is always given in units of:
M
mol
or
(change in concentration over time)
s
Ls
2) If the problem gives initial concentrations and initial rates of reaction, you are finding the rate
law and the rate constant.
Rate Law Form: rate = k[A]m[B]n[C]k ...
Rate Constant (k) is always in units that adjust the overall order of the reaction to the units of the rate of
reaction given in 1) above
a. To find the exponents for the rate law (m, n, k, ...) use the given initial rates and initial
concentrations in fractions as follows:
rate1 k[ A10 ]m [ B10 ]n [C10 ]k
; [ A10 ] means initial concentration (the zero) for first reaction (the one)

0 m
0 n
0 k
rate2 k[ A2 ] [ B2 ] [C 2 ]
The trick is to find experiments in the list where the concentrations of all of the reactants are the same
except one of the reactants. This will isolate one of the reactants and therefore one of the exponents.
Keep doing this in different fractions until all of the exponents have been isolated and determined.
b. Once you have determined all of the exponents, use one of the experiments values to determine the
value of k.
ex: if you determined the rate law to be: plug in the values from one of the experiments for the rate,
the concentration of A raised to the 1st power and B to the 2nd power, and solve for the value of k. The
L2
mol 3
units for the above example will have units of
.
This
will
transform
the
from the
mol 2  s
L3
mol
concentrations raised to the powers in this example to the needed units for all rates of reactions
.
Ls
c. Now you have a framework for the rate of that reaction at the given temperature for any combination
of concentrations. This is what a rate law does: connect concentrations to the rate of a reaction.
3) The Integrated Rate Laws are used to connect concentrations to time. The Integrated Rate
Laws are listed below for single reactants and three common orders:
zero-order rate law: rate  k[A]0 ; zero-order integrated rate law: [ A]t  kt  [ A]0
first-order rate law: rate  k[A]1 ; first-order integrated rate law: ln[ A]t  kt  ln[ A]0
1
1
  kt 
second-order rate law: rate  k[A]2 ; second-order integrated rate law:
[ A]t
[ A]0
a) use the equations above to solve half-life problems, problems that want to know the concentrations
at a certain time, and problems that want to know concentration after a given percent of the reaction has
completed.
NOTE: Remember, the Rate Laws and Integrated Rate Laws are in terms of the disappearance of the
reactant(s). So, if a reaction is 65% complete, you must use 35% of the reactant remaining, not the
65% of the products made so far.
4) If concentration vs. time data is given for a single reactant,
i. the order can be determined by graphing:
a) the concentration vs time: if this is a straight line, the reaction is zero-order
b) the natural log of the concentration vs. time: if this is a straight line, the reaction is first-order
c) one over the concentration vs. time: if this is a straight line, the reaction is second-order
ii. after the order has been determined, the rate constant can be determined by calculating the slope of
the straight-line graph from step i. above.
5) If rate constant vs. temperature data is given,
i. remember to convert to Kelvin
ii. if only two rates and their corresponding temperatures are given, use the form:
 k   Ea  1 1 
  
ln  1  
k
R
 2
 T1 T2 
iii. if more than two pairs of rate constant vs. temperature data has been given, graph it, and use the
slope to calculate Ea, and if requested, use the y-intercept to calculate A. Remember, the slope is NOT
the same as Ea, nor is the y-intercept A, they must both be CALCULATED.
6) Reaction mechanisms:
i. Remember, there are two rate laws:
1) the correct, final rate law based ONLY on the reactants in the FINAL, overall balanced chemical
equation. AND
2) the rate law from the slow step, which must be used to DEVELOP the rate law based on the final
chemical equation.
ii. if there is a fast, equilibrium step before the slow step, you may need to use the fact that the forward
and reverse rates are the same and write an equation setting the rate laws equal and substitute to remove
unwanted reactants from the slow step rate law.