236
Chemical Kinetics
This whole process can be repeated, reversing the roles of A and B, so that a
large known initial concentration [A] e , is used. As a result, a set of equations
identical to Equations 15-13 and 15-14 would be obtained having B in place of
A, and b in place of a. The same is true for Figure 15-5.
Now, having found for both A and B which of the plots in Figures 15-4 or 15-5
gives a straight line, you will know the values of a and b (that is, the order of
each reactant). In addition, the slope of whichever plot in Figures 15-4 or 15-5
gives a straight line will give you k (the rate constant) because you know [B]|
and [A]?. In short,
1. the type of plot that gives a straight line tells the order;
2. the slope of that straight line gives the rate constant, k.
IMPORTANT MISCELLANEOUS COMMENTS
1. In the preceding discussion, we emphasized taking a set of data pairs
(values of [A] at the corresponding times t) and plotting them as (1) log [A]
versus t, (2) 1/[A] versus /, and (3) 1/[A]
2 versus t in order to find which plot
gives a straight line. In many ways, a simpler approach is to use your hand
calculator or a computer and apply the method of least squares (see p 72) to
each of the three possibilities. Whichever combination yields the largest correlation coefficient must be the one that would have given the straight-line plot. At
the same time, you would have found the slope andy intercept of that straight
line, and your problem would be solved without any graphing.
2. If gaseous reactions are studied, it is customary to express concentration
in partial pressures, using atm as the pressure unit. In other words, if A is a gas,
then [A] would be expressed as P A in atm or torr, not moles/liter.
3. The units of k are important. As stated earlier, the units of k for a firstorder reaction are reciprocal time: sec"
1 , min"
1 , hr"
1 , and so on. For secondorder reactions, k has the units of M^'sec"
1 (or some other reciprocal time
unit); for a gaseous reaction the units would be atm^'sec"
1 . For a third-order
reaction, k has the units of M~
2 min~
l (or some other reciprocal time unit) for
reactions in solution, or atm~
2 min~
1 for gaseous reactions. Other units may be
used, but these are the traditional ones; they should always be stated.
4. A modification may be needed when dealing with a second-order or thirdorder plot for a compound involved in a chemical reaction that uses unequal
numbers of moles of reactants and products. In this case, any one of the rate
terms similar to those given by Equation 15-4 can be set equal to k [ AJ
n [B]
6 [C]
c .
When you multiply both sides of the equation by the integer (z) needed to cancel
the fraction in front of whatever term that you chose, you can see that k' in
Equations 15-5, 15-12, 15-13, and 15-14 must also include this integer z. As a
result, A:' = zk[B$, and the slopes of the lines in Figure 15-5 are equal tozA:[B]f
in part (a) and 2zk[B]% in part (b).
Chemical Kinetics
This whole process can be repeated, reversing the roles of A and B, so that a
large known initial concentration [A] e , is used. As a result, a set of equations
identical to Equations 15-13 and 15-14 would be obtained having B in place of
A, and b in place of a. The same is true for Figure 15-5.
Now, having found for both A and B which of the plots in Figures 15-4 or 15-5
gives a straight line, you will know the values of a and b (that is, the order of
each reactant). In addition, the slope of whichever plot in Figures 15-4 or 15-5
gives a straight line will give you k (the rate constant) because you know [B]|
and [A]?. In short,
1. the type of plot that gives a straight line tells the order;
2. the slope of that straight line gives the rate constant, k.
IMPORTANT MISCELLANEOUS COMMENTS
1. In the preceding discussion, we emphasized taking a set of data pairs
(values of [A] at the corresponding times t) and plotting them as (1) log [A]
versus t, (2) 1/[A] versus /, and (3) 1/[A]
2 versus t in order to find which plot
gives a straight line. In many ways, a simpler approach is to use your hand
calculator or a computer and apply the method of least squares (see p 72) to
each of the three possibilities. Whichever combination yields the largest correlation coefficient must be the one that would have given the straight-line plot. At
the same time, you would have found the slope andy intercept of that straight
line, and your problem would be solved without any graphing.
2. If gaseous reactions are studied, it is customary to express concentration
in partial pressures, using atm as the pressure unit. In other words, if A is a gas,
then [A] would be expressed as P A in atm or torr, not moles/liter.
3. The units of k are important. As stated earlier, the units of k for a firstorder reaction are reciprocal time: sec"
1 , min"
1 , hr"
1 , and so on. For secondorder reactions, k has the units of M^'sec"
1 (or some other reciprocal time
unit); for a gaseous reaction the units would be atm^'sec"
1 . For a third-order
reaction, k has the units of M~
2 min~
l (or some other reciprocal time unit) for
reactions in solution, or atm~
2 min~
1 for gaseous reactions. Other units may be
used, but these are the traditional ones; they should always be stated.
4. A modification may be needed when dealing with a second-order or thirdorder plot for a compound involved in a chemical reaction that uses unequal
numbers of moles of reactants and products. In this case, any one of the rate
terms similar to those given by Equation 15-4 can be set equal to k [ AJ
n [B]
6 [C]
c .
When you multiply both sides of the equation by the integer (z) needed to cancel
the fraction in front of whatever term that you chose, you can see that k' in
Equations 15-5, 15-12, 15-13, and 15-14 must also include this integer z. As a
result, A:' = zk[B$, and the slopes of the lines in Figure 15-5 are equal tozA:[B]f
in part (a) and 2zk[B]% in part (b).