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Rate Laws
>> Parts of this equation/concept include:
Rate laws are the relationship between concentration (normally
concentration of reactants) and rate. For the general equation
aA + bB  cC + dD
the rate law has the form
initial rate = k[A]m[B]n
Note that the exponents on the reactant concentration are not the
stoichiometric coefficients. These exponents, as well as the value
of k, can only be determined experimentally. This form specifies
initial rate.
The rate law for a general rate has the same form as the rate law
shown previously. It is possible for products and catalysts to appear
in a rate law. However, unless there is information to the effect,
it is normal to assume that only reactants appear in the rate law.
Reaction order is the exponent on concentration in the rate law.
The order can be discussed in terms of the reactants individually
or as a whole. The "order in A" refers to the exponent on the concentration
of A in the rate law. The "overall order" refers to the sum of the
exponents for all reactant concentrations.
The order is normally a whole number but may be a simple fraction.
>> Example 1
What is the order of each reactant and overall order for the
following reactions?
- 2 H2 + O2
2 H2O rate
= k[H2][O2]
- 2 P + 3 Cl2
2 PCl3 rate
= k[Cl2]2
- HCl + NH3
NH4Cl rate
= k[HCl][NH3]2
Solution:
- The exponent on both hydrogen and oxygen is an understood
1. Therefore this reaction is first order in hydrogen, first
order in oxygen, and second order overall.
- The exponent on the chlorine concentration is 2, so this reaction
is second order in chlorine. It is zero order in phosphorus,
which is why phosphorus does not appear in the rate law. The
reaction is second order overall (2 + 0 = 2).
- The exponent on HCl is an understood 1, so this reaction is
first order in HCl. The exponent is 2 for ammonia, so the reaction
is second order in ammonia. The reaction is third order overall
(1 + 2 = 3).
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The units on a rate constant depend on the overall order of the
reaction. The units are whatever is appropriate for the equation.
Any time units can be used, but seconds are usual. A useful relationship
is that the units on the rate constant are M1–order•time–1.
>> Example 2
What are the units on rate constant for the following rate laws,
assuming the rate is M/s?
- rate = k[A]3
- rate = k
- rate = [R][S]
Solution:
- The order of the reaction is 3. Using the equation M1–order
× time–1, M1–3× s–1
or M–2•s–1.
- The order of the reaction is zero. One minus zero is 1, so
the units are M•s–1.
- The overall reaction order is 2. (One in each R and S.) Using
1 – order, the result is 1–2 = –1. So the units
are M–1•s–1.
>> Example 3
What is the order of the reaction for the following rate constants?
- k = 7.5 s–1
- k = 66.2 M–2•min–1
- k = 2.53 x 102 M–1•s–1
Solution:
The equation of 1 – order = the exponent on concentration
is used.
- Molarity does not appear in the units. This implies that its
exponent is zero, since anything to the zero power is 1. 1 –
order = 0; so order = 1. The rate law is first order.
- The exponent on molarity is –2. 1 – order = –2,
so order = 3. The reaction is third order.
- The exponent on molarity is –1. 1 – order = –1,
so order = 2. This is a second–order reaction.
The value of the rate constant is determined by solving the rate
law for k. This requires knowing the rate and reactant concentration.
The rate constant will vary with temperature but not with time or
concentration. Therefore any values of concentration and the rate
at those concentrations can be used to determine rate. Examples
are shown in the next section.
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| C. Rate Law from Experiments with Varying Concentrations of Reactants |
The order of the reaction is determined by comparing the rate when
reactant concentrations change. The easiest and best choice is to
change only one reactant at a time to determine the effect of that
reactant on rate. The most common method is to set up the rate law
with values from one experiment and compare it (by dividing) to
the values in the rate law from another experiment (where one reactant
concentration has been changed).
Tip: If you use the higher concentration values on the top,
the numbers are easier to work with.
You may use any of the experiments to determine the value of k.
The value will be the same within the appropriate significant
figures.
>> Example 4
For the reaction A + B
C, what is the rate law and value of rate constant based on the
following data.
| Trial |
[A] (M) |
[B] (M) |
Rate of C (M/s) |
| 1 |
0.1 |
0.1 |
9.0 x 10–4 |
| 2 |
0.3 |
0.1 |
8.1 x 10–3 |
| 3 |
0.1 |
0.2 |
1.8 x 10–3 |
Solution:
First, determine the order for each reactant by comparing two
trials where the concentration of only one reactant changes.
Using trials 1 and 2, write the appropriate form of the rate
law for each trial. It helps to write the one with the higher
concentration values first.
| General form: |
rate = k[A]m[B]n |
| Trial 2 (higher [A]) |
(8.1 x 10–3) = k[0.3]m[0.1]n |
| Trial 1 (lower [A]) |
(9.0 x 10–4) = k[0.1]m[0.1]n |
Dividing trial 2 by trial 1, the rate constant (k) and
the concentration of B cancel. Therefore
9 = 3m
There are mathematical ways to solve this problem, but the easiest
is by inspection: 32 is 9; therefore m = 2.
(Note: Really it is! For some reason the instinctive answer
is m = 3, but 32 is, of course, the same as
3 x 3 = 9; 33 = 27. Don't worry. This particular relationship
is normally the only one that tricks your brain in this fashion.)
To determine the order in B, consider two trials where the concentration
of B is the only reactant concentration that changes. This would
be trials 1 and 3. Since trial 3 has the higher concentration,
it is easier if you write that first.
| General form: |
rate = k[A]2[B]n |
(You just determined the value of m.) |
| Trial 3 |
(1.8 x 10–3)
= k[0.1]2[0.2]n |
|
| Trial 1 |
(9.0 x 10–4)
= k [0.1]2[0.1]n |
|
Dividing trial 3 by trial 1, rate constant and concentration
of A cancel. Therefore
2 = 2n
By inspection, n = 1. Therefore the rate law is: rate
= k[A]2[B] (second order in A, first order in
B, third order overall).
To determine the rate constant, use the values from any trial
and solve for k.
Trial 1:
9.0 x 10–4 M/s = k[0.1 M]2[0.1
M]
9.0 x 10–4 M/s = k(1 x 10–3
M3)
0.9 M–2•s–1 =
k
Trial 2:
8.1 x 10–3 M/s = k[0.3 M]2[0.1
M]
0.9 M–2•s–1 =
k
Trial 3:
1.8 x 10–3 M/s = k[0.1 M]2[0.2]
0.9 M–2•s–1 =
k
Any one of the trials can be used to determine the rate constant.
As you can see, all give the same answer.
>> Example 5
What is the rate law and the value of rate constant for the reaction
2 AB2 + C
CB + BA, based on the data below?
| Trial |
[AB2] (M) |
[C] (M) |
Rate of CB (M/min) |
| 1 |
1.0 |
1.0 |
1.4 x 10–7 |
| 2 |
0.5 |
1.0 |
3.5 x 10–8 |
| 3 |
1.0 |
2.0 |
5.5 x 10–7 |
Solution:
The general form of the rate law for this equation is rate =
k[AB2]m[C]n
Comparing trials 1 and 2, the concentration of C stays the same
and the concentration of AB2 changes. Since the concentration
is lower in trial 2, trial 1 should be divided by trial 2.
| General form: |
rate = k[AB2]m[C]n |
| Trial 1 |
(1.4 x 10–7) = k[1.0]m[1.0]n |
| Trial 2 |
(3.5 x 10–8) = k[0.5]m[1.0]n |
The rate constant and concentration of C cancel, so
4 = 2m
2 = m
Comparing trials 1 and 3 (when AB2 is constant). The
concentration of C is higher in trial 3, so
| Trial 3 |
5.5 x 10–7 = k[1.0]2[2.0]n |
| Trial 1 |
1.4 x 10–7 = k[1.0]2[1.0]n |
3.93 = 4 = 2n
2 = n
Since the order is a whole number or simple fraction it makes
sense to round to the obvious value of 4. Therefore the rate law
is rate = k[AB2]2[C]2.
To determine the value of k, use the data from any trial.
Randomly using trial 1,
(1.4 x 10–7) = k[1.0]2[1.0]2
1.4 x 10–7 M–3 min–1
= k
Note that the time unit was minutes for this rate constant. That
was determined from the rate, given in units of M/min.
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