Chapter 13 Worksheet 4 (ws13.4)
Experimental Determination of Rate Laws
Determining Rate Laws Using the Method of Initial Rates
1. Technically, the initial rate is the instantaneous rate at time zero (immediately after mixing the reactants). In practice, it is not possible to measure the rate at time zero. The initial rate can be approximated by measuring the average rate after a very short period time so that very little reactant has been consumed and very little product has formed. As a rule of thumb, you can assume that you are measuring the initial rate if less than 2% of the reactants have been consumed.
If the reaction A = B is run for a very short period of time, you can assume that:
Average rate =
[A] =
[B] =
rate of the reverse reaction =
2. For the generalized reaction, aA + bB = products, the rate law is
rate = V = k[A]x[B]y
Recall that the values of x, y, and k must be determined experimentally. We can do this by measuring the initial rate starting with various concentrations of reactants. Minimally, how many different combinations of [A]0 and [B]0 must you use in order to calculate x, y, and k?
3. For the reaction: 2 NO (g) + Br2 (g) → 2 NOBr (g) the following initial rate data were obtained:
[NO]0 (M) / Initial [Br2]0 (M) / V0 (d[NOBr]/dt; M-s-1)0.10 / 0.20 / 24
0.25 / 0.20 / 150
0.10 / 0.50 / 60
0.35 / 0.50 / 735
a. V = k[NO]x[Br2]y Calculate x, y, and k
b. Calculate the rate of formation of NOBr and the rates of consumption of NO and Br2 when [NO] = 0.50 M and [Br2] = 0.30 M.
4. Sometimes the data is simple enough to determine the reaction orders by inspection. Use the data below to determine the rate law for the generalized reaction: aA + bB = products
V =
Initial [A] (M) / Initial [B] (M) / Initial Rate (-d[A]/dt; M-s-1)0.010 / 0.20 / 5.4 x 10-7
0.020 / 0.20 / 10.8 x 10-7
0.040 / 0.20 / 21.5 x 10-7
0.20 / 0.01 / 2.7 x 10-8
0.20 / 0.02 / 1.1 x 10-7
0.20 / 0.04 / 4.4 x 10-7
x =
y =
overall order =
k =
Initial [A] (M) / Initial [B] (M) / Initial Rate (-d[A]/dt; M-s-1)0.010 / 0.20 / 5.4 x 10-7
0.020 / 0.20 / 21.5 x 10-7
0.040 / 0.20 / 86.4 x 10-7
0.20 / 0.01 / 2.16 x 10-4
0.20 / 0.02 / 2.16 x 10-4
0.20 / 0.04 / 2.16 x 10-4
x =
y =
overall order =
k =
Integrated Rate Laws and Determining Rate Laws Using the Graphical Method
Consider a reaction with only a single reactant: A = products
Rate = -d[A]/dt = k[A]x
For a reaction that is first order with respect to A:
Integrating gives the first order integrated rate law (see derivation at the end of this worksheet):
ln[A] = -kt + ln[A]0
or
(Notice that the ratio is the fraction of A remaining at time t.)
or
[A] = [A]oe-k’t
(The last equation describes the shape of the [A] vs. time curve for a 1st order reaction with only one reactant. This is an example of “exponential decay”. We will see this equation again when we discuss the decay of radioactive nuclei!)
If a reaction is first order, a plot of ______vs. ______will give a straight line with
slope =
y-intercept =
Sketch the graph:
For a reaction that is second order with respect to A:
Integrating gives the second order integrated rate law (see derivation at the end of this worksheet):
If a reaction is second order, a plot of ______vs. ______will give a straight line with
slope =
y-intercept =
Sketch the graph:
For a reaction that is zeroth order with respect to A:
Integrating gives the zeroth order integrated rate law:
[A] = -kt +[A]0
If a reaction is zeroth order, a plot of ______vs. ______will give a straight line with
slope =
y-intercept =
Sketch the graph:
1. You have seen two methods for the experimental determination of rate laws. When would you use the initial rate method and when would you use the graphical method?
Half-life
The half-life of a reaction, t½, is the time required for the concentration of a reactant to drop to one half of its initial value (when [A] = ½ [A]o).
First order half-life equation (see derivation on page 575 of your textbook):
Second order half-life equation (see derivation on page 578 of your textbook):
Zero order half-life equation (see derivation on page 580 of your textbook):
First half-life: Time required to reach 1/2 of the initial concentration.
Second half-life: Time required to go from 1/2 to 1/4 of the initial concentration.
Third half-life: Time required to go from 1/4 to 1/8 of the initial concentration.
1. Sketch graphs of [A] vs. time for a first order and second order reaction and indicate the first 3 half lives. Describe in words the relationships between successive half-lives.
Supplementary Material (Optional)
Derivation of the integrated rate laws (the change of concentration with time)
First order:
Separate variables and integrate both sides of the equation:
ln [A] = -kt + C
If t = 0, then [A] = [A]0 and C = ln [A]0
ln [A] = -kt + ln[A]0
Second order:
Separate variables and integrate both sides of the equation:
If t = 0, then [A] = [A]0 so C =
See if you can derive the zeroth order integrated rate law.
Note: The equations for the zero, first, and second order half-lives are derived in your textbook.
How do you use the integrated rate law method for reactions with more than one reactant?
The trick is to perform the reaction with one of the reactants present in great excess! Then, the concentration of the excess reactant doesn’t change during the reaction and you can treat the reaction as though it has only one reactant!
Consider the generalized reaction with two reactants: aA + bB = products
Rate = -d[A]/dt = k[A]x[B]y
If [B] > [A], then we can assume that the concentration of B at time t is:
[B]t = [B]0 (only a tiny fraction of B is consumed)
and
-d[A]/dt = k’[A]x
where k’ (pseudo rate constant) = k[B]0y
Now the analysis is the same as for a reaction with only one reactant. The slope of the linear graph, gives you the pseudo rate constant. The actual rate constant is calculated from the equation above.
To determine the order with respect to B, just reverse the procedure and do the reaction with a huge excess of A.
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