P E.2 (pg 1 of 7)
Rate Laws − Using data & graphs
Decomposition of Nitrogen Dioxide
Name_____________________Per____
Nitrogen dioxide will decompose into nitrogen oxide and oxygen gas at 300ºC as indicated in the equation below.
2 NO2(g)
→
2 NO(g)
+
O2(g)
Data can be collected as shown in the table below. A graph of the concentration reactants and products at various times during the
course of the reaction at 300ºC is presented below.
Decomposition of nitrogen dioxide
Data from measuring concentrations at
various times for the reaction above.
time (sec)
[NO2]
[NO]
[O2]
0
0.01
0
0
50
0.0079
0.0021
0.0011
100
0.0065
0.0035
0.0018
150
0.0055
0.0045
0.0023
200
0.0048
0.0052
0.0026
250
0.0043
0.0057
0.0029
300
0.0038
0.0062
0.0031
350
0.0034
0.0066
0.0033
400
0.0031
0.0069
0.0035
0.012
0.01
conc
0.008
0.006
0.004
0.002
Check out the graph to appreciate that the loss of reactant and formation
of products is stoichiometrically related.
0
0
100
200
300
400
500
time
The graph of the concentration vs time is not linear, so we can infer that
this reaction is not zero order, but we can not look at the graph and
determine whether the reaction is first or second order. Remember, that we can not know the order from the stoichiometry of the
equation, because we have no way of knowing if the reaction proceeds in singular one-step event (that would thus be bimolecular)
or if the reaction proceeds in a multi-step process.
So without any rate data to which we can apply the initial-rates-method (table logic) of determining the order, what can we do?
We can use the concentration of the reactant vs time data presented in the data table above and manipulate the concentration
1
1
values to ln[NO2] and
and then graph ln[NO2] vs time and also graph
vs time. We can use the resulting two
[NO2 ]
[NO2 ]
graphs to look for which of the graphs presented will give a straight line in order to determine the order of the reaction. These two
graphs are shown below. Which one is linear?
1/[conc]&vs&'me&
ln[conc]#vs#&me#
350.00"
$4.60"
$4.80"
$5.20"
1/conc&
y"="$0.0028x"$"4.7096"
R²"="0.97799"
$5.00"
ln#conc#
y"="0.5541x"+"98.402"
R²"="0.99925"
300.00"
$5.40"
250.00"
200.00"
150.00"
$5.60"
100.00"
$5.80"
50.00"
0"
$6.00"
0"
50"
100"
150"
200"
250"
&me#
300"
350"
400"
450"
50"
100" 150" 200" 250" 300" 350" 400"
'me&
The R2 value, which is the coefficient of determination (you will learn more about R and R2 when you take a statistics course),
which for now, you can think of R2 as a “goodness-of-fit” indicator. The closer the R2 is to 1, the better the data fits to a straight
1
line. Thus you can see that the R2 value is best for
vs time indicating this reaction is second order.
[NO2 ]
P E.2 (pg 2 of 7)
The dimerization of butadiene is studied at 500 K and the following reaction and experimental data is presented below.
2 C4H6(g) → C8H12(g)
a.
b.
c.
Determine the rate law for this reaction.
Determine the rate constant.
Calculate the half-life for this reaction.
(If it is not 1st order, calculate the half-life at the initial concentration.)
[conc]$vs$%me$
0.018$
conc$
0.016$
0.014$
0.012$
0.01$
0.008$
0.006$
0$
1000$
2000$
3000$
4000$
5000$
6000$
%me$
ln[conc]#vs#&me#
$4.00"
ln#conc#
$4.20"
y"="$0.0001x"$"4.1432"
R²"="0.98523"
$4.40"
$4.60"
$4.80"
$5.00"
$5.20"
0"
1000"
2000"
3000"
4000"
&me#
5000"
6000"
7000"
1/[conc]&vs&'me&
1/conc&
1.
Rate Law for Dimerization of Butadiene
150.00"
140.00"
130.00"
120.00"
110.00"
100.00"
90.00"
80.00"
70.00"
60.00"
y"="0.0142x"+"59.234"
R²"="0.99954"
0"
1000" 2000" 3000" 4000" 5000" 6000" 7000"
'me&
7000$
Time (s)
0
195
604
1246
2180
6210
[C4H6(g)] (M)
0.0167
0.0160
0.0150
0.0130
0.0110
0.0068
P E.2 (pg 3 of 7)
2.
Decomposition of Dinitrogen Pentoxide
Dinitrogen pentoxide can be decomposed according to the reaction below.
2 N2O5(sol'n) → 4 NO2(sol'n) + O2(g)
All Trials are performed at 45ºC
Data for this reaction can be collected in different ways. It is presented in two different ways below, instantaneous rates for
various concentrations and concentration at various times. The concentration vs time data is graphed.
Concentration and Initial Rates
a.
What are several methods by which you could
determine the reaction order?
b.
What are several different methods by which you
could determine the rate constant, k?
[N2O5(sol’n)]
(moles/L)
1.0
0.50
0.25
Instantaneous
rate M/sec
0.00060
0.00030
0.00015
Concentration vs Time Data
[N2O5(sol’n)]
Time (s)
(moles/L)
0
1.00
200
0.88
400
0.78
600
0.69
800
0.61
1000
0.54
1200
0.48
1400
0.43
1600
0.38
1800
0.34
2000
0.30
c.
What are a few different methods by
which you could determine the halflife?
conc$
[conc]$vs$%me$
1.1$
1$
0.9$
0.8$
0.7$
0.6$
0.5$
0.4$
0.3$
0.2$
0$
200$
400$
600$
800$ 1000$ 1200$ 1400$ 1600$ 1800$ 2000$
%me$
1/[conc]&vs&'me&
ln[conc]#vs#&me#
3.50"
0.00"
y"="0.0011x"+"0.828"
R²"="0.97606"
y"="$0.0006x"$"0.0093"
R²"="0.9998"
$0.20"
2.50"
ln#conc#
1/conc&
3.00"
2.00"
1.50"
$0.40"
$0.60"
$0.80"
1.00"
$1.00"
0.50"
0"
500"
1000"
'me&
1500"
2000"
$1.20"
0"
500"
1000"
&me#
1500"
2000"
P E.2 (pg 4 of 7)
3.
Decomposition of H2O2
The decomposition of hydrogen peroxide was studied at a particular temperature, and
experimental data is presented in the table to the right.
2 H2O2(g)
a.
→
2H2O(g)
+
O2(g)
What are the two methods that you could use to determine the rate law for this
reaction with the data that has been given?
b.
Write out the rate law for this decomposition reaction.
c.
What are the three methods that you could use to determine the rate constant?
i. Use the easiest of these two methods to determine a value for k.
ii.
Time (s)
0
120
300
600
[H2O2(g)] (mole/L)
1.0
0.91
0.78
0.59
1200
1800
2400
3000
3600
0.37
0.22
0.13
0.082
0.050
What are the units on k?
d.
conc$
[conc]$vs$%me$
Now that you know the order of the
reaction calculate the half-life for this
reaction (if it is not 1st order, calculate the
half-life at the initial concentration)
• Show how the conc vs time graph
confirms the value that you calculated
1.1$
1$
0.9$
0.8$
0.7$
0.6$
0.5$
0.4$
0.3$
0.2$
0$
200$ 400$ 600$ 800$ 1000$ 1200$ 1400$ 1600$ 1800$ 2000$
%me$
1/[conc]&vs&'me&
22.00"
ln[conc]#vs#&me#
20.00"
18.00"
0.00"
ln#conc#
$1.00"
1/conc&
y"="$0.0008x"$"0.0049"
R²"="0.99978"
$0.50"
y"="0.0046x")"0.8472"
R²"="0.87227"
16.00"
14.00"
12.00"
10.00"
$1.50"
8.00"
$2.00"
6.00"
$2.50"
4.00"
$3.00"
2.00"
0.00"
$3.50"
0"
500"
1000"
1500"
2000"
&me#
2500"
3000"
3500"
4000"
)2.00"
0"
500" 1000" 1500" 2000" 2500" 3000" 3500" 4000"
'me&
P E.2 (pg 5 of 7)
1.
Rate Laws − Graphing
ANSWERS
Dimerization of Butadiene
a. The very easiest way to determine the order of this reaction is using the graphs. The graph that is linear will tell us the
order of the reaction. The y = mx + b equations have been displayed on each graph, for which the slope will be useful,
and there is an R2 value which is the coefficient of determination (you will learn more about R and R2 when you take a
statistics course) for now, you can think of it as a “goodness-of-fit” indicator. The closer the R2 is to 1, the better the data
1
fits to a straight line. Thus you can see that the R2 value is best for the graph of
vs.time gives the straightest line
[C4 H 6 ]
and since this is the most linear graph, we know that the reaction is second order.
The rate law would be: rate = k [C4H6]2
b.
With the information available in this question, the rate constant can most easily be determined by determining at the
1
slope of the
vs.time graph, since the slope will equal k. k = 0.0142 L mol−1 sec−1
[C4 H 6 ]
Remember the second order integrated rate law is
1
1
= kt +
which is in the form y = mx + b for which
[C4 H 6 ]t
[C4 H 6 ]o
m (the slope) is equal to k.
c.
Remember that the half life of a second order reaction is dependent on the starting concentration, and thus not
1
1
particularly useful. The second order half life equation is:
= t1/2 and substitute
= t1/2
k[C4 H 6 ]o
0.0142 × [0.0167]o
thus t½ = 4217 sec (for the initial concentration, the half-life if ever changing throughout the reaction.)
Alternatively you could use the second order integrated rate law
1
1
−
= kt1/2
[C4 H 6 ]t [C4 H 6 ]o
which you can substitute in with the initial concentration and half its value,
1
1
along with the rate constant form part (b).
−
= 0.0142t1/2
[0.00835]t [0.0167]o
t½ = 4217 sec
P E.2 (pg 6 of 7)
2.
Rate Laws − Graphing
ANSWERS
Decomposition of dinitrogen pentoxide
a. With information in various forms, you have various options:
i. First you could use the concentration and rates data in the table on the left to “do table logic” by making a ratio of
x
x
rate1 k1 ⎛ conc1 ⎞
0.0006 k1 ⎛ 1 ⎞
rates and concentrations.
= ⎜
=
⎜
⎟
rate2 k2 ⎝ conc2 ⎟⎠
0.0003 k2 ⎝ 0.5 ⎠
k’s cancel, thus 2 = (2)1 thus first order.
ii. You could also use the graph of concentration vs time to draw tangents to the curve, determine the slope of the
tangents − the slope of those tangents would be instantaneous rates at two separate concentrations. Then you could
rate1 k1 conc1
use those rates with their concentrations to make the ratio of rates and concentrations:
=
This is the
rate2 k2 conc2
same as the “table logic method, however, this time you are getting your rate and concentration data from the graph.
iii. Thirdly, you can use the concentration vs time data in the table on the right and manipulate the concentration values
1
to ln[conc] and
and graph both vs time to look for which of the graphs presented will give a straight line to
[conc]
determine the order of the reaction. These graphs are shown with r2 values, and clearly the ln[conc] vs time is most
linear, thus first order.
iv. Lastly you could look at the [conc] vs time graph and “look for” a constant half life. From conc 1 to 0.5 shows a half
life of ~1150, from conc 0.8 to 0.4 also shows ∆time 1500 − 350 = 1150, and from conc 0.6 to 0.3 show ∆time 2000
− 850 = 1150. Constant half life confirms first order kinetics.
b.
You can use several different methods to determine the rate constant after learning that the reaction is first order.
i. Calculate the value for the reaction constant, k by using the integrated rate law for first order reactions with the
formula:
ln[A]t − ln[A]0 = −kt and substituting in time and concentration data from the table on the right.
ln(0.88) − ln(0.78) = −k(200) solve to find k=0.000603 sec−1
ii. You also can determine at the slope of the straight line graph of ln[conc] vs time. The equation is shown on the
graph, and thus the −slope = k = 0.0006 sec−1.
iii. You could use one of the instantaneous rate trials from the left table and substitute back into the rate law:
rate = k [N2O5] substitute: 0.0006 = k (1) thus k = 0.0006 sec−1
iv. You could use the first order half life equation k =
0.693
if you chose to first determine the half life as outlined
t1/2
by methods (ii) or (iii) as outlined in part (c).
c.
There are a few possible methods by which you could determine the half life.
i. Calculate t½ using the half-life formula, having established a rate constant as in part (b).
k=
0.693
0.693
t1/2 =
t1/2
0.0006
t1/2 = 1155sec
ii. Read half-life times from the [conc] vs time graph.
iii. Use the [conc] vs time data in the table to the left to get an approximate half life using
[N2O5] = 1.00 and then [N2O5] = 0.54 for a t½ ≃ 1000 sec (close to 1155).
P E.2 (pg 7 of 7)
3.
Rate Laws − Graphing
ANSWERS
Decomposition of H2O2
a. Determining rate law....
i. You could use the graph of concentration vs time to draw tangents to the curve, determine the slope of the tangents −
the slope of those tangents would be instantaneous rates at two separate concentrations. Then you could use those
rate1 k1 conc1
rates with their concentrations to make the ratio of rates and concentrations:
=
This is the same as
rate2 k2 conc2
the “table logic method, however, without a table with information, you must get your rate and concentration data
from the graph.
ii. However, it would be far easier to manipulate the data and graph it two more times. By looking at the three graphs to
see which one is linear will tell us what is the order of this decomposition reaction. As you can see, the graphs have
been constructed for you and even without the R2 values, it is quite obvious that the ln(conc) vs time graph is linear
indicating this reaction is first order.
b. Rate = k [H2O2]
c.
Determine the rate constant by three different methods, the first one is likely the easiest.
i.
The −slope of the ln[H2O2] vs time graph is equal to the rate constant. k = 0.0008 sec−1
ii. Or you could use the average rate for the first 120 seconds and make the not so perfect assumption that this would
Δ [ H 2O2 ]
be close to the initial instantaneous rate. initial
=~ inital (instantaneous) rate
Δtime
[1.0 − 0.91] = 0.00075M sec−1
120 sec
Rate = k [H2O2]
and then putting this into the rate equation to solve for k.
and so you can substitute
0.00075 Msec−1 = k [1.0]
and thus k ≃ 0.00075 sec−1.
iii. Or if you first chose to determine the half life by the method (ii) in part (d), you could use the first order half life
equation k =
0.693
0.693
and solve for k k =
k = 0.00082 sec−1s
t1/2
850
(Note, I used 850 as an approximate half life from an average of the two values from answer (d i) and (d ii).)
d.
There are three methods you could use to solve for half life, with the given data.
i.
t½ =
0.693
= 866 sec
0.0008s −1
ii. Using the [conc] vs time graph, you can approximate the half life by reading the time required for the concentration
to make any half reduction in concentration to get t½ ≃ 800 sec
iii. Use the [conc] vs time data in the table to the left to get an approximate half life. Not very many of the
concentrations result in a half reduction, but the [H2O2] = 0.78 to [H2O2] = 0.37 is close resulting in a t½ ≃ 900 sec.