# Rate Laws from Graphs of Concentration Versus Time

advertisement ```TOPIC 8 –Determine the order of reactions by graphically methods
Additional Note
Rate Laws from Graphs of Concentration
Versus Time (Integrated Rate Laws)
In order to determine the rate law for a reaction from a set of data consisting of concentration (or the
values of some function of concentration) versus time, make three graphs.
•
•
•
[A] versus t (linear for a zero order reaction)
ln [A] versus t (linear for a 1st order reaction)
1 / [A] versus t (linear for a 2nd order reaction)
The graph that is linear indicates the order of the reaction with respect to A. Then, you can choose the
correct rate equation:
For a zero order reaction,
For a 1st order reaction,
For a 2nd order reaction,
rate = k
rate = k[A]
rate = k[A]2
(k = - slope of line)
(k = - slope of line)
(k = slope of line)
Examples
For a zero order reaction, as shown in the following figure, the plot of [A] versus time is a straight
line with k = - slope of the line. Other graphs are curved for a zero order reaction.
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For a first order reaction, as shown in the following figure, the plot of the logrithm of [A] versus
time is a straight line with k = - slope of the line. Other graphs are curved for a first order reaction.
For a second order reaction, as shown in the following figure, the plot of 1/[A] versus time is a
straight line with k = slope of the line. Other graphs are curved for a second order reaction.
Exercise 8a
1. The following data were obtained on the rate of hydrolysis of 17 % sucrose in 0.99 mol L-1 HCl aqueous
solution at 35 oC.
t / min
9.82
59.60
93.18
142.9
294.8
589.4
Sucrose remaining, %
96.5
80.3
71.0
59.1
32.8
11.1
What is the order of the reaction with respect to sucrose, and what is the value of the rate constant k?
Solving Method
Sucrose is the reactant. Assume this is first order reaction and we need to prove by plot a suitable graph.
From the integrated rate equation for first-order reaction: ln[A] = ln[A]o – kt , by assuming
[A] = concentration of sucrose = % sucrose remaining and
[A]o = initial concentration of sucrose = 100%
So calculate ln [A] at each times, t and then plot graph ln [A] as y-axis versus t (min)as x-axis. Use graph
paper. If the graph is linear or straight line graph, it shows that the reaction is first-order. Calculate the
slope of the graph; the value is equal to the rate constant, k. Give the unit of k.
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2. Methy acetate is hydrolyzed in approximately 1 M HCl at 25 oC. Aliquots of equal volume are removed
at intervals and titrated with a solution of NaOH. Calculate the first-order rate constant k, from the
following volumes of NaOH required to neutralize the aliquot:
t /s
339
1242
2745
4546
∞
VNaOH /mL
26.34
27.80
29.70
31.81
39.81
Solving Method
The volumes reading are show the amount of the product of the reaction. We must change the data to the
reactant reading as follows;
Lets the reaction is:
A → P
At initial t = 0
[A]o
0
At any time, t
[A]o-x
x = Vt
At ∞ (reaction completed) 0
[A]o = V∞
So recalculate the data :
t /s-1
339
1242
2745
4546
[A] = V∞-Vt
39.81-26.34
39.81-27.80
39.81-29.70
39.81-31.81
ln[A] = ln (V∞-Vt)
Then plot ln[A] ( y-axis) versus t (x-axis) on a graph paper. The graph is linear with the slope equal to the
the rate constant, k.
3.
At 504 oC dimettyl ether undergoes the decomposition
(CH3)2O (g) → CH4 (g) + H2 (g) + CO (g)
The total pressure of the reacting system, measured as function of time, t as below:
t /(s)
0
390
777
1195
3155
P /(torr)
312
408
488
562
779
∞
931
(a) Calculate the partial pressures of (CH3)2O and CH4 at each time, t.
(b) Determine the order and the rate constant, k for the decomposition.
Solving Method
By assuming Po= initial pressure, and x are the reduced pressures during the composition at any time, t.
You will find the equation that relate Po with P (Total pressure of the system)
(CH3)2O → CH4 + H2 + CO
At t = 0
Po
0
0
0
At t = t
Po-x
x
x
x
Total pressure, P = Po-x+3x = Po+2x
x = (P-Po) / 2
Find the values of x, which is gives the partial pressures of CH4, and calculate P(CH3)2O, which is equal
to Po-x, put the data in the table below. Then assume the order of the reaction is first-order, plot ln
P(CH3)2O versus t on graph paper, find the slope, m = k ( rate constant) .
t /(s)
P /(torr)
( P − Po )
x=
= PCH 4 / (torr)
2
Po − x = P( CH 3 ) 2 O /(torr )
ln (Po-x)
0
312
0
0
390
408
(408 − 312)
= 48
2
312-48=264
0
5.576
36
777
488
1195
562
3155
779
∞
931
4. The following total pressure measurements as a function of time were obtained for the dimerization of 1,
3-butadiene (C4H6) under constant volume conditions at 326 oC.
t/min
P/ Torr
0
632.0
3.25
618.5
12.18
584.2
24.55
546.8
42.50
509.3
68.05
474.6
Determine the order of the reaction and evaluate the rate constant.
Solving Method
By assuming Po= initial pressure, and x are the reduced pressures during the composition at any time, t.
You will find the equation that relate Po with P (Total pressure of the system)
Lets the reaction is:
2A → A2
At initial t = 0
Po
0
At any time, t
Po-2x
x
Total pressure, P = Po-2x+x = Po-x
x = (Po-P)
PA = Po-2x = Po-2(Po-P) = 2P-Po
Calculate the values of 2P-Po, which is gives the partial pressures of C4H6, and put the data in the table
below. Then assume the order of the reaction is second-order, plot 1/P(C4H6) versus t on graph paper, find
the slope, m = k ( rate constant) .
t/min
0
3.25
12.18
24.55
42.50
68.05
P/ Torr
632.0
618.5
584.2
546.8
509.3
474.6
P(C4H6)=2P-Po/Torr
ln P(C4H6)
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