Posts filed under 'Chemical Kinetics'
If we have a problem that we must determine the rate law of the reaction and we just given by a particular data, the easy step to solve the problem is by taking a plot of ln[reactant] versus time and 1/[reactant] versus time too. If we have a straight line between one of the two of the plots, thus the plot with the straight line is determined the order of the reaction. If the straight line is the plot which ln[reactant] versus time thus the reaction is first order, and the reverse is the second order. Take a look to the sample below:
A student has studied the decomposition of hydrogen peroxide, and he obtained the following data at a given temperature
|
|
[H2O2] M
|
Time (s)
|
|
1.00
|
0
|
|
0.91
|
120
|
|
0.78
|
300
|
|
0.59
|
600
|
|
0.37
|
1200
|
|
0.22
|
1800
|
|
0.13
|
2400
|
|
0.082
|
3000
|
|
0.050
|
3600
|
|
- Is this reaction first order or second order?
- Determine the rate law and the integrated law
- What is the value of the rate constant for the reaction
- What is the half-life for the reaction under the condition of the experiment
To find whether the rate law of this reaction is fisrt order or second order, we should have to make a plot of ln[H2O2] versus time and 1/[H2O2] versus time. If the first graph is a straight line thus the reaction should be fisrt order. If the second graph is a straight line thus the reaction is second order. The data necessary to make these plots are as follows:
|
|
Time (s)
|
1/[H2O2]
|
ln [H2O2]
|
|
0
|
1
|
0
|
|
120
|
1.094
|
-0.094
|
|
300
|
1.282
|
-0.248
|
|
600
|
1.695
|
-0.528
|
|
1200
|
2.703
|
-0.994
|
|
1800
|
4.545
|
-1.514
|
|
2400
|
7.692
|
-2.040
|
|
3000
|
12.195
|
-2.501
|
|
3600
|
20
|
-2.995
|
|
the resulting plots are shown in the figures below:
from the figures above we know that ln[H2O2] versus time is a straight line but 1/[H2O2] versus time is not a straight line. Thus the decomposition of H2O2 is a fisrt order reaction and we can write the rate law as:
rate = k[H2O2]
and the integrated rate law as
ln[H2O2] = -kt + ln[H2O2]o
Since the reaction is first order, the slope of the line equals -k where
slope = Change in y / cahnge in x
slope = -2.995 - (-2.501) / (3600-3000)
slope = -8.233×10-4 s-
k = -(slope) = 8.233×10-4 s-
The half life of the fisrt order reaction difined as
substituting the value of k we have
t1/2 = 0.693 / 8.233×10-4 s-
t1/2 = 841.735 s
May 30th, 2008
In this article we try to find the general formula for the half-life of fecond order reaction. take a look into a general reaction that involve a single reactant and this reaction is a second order in the reactant (A), then the reaction can be writen as
aA -> products
The rate law for the reaction is
Rate = k[A]2
and the integrated second order reaction can be write as
The important meaning of the equation above is a plot of 1/[A] versus t will produce a straight line with the slope of k
when t1/2 is reached the concentration of [A] is equal to 1/2[A] and the equation above can be writen as
1/ (1/2[A]o) = kt1/2 + 1/[A]o
2/[A]o - 1/[A]o = kt1/2
1/[A]o = kt1/2
solving for t1/2 will give the expression for the half life of the second order reaction
This is the general equation to find half life of second order reaction.
May 28th, 2008
We can get general formula for the half-life of a first order reaction that erived from the integrated rate law for the general reaction:
aA -> products
The rate law is
Rate = k[A]
and the integrated fisrt order rate law is
ln[A] = -kt + ln [A]o
the important thing of this equation is that this is the y = mx + c where plot of y versus x is a straight line with the slope is m and the intercept c. Where y is ln[A], x is t, m= -k, and the c = ln[A]o. The plot of ln[A] versus t is the straight line it means that the reaction is first order when the plot is not a straight line the reaction is not first order. This is how we can test whether a reaction is the first order reaction or not.
or we can be expressed in term of a ratio as:
when t1/2 we have the concentration of [A] = 1/2 [A]o then the integrated formula becomes
ln { [A]o / (1/2 [A]o) } = k.t1/2
ln(2) = k.t1/2
because ln (2) is 0.693 then substituting this value to the equation above gives
the equation above is the formula to find half life of first order reaction. Just remember that for the fisrt order reaction the half life does not depend on the concentration
May 28th, 2008
Chemistry Tutor Site
A certain first order reaction has a half life of 20.0 minutes.
a. Calculate the rate constant for this reaction.
b. How much time is required for this reaction to be 75% complete?.
solution
a. Solving for rate constant:
k = 0.693/t1/2k = k = 0.693/20.0 min
k = 3.47 x 10-2 min-.
b. Calculate time
ln( [A]o/[A] ) = kt
the reaction is 755 complete, this means 75% of the reactant has been consumed, and leaving 25% in the original form
This means that :
[A]/[A]o x100 = 25
[A]/[A]o = 0.25 or can be write as: [A]o/[A] = 4.0
ln([A]o/[A]) = ln(4.0) = kt = 3.47 x 10-2 min-
t = ln(4.0) / 3.47 x 10-2 min-
t = 40 min
April 8th, 2008
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