lecture 4 c - reaction kinetics
DESCRIPTION
chemistryTRANSCRIPT
Objectives:
1.Write the rate law for zero order, 1st order and 2nd order reaction
2. Define half-life.3. Draw the respective graphs for the different
order reactions4. Solve quantitative problems.
Integrated Rate Law
Zero Order Reaction
A zero order reaction is a reaction independent of the concentration of reactant.
A product
The rate law is given byrate = k[A]0
rate = k
[A] M
rate
Integrated rate equations
- d[A] = k dt Using calculus, - d[A] = kdt - ∫d [A] = k ∫dt - [A] = kt + c substituting t=0, [A] = [A]0
- [A]0 = k(0)+c c = - [A]0
[A]0 -[A] = kt
Unit of k for zero order reaction M s-1
Graphs for zero order reaction
[A]o – [A] = kt
[A] = -k t + [A]o[A]
t
[A]o
t
[A]o – [A]
y = mx + c
y = m x + c
[A]
rate
Half-life (t½)
Half life (t½) is the time required for the concentration of a reactant to decrease to half of its initial value.
zero order reactionSubstituting t = t1/2 , and [A] = [A]0 into the zeroorder reaction, gives 2 [A]0 - [A] = kt [A]0 – [A]0 = kt1/2
2 Solving for t1/2 gives t1/2 = [A]0 2k
From the rate law, rate = k[A] To obtain the units of k k = rate [A] unit k = M s-1 M = s-1
First Order Reactions
A first order reaction is a reaction whereby its rate depends on the concentration of reactant raised to the first power.
Rate = k[A]y = mx + c
Rate
Ms-1
[A] ,M
For first order reaction,
- d[A]dt
= k[A]
- d[A][A]
= k dt
- d[A][A]
= k dt∫ ∫
- ln [A] = kt + c
substituting
t = 0, [A]=[A]0
- ln [A]0 = k(0) + c c = ln[A]0
-ln [A] = kt – ln[A]0
ln[A]0
[A] = kt
Rate = k[A]
Characteristic graphs of 1st order reaction
ln[A]
ln[A]o
t
ln[A]o
[A]
t
ln[A]o – ln[A] =kt
[A]
tln
[A]0------[A] = kt
ln[A] = - kt +ln[A]o
The reaction 2A B is first order with respect A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ?
ln[A]o – ln[A] = kt
kt = ln[A]0 – ln[A]
t =ln[A]0 – ln[A]
k= 66 s
[A]0 = 0.88 M[A] = 0.14 M
ln[A]0
[A]k
=ln
0.88 M0.14 M
2.8 x 10-2 s-1=
Example
Example Decomposition of H2O2 (aq) is first order, given that k
= 3.66 x 10-3 s-1 and [H2O2 ]o = 0.882 M, determinea) the time at which [H2O2] = 0.600 Mb) the [H2O2 ] after 225 s.
Solution :
a) ktln[H2O2]0
[H2O2]=
[H2O2]=
=ln 0.8820.600
3.66 x 10-3 s-1 x t
ln 1.47 = 3.66 x 10-3 s-1 x t
t =ln 1.473.66 x 10-3
= 105.26 s
b)ln
[H2O2]0
[H2O2]=
[H2O2]= kt
ln0.882[H2O2]
=[H2O2]
= 3.66 x 10-3s-1x 225 s
[H2O2] = 0.387 M
Exercise,
The conversion of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7 X 10-4 s-1 at 500°C.
CH2
CH2 CH2 CH3-CH=CH2
a) If the initial concentration of cyclopropane was 0.25 M, what is the concentration after 8.8 minute. (0.18 M)
b) How long will it take for the concentration of cyclopropane to decrease from 0.25 M to 0.15 M? (13 min)
The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.
t=t½ when [A] = [A]0
2
ln[A]0
[A]0/2k
=t½ln2k
=0.693
k=
What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1?
t½ln2k
=0.693
5.7 x 10-4 s-1= = 1200 s = 20 minutes
How do you know decomposition is first order?units of k (s-1)
Half-life of a first-order reaction
A product
First-order reaction
No. of half-lives [A]o = 8 M
1
2
3
4
4
2
1
1/2
t½ln2k
=
Example The decomposition of ethane C2H6 to methyl radicals is a first order reaction with a rate constant of 5.36 x 10-4 s-1 at 700o C. C2H6 (g) → 2 CH3 (g)Calculate the half life of the reaction in minutes.Solution
t1/2 =ln 2
k
=0.693
5.36 x 10-4
= 1.29 x 103 s= 21.5 min
Problem 2
What is the half-life of a compound if 75% of a given sample of the compound decomposes in 60 min? Assume it is first-order reactions kinetics.
(t1/2=30 min)
ln [A0]/[A] = kt ln [A0]/[A] = kt
t = 60 mint = 60 min[A0] = 1[A0] = 1[A] = 0.25 [A] = 0.25
Second Order ReactionsA second order reaction is a reaction which rate depends on the concentration of one reactant raised to the second power or on the concentration of two different reactants each raised to the first power.
Example A productWhere
Rate = - d[A]dt
= k[A]2
To obtain the units of k
k =rate[A]2
= M/sM2
= M-1 s-1
Using calculus, the following expression can be obtained
1[A]
1-[A]0
= + kt
Unit k
Characteristic graphs for second order reaction
rate
[A]
Rate = k [A]2
rate
[A]2
Graphs for second order reaction
[A]
t
[A]1 + kt=
[A]o
1
1/[A] M-1
1/[A]o
t
1/[A] – 1/[A]o
t
22ndnd –order reaction, r = k[A] –order reaction, r = k[A]22
If [A] doubles,If [A] doubles, rr22 = k (2[A]) = k (2[A])22
= k ( 4 [A]= k ( 4 [A]22 ) ) = 4 k [A]= 4 k [A]22
= 4 r= 4 r
R will increase by 4 times if [A] doublesR will increase by 4 times if [A] doubles
Half life of a second order reaction
1[A]
1[A]0
= + kt
Substituting t = t1/2[A]= [A]o
1[A]0
1[A]0
= + kt1/2
t1/2= 1k[A]0
2
2
Detemination of half-life using graph for second order reaction[A]0
[A]0/2
[A]0/4
[A]0/8
tx 2x 4x
t1/2= 1k[A]0
ExampleIodine atoms combine to form molecular iodine inthe gaseous phase I (g) + I (g) I2(g)
This reaction is a second order reaction , with the rate constant of 7.0 x 109 M-1 s-1
If the initial concentration of iodine was 0.086 M, i) calculate it’s concentration after 2 min.ii) calculate the half life of the reaction if the initial concentration of iodine is 0.06 M and 0.42 M respectively.
Solution :
i)1[A]
1[A]0
= + kt
1[A]
1[0.086]
= + (7.0 x 109 x 2 x 60 )
= 8.4 x 1011
[A] = 1.190 x 10-12 M
ii) [I2] = 0.06 M
t1/2= 1k[A]0
=1
7.0 x 109 x 0.060
= 2.4 x 10-10 s
[I2] = 0.42 M
t1/2= 1k[A]0
7.0 x 109 x 0.042
1=
= 3.4 x 10-10 s
Example,
The following results were obtained from an experimental investigation on dissociation of dinitrogen pentoxide at 45oC
N2O5(g) 2 NO2(g) + ½ O2(g)
time, t/min 0 10 20 30 40 50 60
[N2O5] x 10-4 M 176 124 93 71 53 39 29
Plot graph of [N2O5] vs time, determine
i) The order of the reaction
ii) the rate constant k
Using graph
Solution :
[N2O5] x 10-4 /M
Time ( min)
180
160
140
80
120100
8060
4020
10 20 30 40 50 60 70
Based on the above graph, Time taken for concentration of N2O5 to change from 176 x 10-4 M to 88 x 10-4 M is 20 minTime taken for concentration of N2O5 to change from 88 x 10-4 M to 44 x 10-4 M is also 20 minThe half life for the reaction is a constant and does not depend on the initial concentration of N2O5
Thus, the above reaction is first order
i)
ii) k =ln2
20 min = 0.03 min-1
Summary of the Kinetics of Zero-Order, First-Orderand Second-Order Reactions
Order Rate LawConcentration-Time
Equation Half-Life
0
1
2
rate = k
rate = k [A]
rate = k [A]2
ln[A] = ln[A]0 - kt
1[A]
=1
[A]0+ kt
[A] = [A]0 - kt
t½ln2k
=
t½ = [A]0
2k
t½ = 1k[A]0
Zero order 1st order 2nd orderA product A product A product
r = k [A]0 r = k [A]1 r = k [A]2
[A]
r
[A]
r
[A]2
r
[A]
rUnit k = M s-1 Unit k = s-1 Unit k = M-1 s-1
Integrated rate law Integrated rate law Integrated rate law
[A]0 – [A] = kt ln([A]0 / [A]) = kt 1/[A] – 1/[A]0 = kt
t
[A]
[A]0
t
[A]0 - [A]t
[A]
t
[A]
t
ln[A]
ln[A]0
t
1/[A]
1/[A]0
t
ln([A]0 / [A])
t
1/[A] – 1/[A]0
t1/2 = [A]0/2k
t1/2 = ln2/k
t1/2 = 1/k[A]0