CHEM1612 - PharmacyWeek 11: Kinetics - Half Life
Dr. Siegbert Schmid
School of Chemistry, Rm 223
Phone: 9351 4196
E-mail: [email protected]
Unless otherwise stated, all images in this file have been reproduced from:
Blackman, Bottle, Schmid, Mocerino and Wille, Chemistry, John Wiley & Sons Australia, Ltd. 2008
ISBN: 9 78047081 0866
Lecture 31-3
For the general reaction:
a A + b B + c C … d D + e E ….
rate = k [A]m [C]n
k rate constant (depends only on temperature) m is the order of the reaction with respect to A (or “in” A), n is the order of the reaction with respect to C Overall order of the reaction is = m + n
Reaction orders cannot be deduced from the balanced reaction, but only by experiment.
Rate Law and Reaction Order
Lecture 31-4
Rate = k [A]m[B]n[C]p
If m = 1, reaction is 1st order in A Rate = k [A]1
If [A] doubles, then rate goes up by a factor of ?
If m = 2, reaction is 2nd order in A. Rate = k [A]2
If [A] doubles, then rate goes up by a factor of ?
If m = 0, reaction is zero order in A. Rate = k [A]0
If [A] doubles, then rate goes up by a factor of ?
Interpreting Rate Laws
2
4
1
Lecture 31-5
Consider the reaction:
ClO3-(aq) + 9 I-(aq) + 6 H+(aq) Cl-(aq) + 3 I3
-(aq) + 3 H2O(l)
Expt. [ClO3-] / M [I-] / M [H+] / M Initial Rate / M s-1
1 0.10 0.10 0.10 0.05
2 0.20 0.10 0.10 0.10
3 0.20 0.20 0.10 0.20
4 0.20 0.20 0.20 0.80
Determine the rate law for the above reaction.
General expression: Rate = k [ClO3-]x[I-]y[H+]z; must find x, y, and z
Using Data to Determine Order
Lecture 31-6
Using Data to Determine Order in ClO3
-Expt. [ClO3
-] / M [I-] / M [H+] / M Initial Rate / M s-1
1 0.10 0.10 0.10 0.05
2 0.20 0.10 0.10 0.10
3 0.20 0.20 0.10 0.20
4 0.20 0.20 0.20 0.80
General expression: Rate = k [ClO3-]x [I-]y [H+]z
Determine x:
Compare experiment 1 and 2
1
5.05.0
]20.0[
]10.0[
10.0
05.0
]ClO[
]ClO[
][][]ClO[
][][]ClO[
rate2
rate1
23
13
2223
1113
x
HIk
HIk
x
x
x
zyx
zyx
Pick exptswhere onlyone conc.changes
Lecture 31-7
Using Data to Determine Order in I-Expt. [ClO3
-] / M [I-] / M [H+] / M Initial Rate / M s-1
1 0.10 0.10 0.10 0.05
2 0.20 0.10 0.10 0.10
3 0.20 0.20 0.10 0.20
4 0.20 0.20 0.20 0.80
General expression: Rate = k [ClO3-]x [I-]y [H+]z
Determine y:
Compare experiment 2 and 3
1
5.05.0
]20.0[
]10.0[
20.0
10.0
]I[
]I[
][][]ClO[
][][]ClO[
rate3
rate2
3
2
3333
2223
y
HIk
HIk
y
y
y
zyx
zyx
Pick exptswhere onlyone conc.changes
Lecture 31-8
Using Data to Determine Order in H+Expt. [ClO3
-] / M [I-] / M [H+] / M Initial Rate / M s-1
1 0.10 0.10 0.10 0.052 0.20 0.10 0.10 0.103 0.20 0.20 0.10 0.204 0.20 0.20 0.20 0.80
General expression: Rate = k [ClO3-]x [I-]y [H+]z
Determine z:
Compare experiment 3 and 4
So overall the rate law is:
Rate = k [ClO3-] [I-] [H+]2
2
5.025.0
]20.0[
]10.0[
80.0
2.0
][
]H[
][][]ClO[
][][]ClO[
rate4
rate3
4
3
4443
3333
z
HHIk
HIk
z
z
z
zyx
zyx
Pick exptswhere onlyone conc.changes
Lecture 31-9
ClO3-(aq) + 9 I-(aq) + 6 H+(aq) Cl-(aq) + 3 I3
-(aq) + 3 H2O(l)
rate= k [ClO3-][I-][H+]2
The order is 1 with respect to [ClO3-]
The order is 1 with respect to [I-]The order is 2 with respect to [H+]The overall order of the reaction is (1+1+2)= 4.
Calculate the rate constant, ke.g. Experiment 1
0.05 M s-1 = k (0.10 M)(0.10 M)(0.10 M)2
k = 5.0 x 102 M-3 s-1
(note: units depend on rate law)
Determine the Rate Constant
[ClO3-] / M [I-] / M [H+] / M Initial Rate / M s-1
0.10 0.10 0.10 0.05
Lecture 31-10
Deriving Rate Law and Rate ConstantFrom the following experimental data derive rate law and k for the reaction:
CH3CHO(g) CH4(g) + CO(g)
Expt. [CH3CHO] Rate of loss
(mol/L) (mo/L.sec)
1 0.10 0.020
2 0.20 0.081
3 0.30 0.182
4 0.40 0.318
2
5.025.0
]20.0[
]10.0[
081.0
02.0
]CHOCH[
]CHOCH[
]CHOCH[
]CHOCH[
rate2
rate1
23
13
23
13
x
k
k
x
x
x
x
x
Lecture 31-11
Rate law is: rate = k [CH3CHO]2
Here the rate goes up by a factor of four when initial concentration doubles. Therefore, this reaction is second order.
Now determine the value of k.
From experiment #3 data:
0.182 mol/L·s = k (0.30 mol/L)2
k = 0.182 / (0.3)2 = 2.0 (L / mol·s)
Using k you can calculate the rate at other values of [CH3CHO] at same T.
Deriving Rate Law and Rate Constant
Lecture 31-12
Determine the rate law and value of the rate constant for this reaction:
NO2 (g) + CO (g) NO (g) + CO2 (g)
[NO2] / M [CO] / M Initial Rate / M s-1
1 0.10 0.10 0.0051
2 0.40 0.10 0.080
3 0.10 0.20 0.0052
Homework
Rate = k[NO2]2
k = 0.5 s-1 M-1
Homework
Lecture 31-13
The Iodine Clock Mix different amounts of HIO3 + NaHSO3 + starch. Concentration of reactants is: [beaker I] > [beaker II] >[beaker III].
The following reactions take place consecutively in each beaker:
Starch forms a blackish blue complex with iodine. As the final reaction is the fastest, the colour of the complex only becomes apparent once the sulphite is fully consumed.
The reaction is slowest in the solution with the lowest concentration, as the reaction time is dependent on the concentration.
+ starchBlue-Black complex
Lecture 31-14
Concentration - Time RelationshipsHow long will it take for x moles of A to be consumed?
What is the concentration of reactant as function of time?
For FIRST ORDER REACTIONS the rate law is
For infinitesimal differences, Δ /Δt becomes a differential, so
Integrating both of the expression , gives:
k[A]ΔtΔ[A]
=- =rate
k[A]dt
d[A]
tA
A
t
0
dtkA
dA
0
Lecture 31-15
Integrating gives
[A]t / [A]0 = fraction remaining after time t has elapsed.
Called the integrated first-order rate law.
ktA
A t 0][
][ln
First Order Reactions
ktAA
ktAA
t
t
]ln[]ln[
]ln[]ln[
0
0
[A]t =conc. of A at time = t during exp[A]0 =conc. of A at time t = 0k = rate constantt = time
tA
A
t
0
dtkA
dA
0
Lecture 31-16
Second Order Reactions
For a second-order reaction, with one reactant A only:
Integrating both sides gives:
2][][
Akdt
Adrate
tA
A
t
02
dtkA
dA
0
ktAA t
0][
1
][
1 [A]t =conc. of A at time = t during exp[A]0 =conc. of A at time t= 0k = rate constantt= time
Lecture 31-17
Zero Order Reactions
For a zero order reaction:
kAkdt
Adrate 0][
][
Integrating gives tA
A
t
k 0
dtdA1
0
[A]t = [A]0 – k t[A]t = [A]0 – k t[A]t =conc. of A at time = t during exp[A]0 =conc. of A at time t= 0k = rate constantt= time
Lecture 31-18
Identifying reaction order
ktA
A t 0][
][ln
FIRST ORDER SECOND ORDER ZERO ORDER
ktAA t
0][
1
][
1[A]t = [A]0 – k t[A]t = [A]0 – k t
ln[A]t = -k t + ln[A]0
0][
1
][
1
Akt
A t
Figure from
Silberberg, “C
hemistry”,
McG
raw H
ill, 2006.
Lecture 31-19
14C 12C occurs at a known rate.
How old is a bone if the 14C concentration
is only 9% of the natural concentration and
k = 1.21 x 10–4 y –1 ?
9% concentration:
[14C] / [14C]o = 0.09
Now: ln[14C] / [14C]o = -kt
Therefore: ln(0.09) = -(1.21 x 10–4) x t
Solving for t: -2.407 / -1.21 x 10–4 = t, or t = 20,000 y
Radiocarbon Dating
What happened to Giant Kangaroos?
Lecture 31-20
Half-life for a first-order reaction is independent of initial concentration.
Expression for half-life:
t1/2 = 0.693 / k
Half-life Half-life: time required for concentration of reactant to decrease to
half of its original concentration.
Lecture 31-21
90Sr was released during the Chernobyl explosion in 1986.
Given t1/2 (90Sr) = 28.1 y, what percentage remains today?
Reaction is of type: A B (nuclear decay follows 1st order reaction)
t1/2 = 0.693 / k
Thus, k = 0.693 / 28.1 = 0.02466 y–1
Now, ln([A] / [A]o) = -kt
Therefore, [A] / [A]o = exp(-0.02466 x 25)
[A] / [A]o = 0.54, thus 54% remain today
2011 - 1986 years
Half-life Example
Lecture 31-22
Summary
CONCEPTS Reaction rate as change in concentration of reactants and products Rate law Reaction order Integrated rate laws for first, second, and zero order reactions Half-life relations
CALCULATIONS Calculate average and instantaneous rates from data (with units) Derive rate law and rate constant from experimental data (with units) Use integrated rate laws to calculate [A] at given times Calculation using the half-life relations