CHEM1612 - Pharmacy Week 11: Kinetics - Half Life

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 m = 0, reaction is zero order in A. Rate = k [A]0


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


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 y:


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


Lecture 31-8

Using Data to Determine Order in H+Expt. [ClO3


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


Determine z:


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


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

CHEM1612 - Pharmacy Week 11: Kinetics - Half Life

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