Integrated Rate EquationIntegrated Rate Equation

C. Y. Yeung (CHW, 2009)

p.01

To study Integrated Equation, To study Integrated Equation, e.g. :e.g. :

During the reaction, both [A] and [B] During the reaction, both [A] and [B] decrease!decrease!

p.02

large excess of [B]large excess of [B] should be used.should be used.

[A] = -k’t + [A][A] = -k’t + [A]00 (zeroth order)(zeroth order)

In order to ensure that the decreasing In order to ensure that the decreasing rate is due to decreasing [A], not [B] …rate is due to decreasing [A], not [B] …

i.e. keep [B] as “effectively constant”.i.e. keep [B] as “effectively constant”.

p.03

rate =rate = = k’ [A]= k’ [A]11- d[A]- d[A]dtdt

= k’[A]= k’[A]- d[A]- d[A]

dtdt

First OrderFirst Order RxnRxn (m = 1)(m = 1)

= k’t + C= k’t + C- ln [A]- ln [A]

when t = 0, [A] = [A]when t = 0, [A] = [A]00

C = - ln [A]C = - ln [A]00

- ln [A] = k’t - ln [A]- ln [A] = k’t - ln [A]00

integrated rate eqn. integrated rate eqn. (first order)(first order)

= k’dt= k’dt- d[A]- d[A]

[A][A]

= k’ dt= k’ dtd[A]d[A] 11[A][A]--

ln [A] = - k’t + ln [A]ln [A] = - k’t + ln [A]00

p.04Thus,Thus,

Time t0 t1 t2 t3 ….[A] [A]0 [A]1 [A]2 [A]3 ….ln[A] ln [A]0 ln [A]1 ln [A]2 ln [A]3 ….

ln [A] = - k’t + ln [A]ln [A] = - k’t + ln [A]00

ln [A]ln [A]

tt

ln [A]ln [A]00 slope = - k’slope = - k’

(first order)(first order)

p.05

rate =rate = = k’ [A]= k’ [A]22- d[A]- d[A]dtdt

= k’[A]= k’[A]22- d[A]- d[A]

dtdt

Second OrderSecond Order RxnRxn (m = 2)(m = 2)

= k’t + C= k’t + C [A][A]-1-1

when t = 0, [A] = [A]when t = 0, [A] = [A]00

C = [A]C = [A]00-1-1

[A][A]-1-1 = k’t + [A] = k’t + [A]00-1-1

integrated rate eqn. integrated rate eqn. (second order)(second order)

= k’dt= k’dt- d[A]- d[A]

[A][A]22

= k’ dt= k’ dtd[A]d[A] 11[A][A]22--

p.06Thus,Thus,

Time t0 t1 t2 t3 ….[A] [A]0 [A]1 [A]2 [A]3 ….[A]-1 [A]0

-1 [A]1-1 [A]2

-1 [A]3-1 ….

[A][A]-1-1

tt

[A][A]00-1-1

slope = k’slope = k’

(second order)(second order) [A][A]-1-1 = k’t + [A] = k’t + [A]00

-1-1

p.07

Summary … Summary … 3 Integrated Rate Eqns3 Integrated Rate Eqns

ln [A] = - k’t + ln [A]ln [A] = - k’t + ln [A]00

m = 0m = 0

m = 1m = 1

m = 2m = 2 [A][A]-1-1 = k’t + [A] = k’t + [A]00-1-1

[A] = - k’t + [A][A] = - k’t + [A]00

p.08

p. 76 Q.4 Decomposition of Hp. 76 Q.4 Decomposition of H22OO22

Vol. of 0.10MKMnO4 used / cm3

0 30 0.7505 23.4 0.58510 18.3 0.45815 14.2 0.35520 11.1 0.27825 8.7 0.21830 6.8 0.170

Time/min [H2O2]/M

p.09(a)(a) To show 1To show 1stst order : order :

ln [Hln [H22OO22] = - k t + ln [H] = - k t + ln [H22OO22]]00

0 -0.2885 -0.53610 -0.78215 -1.03620 -1.28225 -1.52630 -1.772

Time/min ln [H2O2]

p.10

Plot ln [HPlot ln [H22OO22] against time] against time

y = -0.0495x - 0.2889

-2.000

-1.800

-1.600

-1.400

-1.200

-1.000

-0.800

-0.600

-0.400

-0.200

0.0000 5 10 15 20 25 30 35

time / min

ln [H

2O2]

The graph gives a straight line, The graph gives a straight line, therefore the reaction is 1therefore the reaction is 1stst order order w.r.t. [Hw.r.t. [H22OO22].].

p.11(b)(b) Expression for the Rate Equation :Expression for the Rate Equation :

rate = k[Hrate = k[H22OO22]]

Calculate “k” :Calculate “k” :slope = -k = -0.0495, slope = -k = -0.0495,

k = 0.0495 min k = 0.0495 min-1-1

Calculate half life Calculate half life (time at which [A] = ½[A](time at which [A] = ½[A]00))::

ln (1/2[Hln (1/2[H22OO22]) = - (0.0495) t + ln [H]) = - (0.0495) t + ln [H22OO22]]00

ln (1/2) = - (0.0495) tln (1/2) = - (0.0495) tt = 14.0 minst = 14.0 mins

p.12At the beginning, [HAt the beginning, [H22OO22] = ] = 3.0 mol dm3.0 mol dm-3-3

When expt. started, [HWhen expt. started, [H22OO22] = ] = 0.750 mol dm0.750 mol dm-3-3

(c)(c) How long the [HHow long the [H22OO22] in the ] in the contaminated bottle?contaminated bottle?

ln (0.750) = - (0.0495) t + ln (3.0)ln (0.750) = - (0.0495) t + ln (3.0)t = 28.0 minst = 28.0 mins

p.13

Expt. 8 Decomposition of HExpt. 8 Decomposition of H22OO22

Flask AFlask A(150cm(150cm33 water) water)

10cm10cm33 1.00 mol dm 1.00 mol dm-3-3 HH22OO22

50cm50cm33 borate buffer borate buffer10cm10cm33 diluted diluted

KMnOKMnO44

start stop watch!start stop watch!

10cm10cm33 sample sample(around 5 mins)(around 5 mins)

10cm10cm33 1.0M 1.0M HH22SOSO44Flask BFlask B

Titrate against Titrate against dilutedilute KMnO KMnO44

(4(4×10×10-3-3 M) M)

What happens in Flasks A and B …?What happens in Flasks A and B …?

p.14

Flask AFlask A

HH22OO22 + 2OH + 2OH-- O O22 + 2H + 2H22O + 2eO + 2e-`-`

MnOMnO44-- + 2H + 2H22O + 3eO + 3e-- MnOMnO22 + 4OH + 4OH--

× 3× 3

× 2× 2

3H3H22OO22 + 2MnO + 2MnO44-- 3O 3O22 + 2H + 2H22O + 2OHO + 2OH-- + 2 + 2MnOMnO22

2H2H22OO22 O O22 + 2H + 2H22OOMnOMnO22

Flask BFlask B

MnOMnO22 is killed by H is killed by H22SOSO44..

2MnO2MnO44-- + 5 + 5HH22OO22 + 6H + 6H++ 2Mn 2Mn2+2+ + 8H + 8H22O + 5OO + 5O22

p.15

Date Treatment …Date Treatment …

For 1For 1stst order rxn, ln [A] = - k t + ln [A] order rxn, ln [A] = - k t + ln [A]00

ln ([A]ln ([A]00/[A]) = k t/[A]) = k t

As vol. of MnOAs vol. of MnO44-- used used [A], [A],

ln (Vln (V00/V) = k t/V) = k t

If a If a straight linestraight line is plotted [ln(V is plotted [ln(V00/V) vs t] /V) vs t] 1st order1st order, and , and slope = kslope = k!!

Next ….Next ….p.16

Activation Energy and Arrhenius Equation Activation Energy and Arrhenius Equation (p.??? )(p.??? )

AssignmentAssignmentp.48 Q.7-11 [due date: 16/2(Mon)] p.48 Q.7-11 [due date: 16/2(Mon)]

Lab report [due date: 23/2(Mon)] Lab report [due date: 23/2(Mon)] p.75 Q.2-3 [due date: 16/2(Mon)] p.75 Q.2-3 [due date: 16/2(Mon)]