Integrated Rate Equation C. Y. Yeung (CHW, 2009) p.01.
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Transcript of Integrated Rate Equation C. Y. Yeung (CHW, 2009) p.01.
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)]