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February 181 0th, 1st and 2nd Order Reaction
14.2 0th, 1st and 2nd Order ReactionFactors Affecting Reaction Rates
Fred Omega GarcesChemistry 201Miramar College
February 182 0th, 1st and 2nd Order Reaction
Outline
Rate Laws: Differential Rate Law Integrated Rate Law
0th Order: Rate = k[A]0 = k [A] = -kt + [Ao]
1st Order: Rate = k[A]1 ln[A] = -kt + ln[Ao]
2nd Order: Rate = k[A]2 1 / [A] = kt + 1 / [Ao]
February 183 0th, 1st and 2nd Order Reaction
Evaluation of Rate Constant and Half-lives
Oth Order:2N2O (g) g 2N2 (g) + O2 (g)
Rate = k[A]0 = k
2nd Order:
NO2 (g) g NO (g) + 1/2 O2 (g)
CH3Cl + OH- gCH3OH + Cl-
Rate = k[A]2
1st Order:14C g 14N + -1e
Rate = k [A] 1
February 184 0th, 1st and 2nd Order Reaction
Zeroth Order
Decomposition: 2N2O D 2N2 + O2Some reactions are independent on concentrations.rxn rate = k [A]0 = k
[A] - [A o] = -ktor [A] = -k t + [A o]
y = mx + b
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− d[A]dt
= k
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d[A] = − kdt∫∫
A
tplot of [A] vs t (time)straight line
Slope = D [A] = -kDt
@ t g te
[A] = 0
Data :[A] Rates1 14 18 1
February 185 0th, 1st and 2nd Order Reaction
Half-LifeHalf-life (t1/2) Time for the concentration of a reactant to decrease to
half its original concentration.
Ao - original concentration@ t1/2 A = Ao/2
The time it takes for the concentration of the original sample to decrease to half its original value.
Consider, t1/2 = 1 hr
. 1st t1/2 2nd t1/2 3rd t1/20 1 hr 2hr 3hr
February 186 0th, 1st and 2nd Order Reaction
Zeroth Order: Half-life
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A = −k ⋅t + Ao
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t1/2
= Ao2k
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@ t1/2
: A = Ao2
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Ao2
= −k ⋅t1 /2
+ Ao
Data :[A] Rates1 14 18 1
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Ao2−Ao = -k ⋅t
1 /2
Metabolism of Alcohol
Rate: 1 oz / 5 hror 0.2 oz /hr
Oth order reaction seldom
occurs except for reactions
which are catalyzed by
enzymes
C C OHH
H
H
HH
Ethanol
C COH
H OHH
Acetic acid
February 187 0th, 1st and 2nd Order Reaction
t
Slope = D ln [A] = -kDt
ln [A]o
ln [A]
First OrderRadioactivity: 14C g 14N + -1erxn rate = -dA/dt = k[A] g dA / [A] = - k dt
Separate:
Integrate:
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d[A]A∫ = - k ⋅ t∫
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ln [A]o[A]
= kt
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ln[A]o - ln[A] = kt
y = mx + b
ln[A] = - kt + ln[A]o
plot of ln[A] vs tstraight line
Data :[A] Rates m/s
1 14 48 8
February 188 0th, 1st and 2nd Order Reaction
First Order: Half-life
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@ t1 /2
: A = Ao2
Carbon Dating: Half-life is independent of reactant conc.
Data :[A] Rates m/s
1 14 48 8
ln[A] = - kt + ln[A]o
ln [A]o
[A]o
2
!
"
##
$
%
&&
= + k ⋅ t1/2
ln 2 = + k t1/2
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t1 /2
=0.693
k\ Half-life is not function of conc.
14N g 14C + 1H
1n *CO2O2
14C - Carbon datingt1/2 = 5730 yr.
*CO2
*CO2
*C
February 189 0th, 1st and 2nd Order Reaction
Radioisotope DatingFor 1st Order Reaction: Half-life is independent of concentration of reactant.
C-14 dating is accurate only up to 50,000yr.
14C ® 14N + -1e (b-emission)
U-238 accurate up to 4.5•109 yr.Based on data, Earth is 4 - 4.5 Billion yrs. old.
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t12
= ln2k
ln[A] = - kt + ln[A]o
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ln [A]o[A]o
2"
# $
%
& '
= k ⋅ t1 /2
http://www.crystalinks.com/oetzi.html3300 BC or 5300 years ago
Otzi The Iceman
February 1810 0th, 1st and 2nd Order Reaction
Calculation of Age Based on t1/2TURIN, Italy -- Almost everything about the Shroud of Turin is mysterious- its age, its authenticity, and the identity
of the bearded man with deep-set eyes whose image is imprinted on the 14-foot length of yellowing linen, still believed by many Christians to be the burial cloth of Jesus. ....as carbon testing done on tiny swatches of the shroud concluded in 1988 -- or to the time of Jesus, the centuries-old fascination with the shroud....
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t12
= ln2k
t12
= 5730 yr
k = ln2t
12
=ln2
5730 yr=
0.6935730 yr
=1.21•10-4 yr−1
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12C14 C
=1012
114 C[ ] = 50.000 ppt based on 12C14 C[ ] = 46.114 ppt today
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Ao = 50.000A = 46.114
" # $
= ln AoA
= kt
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ln50.00046.114
1.21 •10−4 yr−1 = t
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t = 668.6 yr
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1990 - 669 = 1321 ± 50 AD
The Shroud of Turin is a linen cloth over 4 m long. It bears a faint, straw-colored image of an adult male of average build who had apparently been crucified. Reliable records of the shroud date to about 1350, but for these past 600 years it has been alleged to be the burial shroud of Jesus Christ. Numerous chemical and other tests have been done on tiny fragments of the shroud in recent years. The general conclusion has been that the image was not painted on the cloth by any traditional method, but no one could say exactly how the image had been created. Re-cent advances in radiochemical dating methods, however, led to a new effort in 1987–1988 to estimate the age of the cloth. Using radioactive 14 C, the flax from which the linen was made was shown to have been grown between 1260 and 1390 A.D. There is no chance that the cloth was made at the time of Christ.
https://en.wikipedia.org/wiki/Shroud_of_Turin
February 1811 0th, 1st and 2nd Order Reaction
Second Order
Decomposition: NO2 (g) g NO (g) + 1/2 O2 (g)
rxn rate = -dA/dt = k[A]2 g dA / [A]2 = - k dt
Separate:
Integrate:
y = m x +. b
t
plot of 1/[A] vs tstraight line
Slope = D 1/ [A] = kDt
1/ [A]o
1/[A]
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d[A]A2∫ = - k ⋅t∫
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1[A]
= kt + 1[A]o
[A]
t
February 1812 0th, 1st and 2nd Order Reaction
Second Order: Half-life
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@ t1 /2
: A = Ao2
Decomposition of NO2:Data :[A] Rates m/s1 14 168 64
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1[A]
= kt + 1[A]o
NO2 (g) g NO (g) + 1/2O2 (g)
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t1 /2
=1
k[A]o
\ Half-life is a function of conc. & k
In a second-order reaction each successive
half-life is double the preceding one.
Furthermore, the half-life for a second
order reaction is inversely related to the
initial concentration of reactant.
February 1813 0th, 1st and 2nd Order Reaction
In Class ExerciseN2O5 can decomposes to nitrogen dioxide and oxygen gases.Here are some data for the decomposition reaction:Time (min) 0.0 20.0 40.0 60.0 80.0[N2O5] • 10-2M 0.92 0.50 0.28 0.15 0.08
Determine the order and the rate constant by constructing appropriate graphs using the data.
1.What is the rate constant for this reaction?2.What is the overall rate law for this reaction?3.What is the half-life of this reaction?
Time [N2O5] ln[N2O5] 1/[N2O5]0.0 0.0092 -4.68855179520.0 0.0050 -5.29831736740.0 0.0028 -5.87813586260.0 0.0015 -6.50229017180.0 0.0008 -7.13089883
February 1814 0th, 1st and 2nd Order Reaction
Analysis
February 1815 0th, 1st and 2nd Order Reaction
Summary Order Order OrderZero First Second
Rate Law R = k R=k[A] R = k[A]2
Integrated rate law A = -kt + [A]o ln[ A] = -kt + ln[A]o 1/[A]=kt +1/[A]o
Plot (straight line) [A] vs. t ln[A] vs. t 1/[A] vs. t
slope m = -k m = -k m = k
Half-life t1/2 = [A]0/2k t1/2 =0.693/k t1/2 =1/k[A]0
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