factors affecting reaction...

February 181 0th, 1st and 2nd Order Reaction

14.2 0th, 1st and 2nd Order ReactionFactors Affecting Reaction Rates

Fred Omega GarcesChemistry 201Miramar College


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]


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


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


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


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


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


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]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


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


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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

http://www.crystalinks.com/oetzi.html


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

https://en.wikipedia.org/wiki/Shroud_of_Turin


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


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.


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


Analysis


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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