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University of Alberta
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Biochemistry
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BIOCH499A
(12)

Gail Amort- Larson
(11)

School

University of Alberta
Department

Biochemistry

Course Code

BIOCH499A

Professor

Gail Amort- Larson

Description

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Pro•fessi oni•al SCIDEeNCdEline: Office Hours: Bizarro by Dan Piraro
Read
16.5, 16.6
Practice
PS 1 solutions are posted. You can solve all questions.
PS 2 will be on Tuesday or Wednesday 1 2
Radioactive decay – an example of 1 order rxn mass o
ln mass = kt
A 2.00 g sample of P is found to have 0.40 g of P32
2.00 g
after 33.3 days. Calculate the half life of P. ln0.40 g = ln 5.0 = k 33.3 days
st k = ln 5.0 = 0.048 days –1
Radioactive decay is always a 1 order reaction 33.3 days
32
[ P] o
ln [ P] = kt
ln 2
mass t½= = 14.3 days
k
concentration = mol = molar mass
volume volume
mass o The answer is reasonable b/c 33.3 days for 0.4 g falls
M
32 within the time range for 0.5 and 0.25 g.
ln[ P] o= V = lnmass o
[ P] mass mass
M t (days) Mass (g)
V 0 2.0
14.3 1.0
Assuming the volume remains constant, we can use
28.6 0.5
the ratio of masses in this integrated rate law b/c the 42.9 0.25
molar mass and volume will cancel out.
3 4
Zero order See fig. 16.10
A products
Rate = k [A]o = k
[A] d[A]
Rate = – = – = k
t dt
– d[A] = k dt
Solve a differential equation
[A]=[A] o– kt
y = a + bx
slope = – k
Plot [A] vs. t y–intercept = [A]o
b = slope = – k
a = y–intercept = [A]o 5 6
Common examples of zero order N2O mechanism
Surface rxns and enzyme rxns
hot Pt plate
2N 2O (g) 2 N 2 (g)O 2 (g)
The rxn occurs on a hot Pt plate.
In fact it requires the surface for the rxn. If the entire
surface is covered, then the rate of rxn will be
constant even if additional 2O is present.
7 8
Find zero order half life 2 order
[A] o A products
Using [AA]] o– kt and [A] =
2 2
rate = k [A]
[A]o
2 = [A] o kt ½ [A] d[A] 2
Rate = – t = – dt = k[A]
[A]
t½= o dependent on concentration !
2k – d[A2 = k dt
[A]
Solve a differential equation
100% 50% t ½ 1st
1 1
= + kt
50% 25% t ½ 2nd [A] [A] o
and they are not equal y = a + bx
1
Plot [A] vs. t
b = slope = k
1
a = y–intercept =[A] o 9 10
See fig. 16.10 2 C 4 6 (g) C H8 12 (g)
Time (s) C 4 (6) ln [C4HC6[/1 4H 6
0 0.01000 -4.601 500.0
1000 0.00625 -5.075160.0
1800 0.00476 -5.3482

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