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University of Alberta

Biochemistry

BIOCH499A

Gail Amort- Larson

Winter

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The time to apply for summer positions is now, not in
April!
Bizarro by Dan Piraro
Read 16.4,6.5
Practice PS #1, you should be able to solve Q 1 – Q 8
Remember: solving 2 questions per day is better than
all problems in one day! 1 2
To find the order of rxn
Last time
Rate of rxn Measure initial rates (rate at t = 0)
Rate law (equation)
m n
Rate = k [A] [B] Same rxn, different concentrations
rate constant k Initial rat1 = k[A] 1m
m
orders m n, Initial rat2 = k[A] 2
rate k [A] m [A]
So 1 = 1m = 1
rate 2 k [A]2 [A] 2
You know concentrations and rates m
knowing m, one concentration and its rate k
3 4
Data for the decomposition of acetone. Order of rxn with more than one reactant
Rate (M min ) 0.026 0.105 Similar procedure but usually need more data
[Acetone] 0.0500.100 – – –
4ClO 2+ 4 OH 3 ClO + ClO3+ 2 H O 2
find the order of rxn and the rate constant.
Exp [ClO ] [ClO –] Ra(ts –1)
rate2 k [A]2m [A]2m 2
= m = 1 0.010 0.030 6.00 x 10 –4
rate1 k [A]1 [A]1 –3
2 0.010 0.075 1.50 x 10
0.105 0.100 m –2
= 3 0.055 0.030 1.82 x 10
0.026 0.050 2 rate2 k [ClO 2 2[ClO ] –2n [ClO ]–2n
= m – n = –
m 1 rate1 k [ClO 2 1ClO ] 1 ClO ] 1
4.04 = 2 m = 2
1.50 x 10 –3 0.075 n n
Guess or take natural logarithm on both sides –4 = 2.5 = = 2.5 n = 1
6.00 x 10 .030
m ln 4.04
ln 4.04 = ln (2 ) = m ln 2 m= = 2 3 rate3 k [ClO 2 3[ClO ] –3n [ClO ]2 3n
ln 2 = m – n =
1 rate1 k [ClO 2 1ClO ] 1 ClO ]2 1
Rate = k [Acetone] 2 –2 m
1.82 x 10 0.055 m
–1 6.00 x 10 –4 = 30.3 = 0.010 = 5.5
rate1 0.026 M min –1 –1
k = [A] 1m = (0.050 M) 2 = 10.4 M min
ln 30.3
so m = ln 5.5 = 2.01 = 2
Note: units of k will change for different orders 5 6
Integrated rate laws
Rate = k [ClO ]2[ClO ]– 3 rdorder overall Rate = k [A] m
It is more practical to get concentration as a function
–4 –1 of time.
k = ra2e1 – =

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