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BIOCH499A (12)

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School
University of Alberta
Department
Biochemistry
Course
BIOCH499A
Professor
Gail Amort- Larson
Semester
Winter

Description
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]2m 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 ]–2n = 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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