notes03.pdf

7 Pages
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Department
Biochemistry
Course Code
BIOCH499A
Professor
Gail Amort- Larson

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o ‐pm Deadline to nter asses Grants or ki am ach! egistrcan sign up to egun Off Friday 500 has ‐ rize ‐ t . 4th or email alberta.ca st –Fac 5embe rand ISSS A no aluednts Mon‐50 roups anuary The ecruit EEK rants isit our website a alley!CCISo sign up. tudy Fouapply is Fbrua4ry 1 Teams of up to . ariety of mental and physical 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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