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Department
Biochemistry
Course
BIOCH499A
Professor
Gail Amort- Larson
Semester
Winter

Description
If it seems that I am ignoring you in the hallways and corridors, I apologize. It will take me time to recognize your faces and learn your names. I will try though ! Bizarro by Dan Piraro Read 16.7, 17.1, 17.2 Practice PS#2 Lab room assignments are posted online on the lab webpage. 1 2 At equilibrium, the rates for the forward reaction and Case II – A fast step is followed by a slow step. the reverse reaction are equal. Decomposition of ozone, O , 3o oxygen, O , 2s believed to occur via So from step 1 k1[O 3 = k–1O ]2O] k 1 Remember, both the forward and the reverse are Step 1 O 3 (g)  O 2 (g)+sa(g) k–1 elementary! k k [O ] 2 [O] = 1 3 Step 2 O 3 (g) O (g) 2 w2 (g) k–1O ]2 RDS is step 2 Insert the result into the RDS rate Rate RDS= k 2O 3[O] Can’t leave [O] in the rate expression k k [O ]2 No intermediate appears in rate law Rate = k 2O 3[O] = 1 2 3 k–1 [O 2 + – Exceptions: [H ] or [OH ] in solutions could remain if we're unable to find substitute expressions. With the absence of an experimental rate, assume the mechanism and the derived rate are possible. 3 4 Figure 16.22 Energy vs. rxn progress for an exothermic reaction Energy vs. rxn progress for an exothermic reaction having a 2–step mechanism in which the first step is having a 2–step mechanism in which the first step is fast and the second is slow (RDS) slow (RDS) and the second is fast In both diagrams, the intermediates are not at the crest of the curves (that’s where the transition states for each step are). Both diagrams are for exothermic rxns. You could also have an endothermic reaction with products having higher energy than the reactants. 5 6 Fast equilibrium steps Case III – Steady state approximation At equilibrium rate forward = rate reverse A situation can occur in which either: Suppose A + B  2C  No information is known or given about the 2 relative speeds of the rates. Thenk 1[A][B] = k [–1 k [C]2 [products] stoic coeff  When steps have relatively comparable rates The ratio 1 = = stoic coeff k–1 [A][B] [reactants] k1 2NO (g)  N 2 2 (g) The right hand side of the equation is simply the k –1 k 2 definition of the equilibrium constant K. N 2O 2 (g) O 2 (g) 2 NO 2 (g) But 1 equil. step is not much faster than the 2 step. k1 So k–1= Equilibrium constant = K Choose the step that produces the products (usually the last one) Rate of rxn = k [N O ][O ] 2 2 2 2 However you can't leave an intermediate [N O ]2in2 the expression. 7 8 Steady state approximation The formation and disappearance rates are equal After some time the rxn will reach a steady state in k 1NO] = k [N–1 ]2+ 2 [N O2][O2]2 2 which the rate of change of the concentration of the 2 k 1NO] = [N O 2 (2 + k–1O ]2 2 intermediate is constant. k1[NO] 2 Solve for [N 2O 2  [N 2 ]2= k–1 k [2 ] 2 In other words, the rate of formation of the
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