Ch1b09Lecture18

# Ch1b09Lecture18 - 1 Ch1b Lecture 18 Last lecture kinetics...

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Ch1b Lecture 18 February 19, 2009 Last lecture: •kinetics Today ! s lecture: •more kinetics 1 •steady state •polymerization •nucleation •temperature dependence

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Approaches to evaluate the overall reaction rate from a sequence of elementary steps deﬁning a reaction mechanism a. analytical (exact) solution of rate equations - rapidly gets complicated b. numerical solution of rate equations-most useful for simulating full time courses Useful approximate treatments: c. one step is much slower than the rest - the rate determining step d. steady-state approximation - concentration of intermediates are constant 2
H 2 O 2 + I - H 2 O + IO - IO - + H 2 O 2 H 2 O + I - + O 2 2H 2 O 2 2H 2 O + O 2 k 1 k 2 v = " 1 2 d H 2 O 2 [ ] dt = k 1 H 2 O 2 [ ] I " [ ] when the ﬁrst step is much slower than the second, the overall rate is determined by the ﬁrst step since the IO - reacts “immediately” with H 2 O 2 in the second step 3 I - catalyzed decomposition of H 2 O 2 use of rate determining step

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What if k 1 and k 2 are comparable? The steady-state (SS) approximation (assumes the concentration of intermediates are constant) can be quite useful in deriving rate laws for more complex systems. For the I - catalyzed decomposition of H 2 O 2 , there are two intermediates I - and IO - , and the SS approximation is equivalent to: 4 d I " [ ] dt = d IO " [ ] dt = 0 Question: does this imply ? d H 2 O 2 [ ] dt = d O 2 [ ] dt = 0
d I " [ ] dt = " k 1 H 2 O 2 [ ] I " [ ] + k 2 H 2 O 2 [ ] IO " [ ] = 0 # IO " [ ] = k 1 k 2 I " [ ] at SS, the ﬂow through each step is constant (no accumulation of intermediates) 5 H 2 O 2 + I - H 2 O + IO - IO - + H 2 O 2 H 2 O + I - + O 2 2H 2 O 2 2H 2 O + O 2 k 1 k 2 SS approx. v = d O 2 [ ] dt = k 2 H 2 O 2 [ ] IO " [ ] = k 1 H 2 O 2 [ ] I " [ ]

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using the SS approximation, can write the overall reaction rate in terms of the total concentration of iodine [I] tot : IO " [ ] = k 1 k 2 I " [ ] I [ ] tot = I " [ ] + IO " [ ] = k 1 + k 2 k 2 # \$ % & ( I " [ ] I " [ ] = k 2 k 1 + k 2 # \$ % & ( I [ ] tot v = k 1 H 2 O 2 [ ] I " [ ] = k 1 k 2 k 1 + k 2 # \$ % & ( H 2 O 2 [ ] I [ ] tot = k 1 H 2 O 2 [ ] I [ ] tot for k 1 << k 2 = k 2 H 2 O 2 [ ] I [ ] tot for k 2 << k 1 SS approx.
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Ch1b09Lecture18 - 1 Ch1b Lecture 18 Last lecture kinetics...

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