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In the Binary Collision Model we made a good case for the rate expression:

In the Binary Collision Model we made a good case for the rate expression:. R={ ( AB ) 2 < u rel > e -E A /RT } (N A /V)(N B /V). Often, N A /V and N B /V are concentrations in units of molecules per ml. To get these in moles per liter, just multiply by 1000/N 0 !.

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In the Binary Collision Model we made a good case for the rate expression:

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  1. In the Binary Collision Model we made a good case for the rate expression: R={(AB)2 <urel> e -EA/RT} (NA/V)(NB/V) Often, NA/V and NB/V are concentrations in units of molecules per ml. To get these in moles per liter, just multiply by 1000/N0! So the Binary Collision Model predicts R=kRCACB

  2. Basically, the Binary Collision Model predicts a reaction rate that is first order in A, first order in B and second order overall. B) First Order Reactions [will need a model later!] Assume this is first order to get 

  3. = kCA if reaction is 1st order Integrate both sides: + [dCA/CA] = -  kdt Need to find g from initial conditions. lnCAº = - k (0) + g

  4. CAo lnCA CA t t lnCAo  Slope = -k CA = CA0 e-kt

  5. Slope = k ln(CAo / CA) t Equations for first order reactions are very important. In the laboratory almost ALL reactions can be made to APPEAR First Order.

  6. Bonus * Bonus * Bonus * Bonus * Bonus * Bonus Half Life or Half Time

  7. N2O5 2NO2 + 1/2 O2 Rate (mol L-1 s-1) 3 x 10-5 Rate = -d[N2O5]/dt = k[N2O5] 2 x 10-5 1 x 10-5 0 1.0 2.0 [N2O5] (mol L-1) Example of a first order reaction.

  8. Take t0 = 0  C) 2nd Order Kinetics: A  Products -  dCA/[CA]2 = k  dt 1/[CA]= kt + g t = t0 , CA = CA0 Initial conditions

  9. 1/[CA]= 1/[CA0] + kt At t1/2 , CA = (CA0 / 2) and t1/2 = 1 / kCA0

  10. 2) Initial Rates Method 1st order reaction c = c0 - x Where c0 is the initial concentration and x is a function of time, x = x(t). x is simply the amount reacted. If x<<c0dx/dt = const = kc0 (sure to be true if t is small enough!)

  11. CAo CA t Initial part of curve will look like a straight line if t is small! CAo c c/ t = -kCAo  Initial Slope = -kCAo t CA t Blow up first one percent of CA vs t curve Slope NOT Constant dCA/dt = -kCA

  12. Measure c vs.  t for first 1% of reaction. Here, c0>>x,  c  dx and  t  dt  Know c0 , measure ∆c and ∆t,  obtain k nth Order Reactions A + B + C  products a= initial conc of A b= initial conc of B c= initial conc of C

  13. Bonus * Bonus * Bonus * Bonus * Bonus * Bonus

  14. (Note, have kept b, c constant!) (dx/dt)1 and (dx/dt)2 are measured in the laboratory, while a1 and a2 are known quantities. Can do a similar trick for n2 , n3

  15. MechanismConcept 1) Exponents in rate law do not depend on stoichiometric coefficients in chemical reactions. 2)What is the detailed way in which the reactants are converted into products? This is not described by the chemical equation, which just accounts for mass balance. 3) Rate at which reaction proceeds and equilibrium is achieved, depends on the Mechanism by which reactants form products. Elementary Reactions: these are hypothetical constructs, or our guess about how reactants are converted to products. The Mechanism is a set of Elementary Reactions!

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