Reaction Mechanism

A detailed sequence of steps for a reaction

Reasonable mechanism: 1. Elementary steps sum to the overall reaction

2. Elementary steps are physically reasonable

3. Mechanism is consistent with rate law and other experimental observations (generally found from rate limiting (slow) step(s) A mechanism can be supported but never proven NO2 + CO  NO + CO2

2 Observed rate = k [NO2] Deduce a possible and reasonable mechanism

NO2 + NO2  NO3 + NO slow

NO3 + CO  NO2 + CO2 fast

Overall, NO2 + CO  NO + CO2

Is the rate law for this sequence consistent with observation? Yes Does this prove that this must be what is actually happening? No! NO2 + CO  NO + CO2

2 Observed rate = k [NO2] Deduce a possible and reasonable mechanism

NO2 + NO2  NO3 + NO slow

NO3 + CO  NO2 + CO2 fast

Overall, NO2 + CO  NO + CO2 Is the rate law for this sequence consistent with observation? Yes Does this prove that this must be what is actually happening? No! Note the NO3 intermediate product! Arrhenius Equation Temperature dependence of k

 kAe E/RTa

A = pre-exponential factor

Ea= activation energy

Now let’s look at the NO2 + CO reaction pathway* (1D). Just what IS a reaction coordinate? NO2 + CO  NO + CO2

High barrier TS for step 1 Reactants: Lower barrier TS for step 2 NO2 +NO2 +CO

Products Products

Bottleneck NO +CO2 NO +CO2 Potential Energy

Step 1 Step 2

Reaction Coordinate NO2 + CO  NO + CO2

NO NO22 NO NO  3  NO222 CO NO CO In this two-step reaction, there are two barriers, one for each elementary step. The well between the two transition states holds a reactive intermediate.

Potential Energy A reactive intermediate is a compound formed transiently, but Reaction Coordinate absent in the overall reaction.

We need to do better than just using a 1-D energy slice, namely a multidimensional potential energy surface! Different Types of Composite Reactions

Simultaneous: Opposing: A  Y and A  Z A + B Z Y or A z

Consecutive: Consecutive with opposition: A  X  Y A  X Y

In order for the system to be at complete equilibrium, the time derivative of all species concentrations must be zero! Reaction with Three Simultaneous Pathways x 100 k1 [A] k “Branching ratio” A 2 y k [x] : [y] : [z] = k1 : k2 : k3 3 Rate law: z 0 dA time kA  kA   kA  dt 123   Now look at products    (k  k  k ) A dx (k k k )t 123  kA kA  e123 dt 11 0 (k k k )t 123 k (k k k )t AAe  xA1et 1 123 0 kkk 0 123xA k   1 At long times,  kkk 0 123 Kinetic vs Thermodynamic control

x   k1 X K1   Kinetic Control k Y  Z K2 K3 A 2 y k3 z

Thermodynamic Control?

Proportions of various products are determined

using ΔGrxn of all possible reactions. Reaction with Three Simultaneous Pathways

PChem Lab: Quenching of Electronically Excited Naphthalene, A* A* can relax by radiative, non-radiative or collisional quenching decay A*  A + hv (Radiative) A*  A + heat (Non-Radiative) A* + Q  A + Q (Collisional Quenching) Kinetics for Simple Consecutive Reactions

k1 k2 AXZ kk AX12 and Z X 

What are the differential rate equations for A, X and Z?

What are the integrated rate equations for A, X and Z? Kinetics for a Simple Consecutive Reaction

kk AX12 and Z X  What are the differential rate equations for [A], [X] and [Z]? dA   (1) 1  Let [A]t=0 = [A]0 dt kA We can solve (1) directly and [X]t=0 = [Z]t=0 = 0

kt AAe  1  0 Now [X] and [Z] dX    21  (2) dt dZ kX kA   2  (3) dt kX Kinetics for a Simple Consecutive Reaction

 kk kt1 12AAe  AX and Z X  with  0 dX    21  (2) dt becomeskX kA  dX kt    1 21 0 dt kX kAe  dX kt   1 21  a "standard form" o.d.e. with solution ZA AXA0 XeeAdt kX kAe k kt 2 12  kk 0 and 21  ktkt  ke ke21        1 12   00      21  kk Variation of Concentrations for Consecutive Reactions

A  X  Z Kinetics for a Simple Consecutive Reaction

kk AX12 and Z X  ZA  ktkt   21 kt1  ke ke   AAe   1 12 0  0  21  kk

k1 >> k2 k2 >> k1 One More Case: k1 = 10 k2

kk AX12 and Z X 

k1t Quasi-Steady State Approximation A Quasi Rationale

kk AX12  Z

We see that, whenever, k1 > k2, [X] has an initial growth period, but after that, [X] tends to track [A]. This leads one to imagine that we might find a condition where [X] is in a quasi steady state, namely its sources and sinks cancel out each other, and d[X]/dt=0. Try it!