SCH 4U1 Mr. Dvorsky Friday November 15 Reaction Mechanisms SCH 4U1 Mr. Dvorsky Friday November 15

What is a reaction mechanism? In any chemical change, some bonds are broken and new ones are made. Quite often, these changes are too complicated to happen in one simple stage. Instead, the reaction may involve a series of small changes one after the other.

A reaction mechanism describes the one or more steps involved in the reaction in a way which makes it clear exactly how the various bonds are broken and made.

In other words, it is believed that most chemical reactions actually occur as a sequence of elementary steps. The overall sequence is called the reaction mechanism.

Rate-determining Step The overall rate of a reaction (the one which you would measure if you did some experiments) is controlled by the rate of the slowest step. The slow step of a reaction is known as the rate determining step

Let us take the following example: 4 HBr(g) + O2(g)  2H2O(g) + 2Br2(g) This rxn won’t occur in one step. -collision theory: four molecules of HBr and one molecule of O2 would have to collide in proper orientation and with enough energy to break old bonds and form new ones. -it is hard enough for just two molecules to collide!

Example continued: 4 HBr(g) + O2(g)  2H2O(g) + 2Br2(g) Experimental evidence shows adding O2 increases the rate as we would expect. Since four molecules of HBr react for every O2 -we would expect [HBr] to have a larger effect on rate. Not the case. Experiments give the following rate law expression: r=k[HBr][O2]

Example continued: 4 HBr(g) + O2(g)  2H2O(g) + 2Br2(g) Experimental evidence shows adding O2 increases the rate as we would expect. Since four molecules of HBr react for every O2 -we would expect [HBr] to have a larger effect on rate. Not the case. Experiments give the following rate law expression: r=k[HBr][O2]

Rule: the slowest or rate-determining step must be consistent with the rate equation In this case: r=k[HBr][O2]

IN ADDITION: there is a direct link with equation coefficient and exponents in the rate equation. So for r=k[HBr]1[O2] 1 [so one molecule of each must collide] The rate determining step is 1 HBr + 1 O2 -> reaction intermediate [this also says one molecule each must collide]

The take home message of all this If the experimentally determined rate equation is: rate = [molecule X]m[molecule Y]n Then the rate determining step will be: mX + nY -> products or reaction intermediates

A couple of other rules before we can get started in building our own reaction mechanisms: Each step must be elementary; that is to say no more than three reactants (usually 1-2 reactants). The elementary steps must add up to the overall equation.