00:01
So here we're given an example of a reaction mechanism with certain steps.
00:04
So for our first element, these are both all elementary reactions.
00:07
So we can derive our rate equation directly from our elementary reaction.
00:12
We have 2a in equilibrium with a2, and this is a fast step.
00:18
In part two, we have a2 plus e, yield b plus c, and this is a slow step.
00:25
So the rate of reaction in reality is only going to depend on the rate determining step, since whenever you have a change in concentration in one of your rate determining reagents, it's going to lead to this fast reaction to immediately replenish the concentration of everything you lost.
00:43
So the reaction only really depends on the rate of the slow step.
00:47
So since this is an elementary reaction, we can write a rate law directly from the rate equation, which is equivalent to k times the concentration of a2 times e.
00:58
However, this isn't useful since we have the concentration.
01:01
Of an intermediate, since intermediates are produced and consumed in a reaction, and they aren't readily measured.
01:09
So first, we're essentially going to write rate law for the, let's just call this rate constant k1, and the reverse rate constant k -1, then this rate constant k -2...