00:01
For this question, we have a chemical reaction of ch3cbr with hydroxide, forming ch3 coh, and br minus.
00:14
The rate law is known to just be a function of the concentration of ch3cbr, and not a function of hydroxide, and its first order with respect to this reactant.
00:27
So if we go to the mechanisms, for the first mechanism, the slow step, which determines the rate of the reaction, is simply 1, ch3, 3, cbr, going to product, and then a fast second step.
00:46
So the rate law for the slow step for this mechanism is rate is equal to k, multiplied by the concentration of ch3, 3, c, so this mechanism works because its rate law corresponds to the actual rate law.
01:09
These are the same.
01:11
Then let's go to the second mechanism.
01:17
For the second mechanism, it is the, there is only one step.
01:26
The one step is the overall reaction.
01:30
However, if there's just one step, then we assume it's an elementary step and the rate law would be including both reactants, both raised to the first power, because their coefficients are to the first power.
01:52
So this one is yes, but this one is no...