00:01
All right, so this question gives us the reaction to a plus b plus c yields d plus e.
00:14
And so with rate law expressions, we do not really care about the products at all.
00:23
What we're going to do is we're going to write an experimental rate law, and then we're going to use the data in the experiment table and the initial rate to help us figure out how to make our experimental rate law better.
00:44
And so for my experimental rate law, i'm going to start with rate equals the rate constant k, and that's going to be times the concentration of reactive a, and that's going to be raised to some power times the reaction concentration of reactant b raised to some power times the reactant c raised to some power and so what my goal is is i want to use the data from experiment one two three and four to figure out what x, y, and z is.
01:30
The easiest way to to figure out what x, y, and z is, is to look at where does the concentration stay the same for two of the three? so for instance, if we look at experiment two and experiment three, experiment two and experiment three, what you're gonna notice is that the concentration for a changes, the concentration for b stays the same, and the concentration for c stays the same, so that's a great place to start.
02:15
So what i'm going to do here is i'm going to say rate equals, well, the rate for experiment 2 is 9 .6 times 10 to the negative 6 molar per minute, and that's going to equal k times the concentration of a, which in experiment 2, is 0 .40 molar raised to the x, times the concentration of b, which is 0 .30 molar raised to the y, times the concentration of c, 0 .20 molar, raised to the z.
02:55
And i'm going to divide that by this information for experiment 3.
03:00
So in experiment 3, the rate is 2 .4 times 10 to the negative 6 molar per minute equals k.
03:10
The concentration of a is 0 .20 molar.
03:15
The concentration of b is 0 .30 molar raised to the y.
03:20
The concentration of c is 0 .20 molar raised to the z.
03:26
So now what i have is 9 .6 times 10 to the negative 6 divided by 2 .4 is 4.
03:36
4 equals.
03:39
K divided by k cancels.
03:43
0 .4 to the x divided by 0 .2 to the x is 2 to the x, 0 .3 molar raised to the y.
03:59
Divided by 0 .3 molar raised to the y cancels, which is why i want those to be the same.
04:05
0 .20 molar raised to the z, divided by 0 .20 molar raised to the z, which is why i wanted those to be the same.
04:16
Well, here, x equals 2 because 2 to the 2 equals 4.
04:21
So what i'm doing is i'm using this information to take my experimental rate law and start to plug things in.
04:32
Now i know that a is a second rate.
04:38
So i'm going to go through this same process again.
04:41
I'm looking for where does a stay the same, b change, and c stay the same.
04:52
And so if i look at experiment 1 and i look at experiment 3, 3, what i notice is that a stays the same, b changes, and c stays the same.
05:14
So here, i'm going to go through the same exact process.
05:19
I'm going to take, i'm going to put three on top this time because it has a larger concentration of b.
05:32
So i'm going to say the rate of 3 is k times the concentration of a.
05:38
Point two zero molar.
05:40
Now i know that's raised to the second.
05:43
Divided by the concentration of b, 0 .30 molar, raised to the y, times the concentration of c, 0 .20 molar, raised to the z.
05:54
And i know that rate is 2 .4 times 10 to the 6 molar divided by k times 0 .2 molar raised to the second times point two molar raised to the y times point two molar raised to the y times point two molar raised to the z is 2 .4 times 10 to this negative six molar per minute.
06:26
K divided by k cancels.
06:29
0 .20 to the second divided by 0 .20 to the second cancels.
06:36
Divided by point two zero point three divided by point two is one point five so this is one point five to the y point two zero to the second and point two zero to the second cancels and here what we notice is this is one well anything raised to the zero power is one so why actually equals zero which means b is not in my rate law.
07:12
Some teachers will allow you to write the concentration of b to the zero.
07:17
I personally do not.
07:19
I don't want it in your rate law.
07:21
I also teach my students that when we have a concentration change but the rate stays the same, that's evidence of a zero power...