00:01
For this problem, we are given the following reaction, 2x plus z yields products, and we're given the data that i've written down here.
00:12
We are asked for questions a and b.
00:17
I'll get a different color.
00:22
For a and b, we are asked to write the rate law for a, and the rate constant with units for b.
00:41
You do the rate constants, we're going to be using well, actually, just put what our rate law.
00:48
Our rate will be equal to k times x to the x power and z to the y power.
01:01
We're going to use the expression.
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I'm about to write down right now.
01:07
Rate of, it'll be rate of rate over whichever trial.
01:12
We're going to use and kkxx.
01:23
I'm going to fill in something else in a moment here.
01:30
And then we're going to, whichever trials we happen to be using.
01:34
And the first thing we're going to do is we're going to use trials two and three and i'll fill the values in.
01:45
And this is going to give us our concentration, or our order for z.
01:50
Our order in z.
01:55
Let's go ahead and get started.
01:59
I'm just copying these values down from the table.
02:02
I'm going to put two rate of two over the rate of three.
02:11
And i'm just going to go ahead and cross these out.
02:14
You can see that my k's cross out.
02:16
Since i'm using 2 and 3, i've got 0 .50 to the x and 0 .50 to the x.
02:26
That crosses out.
02:27
And for my zs, i have 0 .50 and 0 .75.
02:41
Y and y.
02:44
So, solving for these, i get a 4 over 9 equals 2 over 3.
02:52
This will be to the n, and i can solve for n by taking the log of 4 over 9, divided by the log of 2 or 3.
03:20
And this gives me 2.
03:22
So this is second order in z.
03:26
Second order in z.
03:32
For the second part, we're going to use 1 and 2.
03:36
So i'm going to take the rate of 1 and the rate of 2...