00:01
Hi, so we are to verify if the rate law is need first order in n205 and then calculate the value of the rate constant.
00:11
So take note that the integrated rate law for a first order reaction is given by ln of concentration and time t, be equivalent to negative k multiplied by the time plus ln of the initial concentration of a.
00:31
And slope intercept form equation for straight line is y is equivalent to mx plus b.
00:42
So that means in this equation, an integrated rate law for a first order reaction, y would be l n of concentration at time t, and then the slope m is negative k, and then t would be x, and y intercept b is l n of the initial concentration.
01:00
So to determine if the rate law is first order, we can plot the ln of the concentration of the reactant versus time.
01:13
So we need points here.
01:17
We'll use x and y.
01:19
We will need x and y.
01:21
So for the first row, we have the concentration.
01:26
So why would be equivalent to ln of 0 .1 negative to 0 .1.
01:39
2 .302.
01:41
Let's just use 2 .3.
01:44
So it's easier.
01:51
And then x is 0.
01:55
Y is for the second point ln of 0 .0707.
02:04
It's negative 2 .6 and x would be 50.
02:12
Y would be ln of 0 .05.
02:21
We'll round this up to negative 3 and x is 100.
02:30
Fourth, y is ln of 0 .025, this is negative, 3 .7, x is 200.
02:50
Fifth point, ln of 0 .0 .125 is negative, 4 .4.
03:03
X would be 300.
03:06
Lastly, y would be ln of 0 .0 .625.
03:16
This is negative 5 .1.
03:19
X would be 400.
03:22
Now let's, so at this point, since x are all positive and y are all negative, let's say this is 100, 200, 300, 300, 400, this is positive x.
03:48
And then under y, we have, let's say this is 1, 2, 3, 4, 5.
03:58
Then we have here 50 versus negative y.
04:15
So first point, x is 0 and y is negative 2 .3...