00:02
All right, this is question 63 from chapter 14.
00:07
So in this question, they tell you that you've got the natural log of k plotted against 1 over t, and it's a straight line.
00:14
This is called an erroneous plot.
00:17
The uranus equation plotted so that you get a straight line out of it is the natural log of k is equal to negative ea, or the activation energy, over, or the gas constant times one over the temperature in kelvin plus the natural log of your pre -exponential factor or a.
00:57
So this question tells us that we have a slope or m of negative 7 ,445 kelvin and it wants to know the activation energy.
01:23
Because it tells us a slope, we can equate this arranius equation to a straight line.
01:29
So you get y is equal to mx plus b.
01:37
From there, we can see that this is our slope.
01:48
And from here, we know that we've got this negative 7 ,445 equal to this negative ea over our term.
02:00
So from here, we can say that we have negative 7 -445 kelvin is equal to negative ea over r, which is then equal to negative ea over 8 .314, which is then equal to negative ea over 8 .314, which is the gas constant...