00:01
So this is a question where if you know what's happening, it's really, really easy to solve.
00:07
And so i'll explain where you get the result from once.
00:11
So whenever you see a question that deals with k temperature and activation energy, you should immediately think about the uranus equation, which it says that k, a rate constant for reaction, equals a, e to negative ea, activation energy over rt, a is a constant related to the reaction.
00:33
Ea is our activation energy.
00:35
R is our universal gas constant.
00:38
So, well, basically, what this question is doing, it sets up a linear equation.
00:45
So if we take the natural log of both sides, you'll get that ln of k equals, you'll have the natural log of a plus a negative ea over rt.
01:01
And so this is actually almost in the form that we're told we're told that we graph ln of k on the y axis and 1 over t on the x axis.
01:18
And so remember, our linear form of an equation is y equals mx plus b.
01:27
So we already kind of have this in the right form.
01:29
We said allen of k is on the y axis, one over t is.
01:34
On the x axis...