00:02
In this problem, we are told that for a given reaction, the rate constant k was measured different temperatures.
00:26
And what was concluded is that a graph of natural log of k versus 1 over t produced a straight line, a slope equal to negative 4 ,269 degree kelvin and so this is the magnitude of our, or excuse me, the value of our slope and so we are asked to find the activation energy for this reaction.
01:29
So to do this, let's write down our erroneous equation.
01:33
Write that down a reneous equation and that's where we have our constant k equal to our pre -existing factor so that's going to be a exponent of negative negative e sub a which is our activation energy over or excuse me yeah divided by r t and another form for this would be if you take the natural log you'll simply get that the natural log of rk constant will be equal to negative ea, our activation energy over rt, plus the natural log of our preexisting factor.
02:35
And this essentially is just some constant.
02:41
So now, when the graph was plotted, essentially, what was found was our...