00:01
There's multiple parts to this question.
00:05
First, given the chemical reaction and the rate constant at two different temperatures, you're asked to determine the activation energy.
00:14
The activation energy can be determined by taking the equation in your textbook that has two activation energies and two temperatures as variables and rearranging it to solve for just activation energy.
00:27
That's what i did here, where activation energy is equal to r, natural log of k2 over k1 divided by 1 over t1 minus t2 where these are kelvin temperatures.
00:40
I'll then plug in my values, natural log of the second k value at the higher temperature, divided by the first k value at the lower temperature, and then divide that whole thing by kelvin temperatures.
00:55
We're given temperatures in celsius, so we need to add on 273 to the 100 celsius, and we need to add on 273 to the 130.
01:02
Celsius, and i get ea of 98 ,900 joules per mole, or 98 .9 kilojoules per mole.
01:12
It then wants us to calculate the frequency factor at 373 kelvin.
01:19
So we'll use a different form of the uranus equation.
01:22
The different form of the uranus equation is one over the frequent, it can be rearranged, so one over the frequency factor is equal to e to the negative ea over rt divided by the rate constant.
01:37
Well, ea, i determined up here.
01:39
The temperature they told me to use is 373, and then the k value will be the k value at 373.
01:49
That also is provided...