00:01
Chapter 14, question 6 .2.
00:03
Well, this first part of the question is not very much important.
00:09
The important part, the decomposition of nitric oxide, n02 and 2, n2, is second order, which is not important either, with a rate constant of that at 737 celsius, and this at 947 celsius.
00:28
They want us to calculate the activation energy for the reaction.
00:31
So what we can do is if we write the arrhenius equation for two different ks, what we're going to have is this formula.
00:44
Lnk1 over k2 equals to energy of activation over r, 1 over t2, minus 1 over t1.
01:07
Now, what we have over here is ln.
01:12
For k1, we're going to use this value.
01:18
0 .0796, divided by 4k2, we're going to use this value...