00:03
This is the answer to chapter 15, problem number seven from the smith organic chemistry textbook.
00:10
And this problem asks us to calculate delta h.
00:14
Not for the two propagation steps in the reaction of methane with bromine to form methyl bromide and hydrobomic acid.
00:25
And so that reaction is presented in problem 15 .6.
00:31
So the answer to the problem immediately before this one has this whole mechanism drawn out.
00:37
Okay.
00:39
And so there are actually two propagation steps.
00:45
And so the first one, so we'll say step one, what we have is one of the carbon hydrogen bonds of methane is being broken.
01:03
And we are forming the hbr and this methyl radical.
01:27
Okay.
01:30
So i'll do this in green.
01:33
So this bond is broken.
01:41
You know what? i should draw electrons in.
01:49
So we have a free radical there.
01:54
Okay.
01:54
And so i've arrowed the bond that's going to be broken.
02:00
And this requires 435 kilojoules per mole, 435 kj per mole.
02:12
Okay.
02:14
And then in red on the right side of the arrow here, i will arrow this bond.
02:20
It's the one that's being formed.
02:23
And the formation of that bond is going to release 368 kilojoules per mole.
02:29
So remember when energy is released, we use a negative sign.
02:34
It's 368 kilojoules per mole.
02:38
And then we just need to add those two numbers together.
02:42
And that's going to be our delta h .0.
02:50
And so 435 minus 368 is going to be 67.
03:05
Okay...