00:01
In the given question, we have been provided two reactions, and for these reactions, we have to find the heat of combustion.
00:06
So, delta hc is been asked for the given question.
00:10
Starting with our first reaction, it is the oxidation of methanol.
00:15
Oxidation means reaction with oxygen, and thus it is producing carbon dioxide and two moles of water.
00:22
Now, as we can see, the delta h combustion or the delta h of this reaction will be equals to, delta h formation of carbon dioxide plus 2 into delta h formation of water in liquid state minus delta h of methanol okay delta h formation for methanol okay and for oxygens delta h formation is 0 so we will not use it okay since adding or subtracting 0 will not change our answer.
01:01
Okay.
01:02
So delta hr for this reaction that is in part a.
01:06
So first we need the delta hf of carbon dioxide that is equals to minus 393 .5.
01:13
Okay.
01:14
Plus we have to add the delta hf of water.
01:17
So it has two moles in the reaction.
01:19
So we multiply by two.
01:20
So 2 into minus 285 .83.
01:24
And now we have to subtract the delta hf for methanol.
01:28
Which is equals to minus 239.
01:33
Minus 239 .1.
01:38
So delta h combustion for the given reaction will come out to be minus 7206.
01:45
07 kilojoule per mole.
01:48
So this is our answer for part a.
01:51
And similarly for part b, as we can see, the delta h in case of part b, the reaction is, 2 h4 in liquid form reacts with oxygen gas to form nitrogen gas and 2 moles of water.
02:09
So once again our delta h combustion or delta h of this reaction will be equals to the delta h formation of nitrogen...