00:01
For this question, the chemical reaction is 4 moles of aluminum solid, reacts with 3 moles of o2 gas to produce 2 moles of aluminum oxide al2 -o3 solid.
00:17
So to determine delta h for this process, it'll be 2 times the delta h of formation of aluminum oxide provided as negative 1 ,576.
00:31
0 .4 kilojoules, plus, or actually now minus, three times the delta h of formation of oxygen, which is zero, and four times the delta h of formation of aluminum, which is also zero.
00:52
So for the process, as written, we get a delta h of negative 3 ,000, 152 .8 kilojoules.
01:01
So because it's negative, the process is exothermic.
01:12
So delta g for the reaction is negative, 1 ,500, well, sorry, 3 ,152 .8 kilojoules.
01:23
And because it's negative, let's see, our delta h is negative, and yeah, our delta s is actually negative, because we're going to, our delta h is negative.
01:45
From 3 moles of gas to no moles of gas.
01:49
So, i guess to determine whether it's spontaneous, we need to know what delta s is.
02:02
Huh...