00:02
Hi there.
00:03
In this question, we are trying to calculate the standard enthalpy change for these reactions.
00:09
So, in other words, the standard heat of reaction.
00:18
And to do that, we need to take the sum of the heat of formation for the products and subtract the sum of the heat of formation of the reactants.
00:51
So the sum of the products minus the sum of the products.
00:55
Of the reactants for each of these equations.
00:58
And we need to look up these values in table c.
01:01
So let's get started with letter a.
01:04
In letter a, our product is s -o -3.
01:09
So looking that up, i see that is negative 395 .2.
01:15
Well, in the balanced equation, we have two moles.
01:18
So we need to multiply this value by two.
01:22
And again, that was negative 395 .2.
01:28
And that is kilojoules per mole.
01:30
All right, and i just multiplied it by two moles, so that is going to give me kilojoules in the end.
01:40
All right, that is my only product.
01:42
Now i need to subtract the sum of the heat of formation of the reactants.
01:49
There are two moles of so2.
01:54
So2 is negative 296 .9.
01:59
All right, the other reactant is oxygen, but that is in its elemental form.
02:04
So by definition, its heat of formation is zero.
02:08
So we do not need to include that here.
02:12
So calculating this gives me negative 196 .6 kilojoules.
02:22
And that would be my answer for part a.
02:28
Moving on to part b.
02:31
We have two products there.
02:33
We have mgo and we have liquid water.
02:36
So we need to sum up our products here.
02:39
There is only one mole of hgo and one h2o in the balanced equation.
02:45
So we just need to look up their heats of formation.
02:50
For the mgo, it's 601 .8 kilojoules per mole.
03:01
And for the liquid h2o, it is negative 285 .83.
03:11
All right, those are the products.
03:14
Now we need to subtract the reactants.
03:16
There is only one reactant, that is the mg -oh2.
03:22
There's only one of those.
03:24
It is negative 924 .7 kilojoules per mole...