This problem is from chapter five of the Kotz, Treichel,
Townsend Chemistry and Chemical Reactivity textbook.
So they tell us, suppose you want to know the enthalpy
change-- so the change in total energy-- for the
formation of methane, CH4, from solid carbon as a
graphite-- that's right there-- and hydrogen gas.
So we want to figure out the enthalpy
change of this reaction.
How do we get methane-- how much energy is absorbed or
released when methane is formed from the reaction of--
solid carbon as graphite and hydrogen gas?
So they tell us the enthalpy change for this reaction
cannot to be measured in the laboratory because the
reaction is very slow.
So normally, if you could measure it you would have this
reaction happening and you'd kind of see how much heat, or
what's the temperature change, of the surrounding solution.
Maybe this is happening so slow that it's very hard to
measure that temperature change, or you can't do it in
any meaningful way.
We can, however, measure enthalpy changes for the
combustion of carbon, hydrogen, and methane.
So they're giving us the enthalpy changes for these
combustion reactions-- combustion of carbon,
combustion of hydrogen, combustion of methane.
And they say, use this information to calculate the
change in enthalpy for the formation of
methane from its elements.
So any time you see this kind of situation where they're
giving you the enthalpies for a bunch of reactions and they
say, hey, we don't know the enthalpy for some other
reaction, and that other reaction seems to be made up
of similar things, your brain should immediately say, hey,
maybe this is a Hess's Law problem.
Hess's Law.
And all Hess's Law says is that if a reaction is the sum
of two or more other reactions, then the change in
enthalpy of this reaction is going to be the sum of the
change in enthalpies of those reactions.
Now, when we look at this, and this tends to be the confusing
part, how can you construct this reaction out of these
reactions over here?
And what I like to do is just start with the end product.
So I like to start with the end product, which is methane
in a gaseous form.
And when we look at all these equations over here we have
the combustion of methane.
So this actually involves methane, so
let's start with this.
But this one involves methane and as a
reactant, not a product.
But what we can do is just flip this arrow and write it
as methane as a product.
So if we just write this reaction, we flip it.
So now we have carbon dioxide gas-- let me write it down
here-- carbon dioxide gas plus-- I'll do this in another
color-- plus two waters-- if we're thinking of these as
moles, or two molecules of water, you could even say--
two molecules of water in its liquid state.
That can, I guess you can say, this would not happen
spontaneously because it would require energy.
But if we just put this in the reverse direction, if you go
in this direction you're going to get two waters-- or two
oxygens, I should say-- I'll do that in this pink color.
So two oxygens-- and that's in its gaseous state-- plus a
gaseous methane.
CH4.
CH4 in a gaseous state.
And all I did is I wrote this third equation, but I wrote it
in reverse order.
I'm going from the reactants to the products.
When you go from the products to the reactants it will
release 890.3 kilojoules per moles of the
reaction going on.
But if you go the other way it will need 890 kilojoules.
So the delta H here-- I'll do this in the neutral color-- so
the delta H of this reaction right here is going to be the
reverse of this.
So it's positive 890.3 kilojoules
per mole of the reaction.
All I did is I reversed the order of this
reaction right there.
The good thing about this is I now have something that at
least ends up with what we eventually
want to end up with.
This is where we want to get.
This is where we want to get eventually.
Now, if we want to get there eventually, we need to at some
point have some carbon dioxide, and we have to have
at some point some water to deal with.
So how can we get carbon dioxide, and
how can we get water?
Well, these two reactions right here-- this combustion
reaction gives us carbon dioxide, this combustion
reaction gives us water.
So we can just rewrite those.
Let me just rewrite them over here, and I will-- let me use
some colors.
So if I start with graphite-- carbon in graphite form--
carbon in its graphite form plus-- I already have a color
for oxygen-- plus oxygen in its gaseous state, it will
produce carbon dioxide in its gaseous form.
It will produce carbon-- that's a different shade of
green-- it will produce carbon dioxide in its gaseous form.
And this reaction, so when you take the enthalpy of the
carbon dioxide and from that you subtract the enthalpy of
these reactants you get a negative number.
Which means this had a lower enthalpy, which means energy
was released.
Because there's now less energy in the
system right here.
So this is essentially how much is released.
But our change in enthalpy here, our change in enthalpy
of this reaction right here, that's reaction one.
I'll just rewrite it.
Minus 393.5 kilojoules per mole of
the reaction occurring.
So the reaction occurs a mole times.
This would be the amount of energy
that's essentially released.
This is our change in enthalpy.
So if this happens, we'll get our carbon dioxide.
Now we also have-- and so we would release this much energy
and we'd have this product to deal with-- but we also now
need our water.
And this reaction right here gives us our water, the
combustion of hydrogen.
So we have-- and I haven't done hydrogen yet, so let me
do hydrogen in a new color.
That's not a new color, so let me do blue.
So right here you have hydrogen gas-- I'm just
rewriting that reaction-- hydrogen gas plus 1/2 O2--
pink is my color for oxygen-- 1/2 O2 gas will yield, will it
give us some water.
Will give us H2O, will give us some liquid water.
Now, before I just write this number down, let's think about
whether we have everything we need.
To make this reaction occur, because this gets us to our
final product, this gets us to the gaseous
methane, we need a mole.
Or we can even say a molecule of carbon dioxide, and this
reaction gives us exactly one molecule of carbon dioxide.
So that's a check.
And we need two molecules of water.
Now, this reaction only gives us one molecule of water.
So let's multiply both sides of the equation to get two
molecules of water.
So this is a 2, we multiply this by 2, so this essentially
just disappears.
You multiply 1/2 by 2, you just get a 1 there.
And then you put a 2 over here.
So I just multiplied this second equation by 2.
So I just multiplied-- this is becomes a 1, this becomes a 2.
And if you're doing twice as much of it, because we
multiplied by 2, the delta H now, the change enthalpy of
the reaction, is now going to be twice this.
Let's get the calculator out.
It's now going to be negative 285.8 times 2.
Because we just multiplied the whole reaction times 2.
So negative 571.6.
So it's negative 571.6 kilojoules
per mole of the reaction.
Now, let's see if the combination, if the sum of
these reactions, actually is this reaction up here.
And to do that-- actually, let me just copy and paste this
top one here because that's kind of the order that we're
going to go in.
You don't have to, but it just makes it hopefully a little
bit easier to understand.
So let me just copy and paste this.
Actually, I could cut and paste it.
Cut and then let me paste it down here.
That first one.
And let's see now what's going to happen.
To see whether the some of these reactions really does
end up being this top reaction right here, let's see if we
can cancel out reactants and products.
Let's see what would happen.
So this produces carbon dioxide, but then this mole,
or this molecule of carbon dioxide, is then used up in
this last reaction.
So this produces it, this uses it.
So those cancel out.
Let me do it in the same color so it's in the screen.
This reaction produces it, this reaction uses it.
Now, this reaction right here produces the two
molecules of water.
And now this reaction down here-- I want to do that same
color-- these two molecules of water.
Now, this reaction down here uses those two
molecules of water.
Now, this reaction right here, it requires one molecule of
molecular oxygen.
This one requires another molecule of molecular oxygen.
So these two combined are two molecules of molecular oxygen.
So those are the reactants.
And in the end, those end up as the products
of this last reaction.
So those, actually, they go into the system and then they
leave out the system, or out of the sum
of reactions unchanged.
So they cancel out with each other.
So we could say that and that we cancel out.
And so what are we left with?
What are we left with in the reaction?
Well, we have some solid carbon as graphite plus two
moles, or two molecules of molecular hydrogen yielding--
all we have left on the product side is some methane.
So it is true that the sum of these reactions is exactly
what we want.
All we have left on the product side is the graphite,
the solid graphite, plus the molecular hydrogen, plus the
gaseous hydrogen-- do it in that color-- plus
two hydrogen gas.
And all we have left on the product side is the methane.
All we have left is the methane in the gaseous form.
So it is true that the sum of these reactions-- remember, we
have to flip this reaction around and change its sign,
and we have to multiply this reaction by 2 so that the sum
of these becomes this reaction that we really care about.
So this is the sum of these reactions.
Its change in enthalpy of this reaction is going to be the
sum of these right here.
That is Hess's Law.
So this is the fun part.
So we just add up these values right here.
So we have negative 393.-- no, that's not
what I wanted to do.
Let me just clear it.
So I have negative 393.5, so that step is exothermic.
And then we have minus 571.6.
That is also exothermic.
Those were both combustion reactions, which are, as we
know, very exothermic.
And we have the endothermic step, the reverse of that last
combustion reaction.
So plus 890.3 gives us negative 74.8.
It gives us negative 74.8 kilojoules for every mole of
the reaction occurring.
Or if the reaction occurs, a mole time.
So there you go.
We figured out the change in enthalpy.
And it is reasonably exothermic.
Nowhere near as exothermic as these combustion reactions
right here, but it is going to release energy.
And we're done.