Hess’s law example | Thermodynamics | Chemistry | Khan Academy

Hess’s law example | Thermodynamics | Chemistry | Khan Academy

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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.

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