- [Educator] Bond enthalpies can be used
to estimate the standard change in enthalpy
for a chemical reaction.
Let's use bond enthalpies
to estimate the enthalpy of combustion
of ethanol.
Looking at our balanced equation,
we have one mole of ethanol reacting
with three moles of oxygen gas
to produce two moles of carbon dioxide
and three moles of water in the gaseous state.
To find the standard change in enthalpy
for this chemical reaction,
we need to sum the bond enthalpies
of the bonds that are broken.
And from that, we subtract the sum
of the bond enthalpies of the bonds
that are formed in this chemical reaction.
To figure out which bonds are broken
and which bonds are formed,
it's helpful to look at the dot structures
for our molecules.
So let's start with the ethanol molecule.
We're gonna approach this problem first
like we're breaking all of the bonds in these molecules.
And we're also not gonna worry about units until the end,
just to save some space on the screen.
So looking at the ethanol molecule,
we would need to break a carbon-carbon bond.
So let's go ahead and write this down here.
Right now, we're summing the bond enthalpies
of the bonds that are broken.
So we have one carbon-carbon bond.
So let's write in here,
the bond enthalpy for a carbon-carbon bond.
Next, we have five carbon-hydrogen bonds
that we need to break.
So to this,
we're going to write in here, a five,
and then the bond enthalpy
of a carbon-hydrogen bond.
Next, we have to break a carbon-oxygen single bond.
So we write a one,
and then the bond enthalpy
for a carbon-oxygen single bond.
And then for this ethanol molecule,
we also have an oxygen-hydrogen single bond.
So we'll write in here, a one,
and the bond enthalpy
for an oxygen-hydrogen single bond.
We saw in the balanced equation
that one mole of ethanol reacts
with three moles of oxygen gas.
So to represent the three moles of oxygen gas,
I've drawn in here,
three molecules of O2.
And we can see in each molecule of O2,
there's an oxygen-oxygen double bond.
So we would need to break three oxygen-oxygen double bonds.
So to this,
we're going to add a three times the bond enthalpy
of an oxygen-oxygen double bond.
And this now gives us the sum of the bond enthalpies
for all the bonds that need to be broken.
It takes energy to break a bond.
So the summation of the bond enthalpies
of the bonds that are broken
is going to be a positive value.
And since it takes energy to break bonds,
energy is given off when bonds form.
So next, we're gonna sum the bond enthalpies
of the bonds that are formed.
And notice we have this negative sign in here
because this energy is given off.
So we're gonna write a minus sign in here,
and then we're gonna put some brackets
because next we're going to sum the bond enthalpies
of the bonds that are formed.
In our balanced equation,
we formed two moles of carbon dioxide.
So to represent those two moles,
I've drawn in here,
two molecules of CO2.
And we can see that in each molecule of CO2,
we're going to form two carbon-oxygen double bonds.
So that's a total of four carbon-oxygen double bonds.
So down here,
we're going to write a four times the bond enthalpy
of a carbon-oxygen double bond.
We also formed three moles of H2O.
And in each molecule of water that's drawn here,
we form two oxygen-hydrogen single bonds.
And since we have three moles,
we have a total of six oxygen-hydrogen single bonds.
So to this,
we're going to add six times the bond enthalpy
of an oxygen-hydrogen single bond.
The next step is to look up the bond enthalpies
of all of these different bonds.
For example, the bond enthalpy
for a carbon-carbon single bond is about 348 kilojoules
per mole.
You might see a different value,
if you look in a different textbook.
However, we're gonna go with 348 kilojoules per mole
for our calculation.
And we're gonna multiply this
by one mole of carbon-carbon single bonds.
Next, we look up the bond enthalpy
for our carbon-hydrogen single bond.
And that's about 413 kilojoules per mole
of carbon-hydrogen bonds.
And we're multiplying this by five.
And we continue with everything else
for the summation of the the bond enthalpies
of the bonds broken.
When we do this,
we get positive 4,719 kilojoules.
Next, we do the same thing
for the bond enthalpies
of the bonds that are formed.
So the bond enthalpy
for our carbon-oxygen double bond is 799 kilojoules
per mole,
and we multiply that by four.
The bonds enthalpy for an oxygen hydrogen single bond
is 463 kilojoules per mole,
and we multiply that by six.
When we add these together,
we get 5,974.
So for the final standard change in enthalpy
for our chemical reaction,
it's positive 4,719 minus 5,974,
which gives us negative 1,255 kilojoules.
Notice that we got a negative value
for the change in enthalpy.
And that means the combustion of ethanol
is an exothermic reaction.
And 1,255 kilojoules of energy are given off
for the combustion of one mole
of ethanol.
Also notice that the sum of the bond enthalpies
of the bonds formed,
which is 5,974,
is greater than the sum of the bond enthalpies
of the bonds broken,
which is 4,719.
And since we're subtracting a larger number
from a smaller number,
we get that negative sign
for the change in enthalpy.
If the sum of the bond enthalpies
of the bonds that are broken,
if this number is larger than the sum
of the bond enthalpies
of the bonds that have formed,
we would've gotten a positive value
for the change in enthalpy.
And that would be true for an endothermic reaction.
We did this problem,
assuming that all of the bonds
that we drew in our dots structures were broken
and all of the bonds
that we drew in the dot structures were formed.
However, if we look closely to dots structures
or just look closely to what we wrote here,
we show breaking one oxygen-hydrogen single bonds over here,
and we show the formation
of six oxygen-hydrogen single bonds over here.
So we could have just canceled out one
of those oxygen-hydrogen single bonds.
So we could have canceled this out.
And instead of showing a six here,
we could have written a five times the bond enthalpy
of an oxygen-hydrogen single bond.
We still would have ended up with the same answer
of negative 1,255 kilojoules.
So if you look at your dot structures,
if you see a bond that's the same on the reactant side
and the same on the product side,
you don't have to show the breaking
and forming of that bond.
You can make the problem a little bit shorter,
if you want to.
Finally, let's show how we get our units.
So to get kilojoules as your final answer,
if we go back up to here,
we wrote a one times 348.
The one is referring to breaking one mole
of carbon-carbon single bonds.
And the 348, of course,
is the bond enthalpy
for a carbon-carbon single bond.
So this was 348 kilojoules per one mole
of carbon-carbon single bonds.
When you multiply these two together,
the moles of carbon-carbon single bonds cancels
and this gives you 348 kilojoules.
And so, that's how to end up
with kilojoules as your final answer.
You also might see kilojoules per mole of reaction
as the units for this.
And, kilojoules per mole reaction means
how the reaction is written.
So for the combustion
of one mole of ethanol,
1,255 kilojoules of energy are released.
To get kilojoules per mole of reaction as our units,
the balanced equation had a one as the coefficient
in front of ethanol.
Therefore, you're breaking one mole
of carbon-carbon single bonds per one mole
of reaction.
So we can use this conversion factor.
Now, when we multiply through the moles
of carbon-carbon single bonds,
cancel and this gives us 348 kilojoules per mole
of reaction.