- [Instructor] Enthalpy of a formation
refers to the change in enthalpy
for the formation of one mole
of a substance from the most stable form
of its constituent elements.
Change in enthalpy is symbolized
by delta H and the f stands for formation.
And the superscript nought refers to the fact
that everything is under standard state conditions,
which refers to atmospheric pressure
of one atmosphere and a specified temperature
that is usually 25 degrees Celsius.
So when we're thinking about standard enthalpy of formation,
we're thinking about the elements
and the state that they exist under standard conditions.
So the elements have to be in their standard states.
So let's think about forming one mole of carbon dioxide.
So carbon dioxide is composed of the elements
carbon and oxygen.
And under standard conditions,
the most stable form of carbon is graphite.
So we're gonna write carbon in the solid state
and we're gonna write graphite over here.
And next, when you think about the most stable form
of oxygen under standard conditions.
And so at one atmosphere, so atmospheric pressure
and room temperature of 25 degrees Celsius,
the most stable form of oxygen is oxygen gas.
So we can go ahead and write in here O2.
And since we're forming one mole of carbon dioxide
from the elements that make up carbon dioxide
in their most stable form under standard conditions,
the change in enthalpy for this
would be the standard enthalpy of formation.
So we have our subscript f
and our superscript nought indicate standard conditions.
The change in enthalpy for the formation
of one mole of CO2 is equal to negative 393.5 kilojoules
per one mole of carbon dioxide.
Let's look at some more equations showing the formation
of one mole of a substance.
For example, let's look at the equation
showing the formation of one mole of water.
So water is composed of hydrogen and oxygen
and the most stable forms of those two elements
under standard conditions are hydrogen gas and oxygen gas.
And for the coefficients to make one mole of water,
we need a 1/2 as our coefficient in front of O2.
The standard change in enthalpy of formation
for the formation of one mole of water
is negative 285.8 kilojoules per mole.
We can do the same thing for the formation of one mole
of methane CH4.
We already know that the most stable form
of carbon is graphite
and the most stable form of hydrogen is hydrogen gas.
And the standard change in enthalpy of formation
for the formation of one mole of methane
is equal to negative 74.8 kilojoules per mole.
Next, let's think about forming one mole of oxygen gas.
Well, we're forming the oxygen gas
from the most stable form of oxygen
under standard conditions,
which is also diatomic oxygen gas, O2.
So we're not changing anything we're going from O2 to O2.
And since there's no change, there's no change in enthalpy.
Therefore, the standard enthalpy
of formation is equal to zero.
And this is true for the most stable form of any element.
The standard enthalpy of formation
of the most stable form of any element is zero
since you'd be making it from itself.
Standard enthalpies of formation and kilojoules per mole
are often found in the appendices of many textbooks.
And if you look in the appendix of a textbook,
you'll see the standard enthalpy of formation
for diatomic oxygen gas, O2, is equal to zero.
Ozone, which is O3, also exists under standard conditions.
However, it's not the most stable form of oxygen
under standard conditions and therefore,
its standard enthalpy formation is not zero, it's 142.3.
Graphite is the most stable form of carbon
under standard conditions.
Therefore, it has a standard enthalpy
of formation of zero, but of course,
diamond also exists under standard conditions
but it's not the most stable form.
So its standard enthalpy formation is not zero,
it's 1.88 kilojoules per mole.
Enthalpies of formation can be used to calculate
the change in enthalpy for a chemical reaction.
We can do this by using the following equation.
The standard change in enthalpy for a chemical reaction
is equal to the sum of the standard enthalpies of formation
of the products minus the sum of the standard enthalpies
of formation of the reactants.
Let's say our goal is to find the standard change
in enthalpy for the following chemical reaction.
So we have one mole of methane
reacting with two moles of oxygen
to form one mole of carbon dioxide and two moles of water.
The first thing we need to do is sum
all the standard enthalpies of formation of the products.
So if we look at our two products over here
and we'll start with one mole of carbon dioxide.
So let's go ahead and write this down here.
We have one mole of carbon dioxide
and the standard molar enthalpy of carbon dioxide
we've already seen as negative 393.5 kilojoules
per mole of carbon dioxide.
So we're gonna multiply one mole of carbon dioxide
by negative 393.5 kilojoules per mole of carbon dioxide.
Our other product is two moles of water.
So we're going to add this to the other ones.
We have two moles of H2O.
And the standard enthalpy of formation of H2O
is negative 285.8.
So we're gonna multiply this
by negative 285.8 kilojoules per mole.
So moles cancel out and we get negative 393.5 kilojoules.
And then for the other one, moles cancel out again.
And this would be plus negative 571.6 kilojoules,
which is equal to negative 965.1 kilojoules.
So that's the sum of all
of the standard enthalpies of formation of our products.
Next, we need to sum the standard enthalpies
of formation of our reactants.
So the two reactants that we have are methane and oxygen
and we have one mole of methane.
So let's go ahead and write that in here.
So we have one mole of methane.
The standard molar enthalpy of formation of methane
is negative 74.8 kilojoules per mole.
So we're multiplying one mole
by negative 74.8 kilojoules per mole.
Our other reactant is oxygen.
And we know that diatomic oxygen gas
has a standard enthalpy of formation of zero.
So we could go ahead and write this in
just to show it.
So we have two moles of oxygen
but we're multiplying that number by zero.
Some moles cancel and give us negative 74.8 kilojoules.
And we're adding zero to that.
So negative 74.8 kilojoules
is the sum of all the standard enthalpies of formation
of our reactants.
So to find the standard change in enthalpy for our reaction,
we take the summation of the enthalpies of formation
of our products, which was negative 965.1 kilojoules.
And from that, we subtract the sum
of the standard enthalpies of formation of the reactants,
which we found was negative 74.8 kilojoules.
So negative 965.1 minus negative 74.8
is equal to negative 890.3 kilojoules.
For the unit, sometimes you see kilojoules,
sometimes you see kilojoules per mole,
and sometimes you see kilojoules per mole of reaction.
And what kilojoules per mole a reaction means
is how the balanced equation is written.
For this balanced equation,
we're showing the combustion of one mole of methane.
So combusting one mole of methane
releases 890.3 kilojoules of energy.
So that's what kilojoules per mole of reaction
is referring to.
Let's go back to the step
where we summed the standard enthalpies of formation
of the products to see how we could actually get kilojoules
per mole of reaction as our units.
To do this, we need to use a conversion factor.
For how the equation is written,
we're producing one mole of carbon dioxide.
So we can use as a conversion factor,
there's one mole of carbon dioxide per one mole of reaction.
We can do the same thing for our other product,
which is water.
For how the equation is written,
we're forming two moles of water.
So our conversion factor can be there are two moles of water
for every one mole of reaction.
Next, moles of carbon dioxide cancels out
and moles of water cancel out.
And this gives us kilojoules per mole of reaction
as our units.
It's a little more time-consuming
to write out all the units this way.
So often, it's faster to do it the first way
and add in these units at the end.