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Course: AP®︎/College Chemistry > Unit 6
Lesson 7: Enthalpy of formationEnthalpy of formation
The standard enthalpy of formation of a substance is the enthalpy change that occurs when 1 mole of the substance is formed from its constituent elements in their standard states. A pure element in its standard state has a standard enthalpy of formation of zero. For any chemical reaction, the standard enthalpy change is the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants. Created by Jay.
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For any chemical reaction, the standard enthalpy change is the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants.
How does this relate to the definition stating enthalpy of reaction is sum of enthalpies of products minus enthalpies of reactants ?(5 votes)
I'm confused by the explanation of what "kilojoules per mole of reaction" means at8:00. Can anyone explain it further?(1 vote)
It's the unit for enthalpy commonly used. The kilojoules part is easy enough to understand since it's a unit of energy but the moles part of the unit is introduced because the amount of energy released (or absorbed) by the reaction varies by how much of your reactants you have. So if you just have 1 mole of methane (CH4) then the reaction will release -890.3 kJ of heat, but you had 2 moles of methane then the reaction will release twice that initial amount of heat, or 1780.6 kJ.
Hope that helps.(5 votes)
How are you able to get an enthalpy value for a equation with enthalpies of zero? What values are you using to get the first examples on the slides?(1 vote)
I don't understand how the standard enthalpies of formation (SEF) of products and reactants is relevant when considering the enthalpy of a reaction. Say you have a combustion reaction. The SEF of carbon dioxide, like the video said, is derived from reacting graphite with oxygen. However, in a combustion reaction, you're forming carbon dioxide from a hydrocarbon and oxygen; there is no graphite involved. So how would the SEF of carbon dioxide or really any species involved be relevant when the species are not being formed from constituent elements? Also, does the standard enthalpy change equal the bond enthalpy for a reaction?(1 vote)
Is bond enthalpy considered here?(1 vote)
no, since bond enthalpy is usually defined as the energy required to break a bond and enthalpy of formation is defined as the making of the compound, i.e., making the bond. So, you can say that they are actually opposite(1 vote)
- How do we define 1 mole of reaction?
I mean, we can use methane as a reference, but so can we use oxygen, can't we?(1 vote)- 1 mole of reaction is equal to the amount of reaction when the numbers of moles of the reactants (and therefore products) are equal to their coefficients in the equation. This means that when you change the coefficients (by multiplying each side by the same number) you get a different enthalpy! In this example, if you used oxygen as a reference, your enthalpy value would be 1/2 the enthalpy if you used methane as a reference.(1 vote)
- When writing the chemical equation for water we are told that two molecules of hydrogen reacts with a molecule of oxygen.Why do i see chemical equations where a molecule of hydrogen reacts with half of an oxygen molecule? do i need a refresher on the laws of chemical combination or I'm just getting really confused?(1 vote)
- Standard enthalpy of formation is defined as the change in enthalpy when one mole of the compound forms from its constituent elements in their stand states. The key being that we're forming one mole of the compound. So if we were forming water from hydrogen and oxygen using whole number coefficients as we would normally it would look like: 2H2 + O2 → 2H2O. But since we're only interested in forming one mole of water we divide everything by 2 to change the coefficient of water from 2 to 1. So now it becomes: H2 + (1/2)O2 → H2O which yields a ΔHf° of -241.8 kJ/mol. This second reaction isn't actually happening, it just conforms to the definition. It's convenient that it's defined the way it is though since producing one mole means that using the enthalpy of formation of water to calculate the enthalpy of a reaction with water means that we only have to multiply this -241.8 kJ/mol value by the coefficient of water in the reaction we're studying.
Hope that helps.(1 vote)
if the standard enthalpy of formation of O₂ is zero why then in the formation of H₂O the oxygen on the reactants side doesn't have the standard enthalpy of formation as zero and the same with hydrogen if its standard formation of enthalpy is also zero then the standard formation of the water should also be zero?(1 vote)- How to calculate the heat of formation of triethylaluminum ?(1 vote)
- if the equation for standard enthalpy change is like A = B - C, for reaction change, product change, and reactant change in that order, how do you rearrange it to get B = A - C to solve for the product change?(1 vote)
- Well, I mean if you’re starting with A = B – C, and you’re solving for B, then all you have to do is add C to both sides. And then you get B = A + C.(0 votes)
8:00Video transcript
- [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.