Enthalpy of formation | Thermodynamics | AP Chemistry | Khan Academy

Enthalpy of 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. The superscript naught 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. The elements have to be in their standard states. So let's think about forming one mole of carbon dioxide. Carbon dioxide is composed of the elements carbon and oxygen. Under standard conditions, the most stable form of carbon is graphite, so we're going to write carbon in the solid state and we're going to write graphite over here.

Next, when you think about the most stable form of oxygen under standard conditions, at one atmosphere of 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. 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. Thus, we have our subscript F and our superscript naught to 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. 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. For the coefficients to make one mole of water, we need a one-half 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. 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. 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 in kilojoules per mole are often found in the appendices of many textbooks. 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 of 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 of 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. If we look at our two products over here, 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 is negative 393.5 kilojoules per mole of carbon dioxide.

So we're going to 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 one: so we have two moles of H₂O, and the standard enthalpy of formation of H₂O is negative 285.8, so we're going to multiply this by negative 285.8 kilojoules per mole. Moles cancel out, and we get negative 393.5 kilojoules, and if the other one moles cancel out again, this would be plus negative 571.6 kilojoules, which is equal to negative 965.1 kilojoules. So that's the sum of all the standard enthalpies of formation of our products.

Next, we need to sum the standard enthalpies of formation of our reactants. 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; so 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.

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 units, sometimes you see kilojoules, sometimes you see kilojoules per mole, and sometimes you see kilojoules per mole of reaction. What kilojoules per mole of 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 that 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 cancel 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.