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## AP®︎/College Chemistry

### Course: AP®︎/College Chemistry > Unit 6

Lesson 6: Hess's law# Hess's law

AP.Chem:

ENE‑3 (EU)

, ENE‑3.C (LO)

, ENE‑3.C.1 (EK)

, ENE‑3.D (LO)

Hess's law states that if a process can be expressed as the sum of two or more steps, the enthalpy change for the overall process is the sum of the Δ

*H*values for each step. To use Hess's law, two principles must be understood: one, if an equation is reversed, the sign of the Δ*H*value is also reversed. Two, if an equation is multiplied by a coefficient, the Δ*H*value is multiplied by the same coefficient. Created by Jay.## Want to join the conversation?

- At5:00, how come the answer is not +226.8 kJ/molrxn? Isn't +1299.6 + (-787) + (-285.8)= 226.8 and not -226.8?(9 votes)
- Good eye, yeah they did make a math error. I checked around just to make sure and the reaction is endothermic so the enthalpy should be a positive value.(6 votes)

- Why exactly is reversing equations necessary?(1 vote)
- We want the smaller reactions 1-3 to add in such a way that they make the overall reaction we began with. We need C2H2 as a product and the only way to get that from reaction 1 is to reverse it.(1 vote)

- Getting to a point where you can utilize Hess's Law just seems to convenient. How do chemists find equations that cancel out perfectly to get to that point?(1 vote)
- Chemists have recorded the enthalpy changes for most simple reactions using various calorimetry experimental techniques. Using Hess’ law we only need to experimentally determine the enthalpy of a relatively small number of the most simple reactions which then allows us to calculate the enthalpies of vast number of complex reactions.

For example here the actual reaction we want to know the enthalpy of is reacting solid carbon (graphite) with hydrogen gas to produce acetylene. So, we need a few simpler reactions which has graphite, hydrogen gas, and acetylene. And the three lower reactions involving those chemicals which we know the enthalpies of are about as simple as you can get in an experimental setting. All that is happening with those lower equations is that we’re reacting the desired chemicals with oxygen gas (essentially burning them) and observing the enthalpy change.

Hope that helps.(1 vote)

## Video transcript

- [Instructor] Hess's law
states that the overall change in enthalpy for a chemical
reaction is equal to the sum of the enthalpy changes for each step. And this is independent of the path taken. So it doesn't matter what
set of reactions you use. If you add up those reactions
and they equal the reaction that you're trying to find, you can also sum the enthalpies to find the enthalpy
change for the reaction. As an example, let's say we're
trying to find the change in enthalpy for the reaction
of carbon with hydrogen gas to form C2H2, which is acetylene. We can calculate the change
in enthalpy for the formation of acetylene using these
three reactions below. Our approach will involve
looking at these three reactions and comparing them to
the original reaction to see if we need to change anything. For example, if we look at reaction one, there's one mole of acetylene on the left side of the equation. And if we compare that
to the original reaction, there's one mole of
acetylene on the right side of the equation. So we need to reverse equation one to make it look more like
our original reaction. To save time, I've gone ahead
and reversed equation one. So you can see, I did that down here. Looking at the original
equation for equation one, here where the products and now we've made those
products the reactants. And what were the reactants
over here for equation one have now become the products. The change in enthalpy for equation one is -1,299.6 kilojoules
per mole of reaction. Kilojoules per mole reaction
just means how the reaction is written in the balanced equation. And since we reversed equation one, we also need to reverse
the sign for Delta H. So instead of this being a negative, instead of this being a negative, we're gonna go ahead and
change this into a positive. And also let's go ahead and
cross out the first equation. So we don't get confused. Next, we look at equation two and we compare it to our original. For equation two there's
one mole of solid carbon on the left side and looking
at our original reaction, there's two moles of
carbon on the left side. So to get equation two, to
look like our original equation we need to multiply everything
through, by a factor of two. So we're gonna multiply
everything in equation two by a factor of two. To save some time, I have
gone ahead and written out what we would get. We would get two carbons plus two O2s goes to 2CO2. The change in the
enthalpy for the formation of one mole of CO2 was -393.5 kilojoules
per mole of reaction. But now we're forming two moles of CO2. And since we multiplied
the equation through by a factor of two, we also need to multiply the change in enthalpy by a factor of two as well. And also let's go ahead and cross out this first version here because now we're
forming two moles of CO2. Next, we look at equation three and we can see there's
one mole of hydrogen gas on the left side of the equation which matches the original reaction which also has one mole of
hydrogen gas on the left side. So we don't need to do
anything to equation three. And since we're not doing
anything to the equation, we're also not gonna do anything to the change in the enthalpy. So it's gonna stay -285.8
kilojoules per mole of reaction. Next we add up all of our
reactants and products. So we have two CO2 plus
H2O plus 2C plus 2O2 plus H2 plus one half O2. So those are all written
down here for our reactants. And then for the products, let
me just change colors here. We have C2H2 plus 5O2 plus 2CO2 plus H2O. And so those are written
over here for the products. Next we see what we can cancel out. There's 2CO2 on the left side and there's 2CO2 on the right side. So those cancel out. There's one water on the left
and one water on the right. And there's 2O2s plus one half O2 which is 2.5O2s or five halves O2s. So the oxygen's cancel
out on both sides as well. And we can see we're left with
two carbons plus one hydrogen goes to form one C2H2 which is the same as
our original equation. Since we were able to add up our equations and get the overall equation, according to Hess's law, we should also be able to
add the changes in enthalpies for these steps to get
the change in the enthalpy for the overall reaction. If we look at the changes in enthalpy for the individual steps, we had +1299.6 for the first equation. And so that's up here. For the second equation we
had negative 393.5 times two, which is -787. And for our third equation, we had -285.8. So that's -285.8. When we add everything together we get +226.8 kilojoules
per mole of reaction. So for the formation of
one mole of acetylene from two moles of carbon
and one mole of hydrogen the change in enthalpy for this reaction is equal to +226.8 kilojoules
per mole of reaction.