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Unit 6

6.9 Hess's Law

5 min readβ€’june 1, 2021

Anika P

dalia

Dalia Savy

fiveable-dylan

Dylan Black


AP ChemistryΒ πŸ§ͺ

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State Functions and Pathway Dependency

What is "Pathway Dependency"?

A pathway is the route that a process takes. Typically, pathway dependency/independency is applied to functions. Essentially, if the result of something depends on the order in which you did it, it is pathway dependent and if it is not, it is pathway independent.
For example, the change in elevation of a mountain climber is pathway independentπŸ§—. You could walk in a straight line or a zig-zagged line up the mountain but regardless, you both end up at the top of the mountain⛰️.

State Functions and Pathway Independency

A state function is a property whose value does not depend on the path taken to reach that specific value. In basic terms, state functions are pathway independent, just like our elevation example.
A pathway dependent function is called a path function, though we'll only be dealing with state functions.
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FGC5.14.png?alt=media&token=e0ad6bee-dd22-453e-b96d-1a7f6b01bae7

Hess's Law

Hess's Law is stated as simply: enthalpy is a state function. No matter what way you go from reactants to products, you will end up with the same enthalpy of reaction for the reaction. This means that if we know the enthalpy of formation for different reactions and you can manipulate them to get a single reaction, and you can find the enthalpy of reaction of that single reaction. It's like magic✨, and it's kinda fun!

Rules of Hess's Law

  1. When a reaction is reversed, the enthalpy change stays constant in magnitude but becomes reversed in mathematical sign (Flipping the reaction flips the sign on Ξ”HπŸ”).
  2. If an equation is multiplied by n, Ξ”H has to also be multipliedβœ–οΈ by n.
  3. When two (or more) reactions are added to obtain an overall reactionβž•, the individual enthalpy changes of each reaction are added to obtain the net change of enthalpy of the overall reaction Ξ”H).

Hess's Law Example Problem

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F4f729b6c682a4695ee56e10afd2c6ec8.png?alt=media&token=05e0b540-c5ae-499f-bfb1-3adb458a9c8f
Our general strategy is going to be manipulating these three reactions and then adding them together to form the overall reaction. This is using Hess's Law to find the enthalpy.

Thought Process

When I approach a Hess's Law question, I always find the reactants and products in the given thermochemical equations. This is what I would noteπŸ€”:
  • C(s) is on the reactants side like it is supposed to be, but there should be 2 atoms of carbon so we have to manipulate the second equation.❌
  • H2 is on the reactants side like it is supposed to be, so we shouldn't manipulate the third equation.βœ”οΈ
  • C2H2 is on the reactants side when it should be on the products side, so we have to manipulate the first equation.❌

Steps to Take

Step 1: Flip reaction 1 to get H2O on the other side. If we flip the reaction, we also have to flip the sign on the enthalpyπŸ”.
Step 2: Multiply reaction 2 to get 2 C. If we multiply the reaction, we have to multiply the enthalpy by 2 tooβœ–οΈ.
Whatever you do to the equation, you have to do the same to the enthalpy.

Results of Steps

After doing these things, we are left with:
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F60aa54bd81001078c48320765eb97642.png?alt=media&token=8e0665b4-e449-4bd0-b068-ae560ccaadee
Now the equations are set up properly to create the overall equation, but what happens with the extra compounds?
You could think of these as spectator compounds; they cancel out since they are present on both sides of the equation when all 3 equations are added together.
With this being said, the O2s cancel out, H2O cancels out, and the CO2s cancel out.
πŸ‘‰We then add the reactions together to find the original reaction and that Ξ”H = +1299.5 kJ + (βˆ’787 kJ) + (βˆ’285.8 kJ) = +226.7 kJ.

Example Problem #2

Calculate the value of Ξ”Hreaction for the following reaction using the listed thermochemical equations.
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-t3VKf0hPfNlv.JPG?alt=media&token=a1ec4325-5f0e-4fef-ab59-7480bd7bc5a9

Thought Process

Let's note a few thingsπŸ€”:
  • P4O10 is on the wrong side of the equation, so we have to manipulate equation 2❌.
  • PCl5 is also on the wrong side of the equation and it doesn't have a coefficient of 6, so we have to manipulate equation 3❌.
  • Cl3PO is on the right side of the equation, but it doesn't have a coefficient of 10 so we have to manipulate equation 4❌.

Steps to Take

This problem has a lot more steps! Let's go through the manipulations we have to make:
Step 1: Flip equation 2 to get P4O10 on the other side. If we flip the equation, we also have to flip the sign on the enthalpyπŸ”.
Step 2: Flip equation 3 to get PCl5 on the other side and multiply equation 3 by 6 so PCl5 has a coefficient of 6. Doing this, we have to multiply the enthalpy by 6 and negate it (two steps in one!)βœ–οΈπŸ”.
Step 3: Multiply equation 4 by 10 so that Cl3PO has a coefficient of 10. Doing this, we also have to multiply the enthalpy by 10βœ–οΈ.

Results of Steps

After doing these things, we are left with:
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-1DRfMD2vSA7W.JPG?alt=media&token=2f09365b-e10b-4687-ae2e-caf34248b1d8
Now we have everything on the right side of the equation and they all have the proper coefficients. Our last step is to check if all the spectators cancel out, so let's count them up!
CompoundReactant SideProduct SideManipulation Necessary?
P41/41βœ”οΈ
Cl23/26βœ”οΈ
PCl3107βœ”οΈ
O255❌
Luckily, all 3 compounds that we have to balance are in equation 1, meaning we now have to manipulate it in order for the equations to add up to the overall equation.
πŸ€”What should be multiply equation 1 by? In order to get the same amount of each compound, we have to multiply equation 1 by 4 and its enthalpy by 4!
https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2F-PwR0GDCFUsxY.JPG?alt=media&token=e428934b-4428-446a-a5c2-45d7fb1feb5b
Now everything cancels out and we are left with the overall equationπŸ₯³! Last step is to add up all the enthalpies.
πŸ‘‰Ξ”H = -1225.6 kJ + 2967.3 kJ + 505.2 kJ + (βˆ’2857 kJ) = -610.1 kJ.

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