1. Enthalpy and Hess’s Law

2. Enthalpy – Greek for “to warm”
Enthalpy is defined as the change in heat energy in a reaction.
Knowing the enthalpy of a reaction we can determine how much energy changes, as well as if it is endothermic or exothermic.
Similar to Q in the specific heat equation.
Change in Enthalpy is DH (delta H)

3. How did you get to Grandma’s House?
Start
Finish
Both lines accomplished the same result, they went from start to finish. Net result = same.
A State Function: The pathways in a chemical reaction are different, but the amount of energy required to do remain the same. .

4. State Functions Enthalpy is a state function.
This means it doesn’t matter how you arrived at the products (by whatever chemical reaction or reactions), the value will be the same.
It doesn’t matter if the reaction is one step, or several steps, the answer (Enthalpy) will be the same.

5. Enthalpy Graphs If energy is released, the reaction is exothermic.
That means that the heat energy of the products is less than the reactants.
We end up with less than what we started.
DH (Enthalpy) is negative

6. Enthalpy Graphs If energy is absorbed, the reaction is endothermic.
That means that the heat energy of the products is greater than the reactants.
We end with more energy than we started with.
DH is positive

7. H = Hproducts − Hreactants
Calculating Enthalpy
As a chemical reaction occurs, there is a change in heat energy as the reactants become the products.
To determine this value (H) you subtract the enthalpy of the products minus reactants.
H = Hproducts − Hreactants
2 H2 + O2  2 H2O DH = kJ
What can we learn from this reaction?

8. Calculating Enthalpy 2 H2(g) + O2(g)  2 H2O(l) DH = -483.6 kJ
This reaction is highly exothermic. DH is a negative value meaning heat is released.
Stoichiomety is important here. This is the amount of energy when 2.0 moles of Hydrogen gas react.
What if only 1.0 mole of H2 gas was used? How much energy would be released?

9. Practice Problems
1) Calculate the enthalpy of the reaction when g of methane gas is burned in this reaction.
CH O2  CO H2O DH = -890 kJ
If the above reaction was followed, and -2,225 kJ of energy was released, how many grams of water would be produced?

10. Standard Heat of Formation
The Heat of Formation is the change in energy that happens when 1.00 mol of a compound is formed.
Elements in their natural form, how they exist normally, are given a value of 0.0 kJ/mol. They are not “formed” from anything.
Any other compound you will need some help from your chart. (Pass out Charts!)

11. 2 KCl + 3 O2  2 KClO3 Practice Problem #1 H = Hproducts − Hreactants
Determine the heat of formation for the following:
2 KCl + 3 O2  2 KClO3
H = Hproducts − Hreactants
H = 2(-391.2) – (2(-436.5) + 3(0))
H = ( ) – (- 873)
H = 90.6 kJ/mol *What does a positive value mean?

12. Practice Problem #2
Translate and write this equation: Calcium carbonate decomposes at high temp. to form carbon dioxide and calcium oxide:
Then try to determine the change in enthalpy for the reaction.

13. H = Hproducts − Hreactants
CaCO3  CO2 + CaO
Determine the Heat of formation for each compound in the formula. Use your charts!
* The values from the chart, not solving yet!*
(-1207 kJ/mol)  (-393.5) + (-635.5) CaCO CO CaO
H = Hproducts − Hreactants
(-1029 ) – (-1207) = 178 kJ/mol

14. Worksheet Practice

15. A Few Important Guidelines
1) Enthalpy is a proportional property. Double the molar amount, and it will double the amount of heat energy.
2) The enthalpy change for the reverse reactions is equal in magnitude, but opposite in sign H2(g) + O2(g)  2 H2O(l) DH = kJ H2O(l)  2 H2(g) + O2(g) DH = kJ
3) Enthalpy depends on the state of matter. All states are not equal H2O(l)  2 H2O(g) DH = +88 kJ

16. Hess’s Law
Since Enthalpy is a state function, we can solve for the heat energy through many different reactions. Regardless of any intermediates, the total energy will the same for compounds formed.
Lets say we want to determine the amount of heat energy exchanged in the production of Nitrogen Dioxide?
Well, can’t we just look at our tables and find the DH for NO2? Sadly No…..

17. The problem is that NO2 is not created directly from its elements.
The table is useful, but only when we talk about molecules and compounds created directly from their elements.
NO(g) + ½ O(g)  NO2(g)
Wait…is that a ½ coefficient? But that can’t happen right….

18. ½ Coefficients?
Yes and no. If you are speaking in terms of ½ a molecule, then no. You can’t have half a molecule just like a chicken can’t lay half an egg.
However, if you are speaking in terms of moles, then you can have half a mole, just like you order half a dozen doughnuts.
We are going to think in this context.

19. Back on track
NO(g) + ½ O2(g)  NO2(g) H = ?
I don’t know the enthalpy for the above reaction, but I do know the following: ½ N2 + ½ O2  NO H = kJ ½ N2 + O2  NO2 H = 33.2 kJ
Those enthalpy values came from your sheet. I can use them because their products are made up of their natural elements.
I need to manipulate these two equations so that I can get the original equation.

20. Helpful Hint: Focus on Compounds!
NO(g) + ½ O2(g)  NO2(g) H = ?
½ N2 + ½ O2  NO H = kJ ½ N2 + O2  NO H = 33.2 kJ
Looking at the original equation, I need NO(g) to be a reactant, but in the given equations it is on the product side.
BUT, I know I can reverse the reaction. And when the reaction is reversed…

21. Flip the Equation, Flip the Sign
NO(g) + ½ O2(g)  NO2(g) H = ?
NO  ½ N2 + ½ O H = kJ ½ N2 + O2  NO H = 33.2 kJ
The sign has to flip, and becomes a negative.
NO2(g) is on the proper side because in the original equation, it needs to be a product.
Let’s add up our manipulated equations and see it equals the original equation.

22. Helpful Hint: Focus on Compounds!
NO + ½ N2 + O2  ½ N2 + ½ O2 + NO2
Cross out things that will cancel out….
Combine like terms…..
And we have the desired equation. Now what? NO(g) + ½ O(g)  NO2(g)

23. The answer!
Now that you have the correct set up of the equations, simply add the enthalpy change for each equation together. Remember that one of them was flipped.
NO  ½ N2 + ½ O2 H = kJ ½ N2 + O2  NO2 H = 33.2 kJ
NO(g) + ½ O2(g)  NO2(g) H = kJ
This process is called Hess’s Law! This is how we calculate enthalpy for compounds not created directly from elements.

24. N2(g)+2 O2(g) 2 NO2(g) H = - 66.4 kJ
Using the given equations, determine the enthalpy change for the following:
2 NO2(g) N2O4(g) H = ?
Given:
N2(g)+2 O2(g) 2 NO2(g) H = kJ
N2(g)+2 O2(g)  N2O4(g) H = kJ
To get this equation we need to manipulate the above equations. What needs to happen?

25. 2 NO2(g) N2O4(g) H = 57.24 kJ 2 NO2(g)  N2(g)+2 O2(g) H = 66.4 kJ
Flip the equation, flip the sign:
2 NO2(g)  N2(g)+2 O2(g) H = 66.4 kJ
N2(g)+2 O2(g)  N2O4(g) H = kJ
2 NO2(g) + N2(g)+ 2 O2(g)  N2(g)+2 O2(g) + N2O4(g)
After canceling, you have the original equation. Simply add the manipulated enthalpies.
2 NO2(g) N2O4(g) H = kJ

26. Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: N2H4(l)  +  H2(g)  →  2NH3(g) ΔH = ?
N2H4(l) + CH4O(l) → CH2O(g) + N2(g) + 3H2(g)  ΔH = -37 kJ N2(g)  +  3H2(g)  →  2NH 3(g)                                     ΔH = -46 kJ CH4O(l)  →  CH2O(g) +  H 2(g)                                   ΔH = -65 kJ

27. N2H4(l) + H2(g) → 2NH3(g) ΔH = -18 kJ
N2H4(l) + CH4O(l) → CH2O(g) + N2(g) + 3H2(g)  ΔH = -37 kJ N2(g)  +  3H2(g)  →  2NH 3(g)                                     ΔH = -46 kJ CH2O(g) +  H 2(g) →  CH4O(l)                             ΔH = 65 kJ
N2H4(l)  +  H2(g)  →  2NH3(g) ΔH = -18 kJ

28. You want to do a really hard one…
4 NH3(g) + 5 O2(g)  4 NO(g) H2O(g)
N2(g) + O2(g)  2NO(g) H = kJ
N2(g) + 3H2(g)  2NH3(g) H = kJ
2H2(g) + O2(g)  2H2O(g) H = kJ