1. Section 7.4—Energy of a Chemical Reaction
What’s happening in those hot/cold packs that contain chemical reactions?

2. Enthalpy of Reaction
Enthalpy of Reaction (DHrxn) – Net energy change during a chemical reaction
+DHrxn means energy is being added to the system—endothermic
-DHrxn means energy is being released from the system—exothermic

3. Enthalpy of Formation
Enthalpy of Formation (Hf) – Energy change when 1 mole of a compound is formed from elements in their standard states
Heat of formation equations:
H2 (g) + ½ O2 (g)  H2O (g)
C (s) + O2 (g)  CO2 (g)
A table with Enthalpy of Formation values can be found in the Appendix of your text
Be sure to look up the correct state of matter:
H2O (g) and H2O (l) have different Hf values!

4. Enthalpy of Formation & Enthalpy of Reaction
Breaking apart reactants is the opposite of Enthalpy of Formation.
Forming products is the Enthalpy of Formation.
The overall enthalpy of reaction is the opposite of Hf for the reactants and the Hf for the products
Reactants are broken apart and Products are formed.
DHrxn = sum of Hf of all products – the sum of Hf reactants

5. CH4 (g) + 2 O2 (g)  2 H2O (g) + CO2 (g)
Example
Example:
Find the DHrxn for:
CH4 (g) + 2 O2 (g)  2 H2O (g) + CO2 (g)
Hf (kJ/mole)
CH4 (g)
-74.81
O2 (g)
H2O (g)
-241.8
CO2 (g)
-393.5

6. CH4 (g) + 2 O2 (g)  2 H2O (g) + CO2 (g)
Example
Example:
Find the DHrxn for:
CH4 (g) + 2 O2 (g)  2 H2O (g) + CO2 (g)
Hf (kJ/mole)
CH4 (g)
-74.81
O2 (g)
H2O (g)
-241.8
CO2 (g)
-393.5
DHrxn = kJ

7. 2 CH3OH (l) + 3 O2 (g)  2 CO2 (g) + 4 H2O (l)
Let’s Practice #1
Example:
Find the DHrxn for:
2 CH3OH (l) + 3 O2 (g)  2 CO2 (g) + 4 H2O (l)
Hf (kJ/mole)
CH3OH (l)
-238.7
O2 (g)
H2O (l)
-285.8
CO2 (g)
-393.5

8. 2 CH3OH (l) + 3 O2 (g)  2 CO2 (g) + 4 H2O (l)
Let’s Practice #1
Example:
Find the DHrxn for:
2 CH3OH (l) + 3 O2 (g)  2 CO2 (g) + 4 H2O (l)
Hf (kJ/mole)
CH3OH (l)
-238.7
O2 (g)
H2O (l)
-285.8
CO2 (g)
-393.5
DHrxn = kJ

9. Enthalpy & the Balanced Equation
The Enthalpy of Reaction can be used along with the balanced chemical equation to solve for heat or masses.
The coefficients in front of the substance in a balanced equation can be called a mole amount for that substance. So for ex. 2H2 + 1O2  2H2O This can be read as 2 moles of hydrogen needs 1 moles of oxygen to create 2 moles of water. We will need these mole values to solve for enthalpy!
Steps for Solving for Mass
Write down the heat given
Use the balanced equation to go from the heat given to the moles of the substance you need to find.
Convert from moles to grams.
Steps for Solving for Heat
Write down the mass given
Convert the mass to moles
Use the balanced equation to go from moles to the heat from the reaction.

10. Example
Example:
If 1275 kJ is released, how many grams of B2O3 is produced?
B2H6 (g) + 3 O2 (g)  B2O3 (s) + 3 H2O (g)
DH = kJ

11. Example
Example:
If 1275 kJ is released, how many grams of B2O3 is produced?
B2H6 (g) + 3 O2 (g)  B2O3 (s) + 3 H2O (g)
DH = kJ
DH = kJ (negative because it’s “released”)
From balanced equation: kJ = 1 mole B2O3
Molar mass: 1 mole B2O3 = g B2O3
-1275 kJ 1
mole B2O3
69.62
g B2O3
= ________ g B2O3
43.62
-2035 kJ 1
mole B2O3

12. B2H6 (g) + 3 O2 (g)  B2O3 (s) + 3 H2O (g)
Let’s Practice #2
If you need to produce 47.8 g B2O3, how many kilojoules will be released?
B2H6 (g) + 3 O2 (g)  B2O3 (s) + 3 H2O (g)
DH = kJ

13. B2H6 (g) + 3 O2 (g)  B2O3 (s) + 2 H2O (g)
Let’s Practice #2
If you need to produce 47.8 g B2O3, how many kilojoules will be released?
B2H6 (g) + 3 O2 (g)  B2O3 (s) + 2 H2O (g)
DH = kJ
From balanced equation: kJ = 1 mole B2O3
Molar mass: 1 mole B2O3 = g B2O3
47.8 g B2O3 1
mole B2O3
-2035 kJ
= ________ kJ
-1397
69.92
g B2O3 1
mole B2O3