Explanation:
1. 4Al + 3MnO₂ → 2Al₂O₃ + 3Mn
The reactants are Al and MnO₂. The products are Al₂O₃ and Mn.
We need to flip reaction b, triple it, and add it to reaction a to get the above reaction.
a: 4Al + 3O₂ → 2Al₂O₃, ΔH = -3352 kJ
b: Mn + O₂ → MnO₂, ΔH = -521 kJ
-3b: 3MnO₂ → 3Mn + 3O₂, ΔH = 1563 kJ
a−3b: 4Al + 3MnO₂ → 2Al₂O₃ + 3Mn, ΔH = -1789 kJ
2. Use the same logic as question 1.
a + 2b + c
ΔH = -135.4 kJ + 2(-184.6 kJ) + 74.8 kJ
ΔH = -429.8 kJ
3. -a + b
ΔH = -(-296.8 kJ) + -198.7 kJ
ΔH = 98.1 kJ
4. 2(1) + (2) − (3)
ΔH = 2(131.3 kJ) + -41.2 kJ − 206.1 kJ
ΔH = 15.3 kJ
5. The heat of formation is the heat absorbed/released during the production of a compound from its elements.
So if a compound is on the product side (the right side of the arrow), use the heat of formation from the table.
If a compound is on the reactant side (the left side of the arrow), multiply the heat of formation by -1.
Don't forget to multiply by the coefficient.
Compounds with only one element have no heat of formation.
a. ΔH = -635.09 kJ + -393.509 kJ − (-1206.9 kJ)
ΔH = 178.301 kJ
b. ΔH = -167.2 kJ − (-74.9 kJ)
ΔH = -92.3 kJ
c. ΔH = 4(33.2 kJ) + 6(-285.8 kJ) − 4(-45.90 kJ)
ΔH = -1398.4 kJ
d. ΔH = -858.6 kJ − 2(-167.2 kJ)
ΔH = -524.2 kJ
e. ΔH = 2(33.2 kJ) − 2(90.29 kJ)
ΔH = -114.18 kJ
