section 7.4—energy of a chemical reaction what’s happening in those hot/cold packs that contain...
TRANSCRIPT
Section 7.4—Energy of a Chemical Reaction
What’s happening in those hot/cold packs that contain chemical reactions?
Enthalpy of Reaction
Enthalpy of Reaction (Hrxn) – Net energy change during a chemical reaction
+Hrxn means energy is being added to the system—endothermic-Hrxn means energy is being released from the system—exothermic
Enthalpy of Formation
Enthalpy of Formation (Hf) – Energy change when 1 mole of a compound is formed from elemental 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!
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.
Breaking apart reactants is the opposite of Enthalpy of Formation.
Forming products is the Enthalpy of Formation.
reactHprodHH ffrxn
Hrxn = sum of Hf of all products – the sum of Hf reactants
Enthalpy of Formation & Enthalpy of Reaction
This is not the way a reaction occurs—reactants break apart and then rearrange…remember Collision Theory from Chpt 2! But for when discussing overall energy changes, this manner of thinking is acceptable.
Example
Example:Find the Hrxn for:
CH4 (g) + 2 O2 (g) 2 H2O (g) + CO2 (g)
Hf (kJ/mole)
CH4 (g) -74.81
O2 (g) 0
H2O (g) -241.8
CO2 (g) -393.5
Example
Example:Find the Hrxn for:
CH4 (g) + 2 O2 (g) 2 H2O (g) + CO2 (g)
Hf (kJ/mole)
CH4 (g) -74.81
O2 (g) 0
H2O (g) -241.8
CO2 (g) -393.5
reactHprodHH ffrxn
mole
kJmolemolekJmole
molekJmolemole
kJmoleH rxn
0281.741
5.39318.2412
Hrxn = -802.29 kJ
Let’s Practice #1
Hf (kJ/mole)
CH3OH (l) -238.7
O2 (g) 0
H2O (l) -285.8
CO2 (g) -393.5
Example:Find the Hrxn for:
2 CH3OH (l) + 3 O2 (g) 2 CO2 (g) + 4 H2O (l)
Let’s Practice #1
Hf (kJ/mole)
CH3OH (l) -238.7
O2 (g) 0
H2O (l) -285.8
CO2 (g) -393.5
Example:Find the Hrxn for:
2 CH3OH (l) + 3 O2 (g) 2 CO2 (g) + 4 H2O (l)
reactHprodHH ffrxn
Hrxn = -1453 kJ
mole
kJmolemolekJmole
molekJmolemole
kJmoleH rxn
037.2382
8.28545.3932
Enthalpy & Stoichiometry
The Enthalpy of Reaction can be used along with the molar ratio in the balanced chemical equation
This allows Enthalpy of Reaction to be used in stoichiometry equalities
Example:If 1275 kJ is released, how many grams of B2O3 is
produced?B2H6 (g) + 3 O2 (g) B2O3 (s) + 2 H2O (g)
H = -2035 kJ
Example
Example
-1275 kJ
kJ
mole B2O3 = ________ g B2O3
-2035
143.62
mole B2O3
g B2O3
1
69.62
H = -1275 kJ (negative because it’s “released”)From balanced equation: -2035 kJ = 1 mole B2O3
Molar mass: 1 mole B2O3 = 69.62 g B2O3
Example:If 1275 kJ is released, how many grams of B2O3 is
produced?B2H6 (g) + 3 O2 (g) B2O3 (s) + 2 H2O (g)
H = -2035 kJ
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) H = -2035 kJ
Let’s Practice #2
47.8 g B2O3
g B2O3
mole B2O3 = ________ kJ
69.92
1-1397
mole B2O3
kJ
1
-2035
From balanced equation: -2035 kJ = 1 mole B2O3
Molar mass: 1 mole B2O3 = 69.62 g B2O3
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) H = -2035 kJ