Transcript Chemistry You Need to Know

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 (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
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!
Enthalpy of Formation & Enthalpy of Reaction
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.
The overall
enthalpy of
reaction is the
opposite of Hf
for the reactants
and the Hf for
the products
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.
DHrxn   H f prod   H f react
DHrxn = sum of Hf of all products – the sum of Hf reactants
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)
0
H2O (g)
-241.8
CO2 (g)
-393.5
Example
Example:
Find the DHrxn for:
CH4 (g) + 2 O2 (g)  2 H2O (g) + CO2 (g)
Hf (kJ/mole)
DHrxn   H f prod   H f react

CH4 (g)
-74.81
O2 (g)
0
H2O (g)
-241.8
CO2 (g)
-393.5


 1m ole 393.5 kJ
m ole
m ole
 2m ole 0 kJ m ole
 1m ole 74.81kJ
m ole
DH rxn  2m ole 241.8 kJ
DHrxn = -802.29 kJ
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)
0
H2O (l)
-285.8
CO2 (g)
-393.5
Let’s Practice #1
Example:
Find the DHrxn for:
2 CH3OH (l) + 3 O2 (g)  2 CO2 (g) + 4 H2O (l)
Hf (kJ/mole)
DHrxn   H f prod   H f react

CH3OH (l)
-238.7
O2 (g)
0
H2O (l)
-285.8
CO2 (g)
-393.5



 4m ole 285.8 kJ
m ole
m ole


 2m ole 238.7 kJ
 3m ole 0 kJ
m ole
m ole
DH rxn  2m ole 393.5 kJ
DHrxn = -1453 kJ
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
Example:
If 1275 kJ is released, how many grams of B2O3 is
produced?
B2H6 (g) + 3 O2 (g)  B2O3 (s) + 2 H2O (g)
DH = -2035 kJ
Example
Example:
If 1275 kJ is released, how many grams of B2O3 is
produced?
B2H6 (g) + 3 O2 (g)  B2O3 (s) + 2 H2O (g)
DH = -2035 kJ
DH = -1275 kJ (negative because it’s “released”)
From balanced equation: -2035 kJ = 1 mole B2O3
Molar mass: 1 mole B2O3 = 69.62 g B2O3
-1275 kJ
1
mole B2O3
-2035
kJ
69.62
1
g B2O3
mole B2O3
= ________
43.62 g B2O3
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 = -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)
DH = -2035 kJ
From balanced equation: -2035 kJ = 1 mole B2O3
Molar mass: 1 mole B2O3 = 69.62 g B2O3
47.8 g B2O3
1
mole B2O3
69.92
g B2O3
-2035
1
kJ
mole B2O3
= ________
-1397 kJ