4/20/14
ENTHALPY
p  Enthalpy
= a type of chemical energy
(thermodynamic potential), sometimes
referred to as “heat content”
Thermochemistry
Part 3 - Enthalpy and
Thermochemical Equations
Enthalpies of Reaction
Enthalpies of Reaction
p  The
enthalpy change (in kJ/mol) that
accompanies a chemical reaction is called
the enthalpy of reaction
n 
Δ Hrxn
n 
Also called heat of reaction
p  If
n 
Δ Hrxn = negative
exothermic
p  heat is given off
p  Under conditions of constant pressure,
q = ΔH < 0 (negative sign)
R
H
P
time
Enthalpies of Reaction
p  If
n 
Thermochemical Equations
p  Thermochemical
equations are balanced
chemical equations that show the
associated enthalpy change (ΔH)
Δ Hrxn = positive
endothermic
p  heat must be added
p  Under conditions of constant pressure,
q = ΔH > 0 (positive sign)
H
n 
n 
balanced equation
enthalpy change (ΔHrxn)
P
R
time
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Thermochemical Equations
p 
An example of a thermochemical equation:
CH4(g) + 2O2(g)
p 
CO2(g) + 2H2O(l) ΔH = -890. kJ
The coefficients in the balanced equation show the
# moles of reactants and products that produced
the associated ΔH.
n 
If the number of moles of reactant used or product formed
changes, then the ΔH will change as well.
Thermochemical Equations
p  The
thermochemical equation for burning
1 mole of CH4 (g):
CH4(g) + 2O2(g)
p 
For the following reaction:
CH4
(g)
+ 2O2
(g)
CO2
(g)
+ 2H2O
(l)
-890. kJ
1 mol CH4
-890. kJ
2 mol O2
-890. kJ
1 mol CO2
-890. kJ
2 mol H2O
ΔH = -890. kJ
Rules of Thermochemistry
Rule #1 - The magnitude of ΔH is directly
proportional to the amount of reactant
consumed and product produced.
CO2(g) + 2H2O(l) ΔH = -890. kJ
n 
When 1 mole of CH4 is burned, 890. kJ of heat
are released.
n 
When 2 moles of CH4 are burned, 1780. kJ of
heat are released.
Rules of Thermochemistry
Example 1:
H2 + Cl2 → 2HCl ΔH = - 185 kJ
Calculate ΔH when 1.00 g of Cl2 reacts.
This thermochemical equation says that there is
185 kJ of energy released per how many moles
of Cl2? (look at the balanced equation)
For every 1 mole (coefficient) Cl2 = -185 kJ (released)
ΔH = 1.00 g Cl2 x
Thermochemical Equations
Rules of Thermochemistry
Example 2: When an ice cube weighing
24.6 g of ice melts, it absorbs 8.19 kJ of
heat. Calculate ΔH when 1.00 mol of solid
water melts.
ΔH = 1.00 mol x 18.02 g x 8.19 kJ = 6.00 kJ
1 mol
24.6 g
1 mol Cl2 x -185 kJ = -2.61 kJ
1 mol
70.90 g Cl2
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Rules of Thermochemistry
Example 3: Methanol burns to produce
carbon dioxide and water:
2CH3OH + 3O2 → 2CO2 + 4H2O + 1454 kJ
What mass of methanol is needed to
produce 1820 kJ? Does “produce” mean exo or endo? + or – sign?
Rules of Thermochemistry
Example 4: How much heat is produced
when 58.0 liters of hydrogen (at STP) are
also produced?
Zn + 2HCl → ZnCl2 + H2 + 1250 kJ
​−1250 𝑘𝐽/22.4 𝐿 𝐻2 𝐻
=2 ​𝑥/58.0 ​−1454 𝑘𝐽/2 𝑚𝑜𝑙 𝐶𝐻
​−1820 3𝑂𝐻 𝑘𝐽/𝑛
= 𝑚𝑜𝑙 𝐶𝐻3𝑂𝐻 𝑛=2.5034 𝑚𝑜𝑙
𝑥=−𝟑𝟐𝟒𝟎 𝒌𝑱
×​32.05 𝑔/1 𝑚𝑜𝑙 𝐶𝐻3𝑂𝐻 =
80.2 𝒈
Rules of Thermochemistry
Rule #2 - ΔH for a reaction is equal in
the magnitude but opposite in sign to
ΔH for the reverse reaction.
Rules of Thermochemistry
2 H2O2 (l)
2 H2O (l) + O2(g)
2 H2O (l) + O2(g)
2 H2O2 (l)
ΔH = -196 kJ
ΔH = +196 kJ
(If 6.00 kJ of heat absorbed when a mole of ice
melts, then 6.00 kJ of heat is given off when 1.00
mol of liquid water freezes)
Remember
“elephant’s
toothpaste?”
Rules of Thermochemistry
Example:
Given:
H2 + ½O2 → H2O ΔH = -285.8 kJ
Flip eq’n and x 2
…so do the same to ∆H
Calculate ΔH for the equation:
2H2O → 2H2 + O2
ΔH = 2 x –(-285.8 kJ)
Rules of Thermochemistry
Rule #3) The value of ΔH for a reaction
is the same whether it occurs in one step
or in a series of steps.
ΔH for the overall equation is the sum of
the ΔH’s for the individual equations:
Hess’s Law: ΔH = ΔH1 + ΔH2 + …
ΔH = 571.6 kJ
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Hess’s Law: ΔH = ΔH1 + ΔH2 + …
Hess’s Law: ΔH = ΔH1 + ΔH2 + …
Example 1:
Calculate ΔH for the reaction:
C + ½O2 → CO
÷2
flip
Given:
1) C + ​𝟏O2 → CO2
2) 2CO/𝟐 + O2 → 2CO2
2) CO2 → CO +
+283.0 kJ
C + ½O2 → CO
​1/2 O2
Example 2: Find the heat of reaction
(enthalpy) for the following reaction
NO + ½O2 → NO2
ΔH = ?
ΔH = -393.5 kJ
ΔH = -566.0 kJ
ΔH =
ΔH = -110.5 kJ
flip
Given the following equation….
½N2 + ½O2 → NO ΔH = +90.4 kJ
½N2 + ​𝟏O2 → NO2
ΔH = +33.6
/
NO → ½N
2 + ½O2
𝟐
ΔH = -90.4 kJ
NO + ½O
2 → NO2 ΔH = -56.8 kJ
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