Thermochemistry Podcast #2
Heats of Reaction
Heats of Formation
Hess’ Law
Heat of Solution
When a substance dissolves in water energy can be
absorbed or released.
This is quantity of energy is called the molar heat of
solution and is represented by __________
Energetically, solution formation is like a chemical
reaction.
Sample Problem #1
What is the heat change when 80 g of NaOH
dissolves in water? (DHsol=-44.51 kJ/mol)
From the data presented below, determine the heat of
solution for bengiovannium (molar mass = 217 g/mol).
Heat of Formation
DHf is the energy change during formation of
the substance from its elements. Heats of
formation are usually listed on a table.
(pg 975 in your book)
DHf for elements = ___________
Heat of Reaction
DHr can be determined using this equation:
A __________ DHr indicates an ___________ reaction
A __________ DHr indicates an ___________ reaction
Hess’ Law
Hess's Law states that "the change in enthalpy for any
chemical reaction is constant, whether the reaction
occurs in one step or in several steps."
Thermochemical data, therefore, may be treated
algebraically.
Heat of Reaction Sample Problem
H2S(g) 4F2(g)  2HF(g) + SF6(g)
Determine the DHr for the reaction above.
DHf(H2S) = -21 kJ/mol
DHf(HF) = -273 kJ/mol
DHf(SF6) = -1220 kJ/mol.
Pg 572
The enthalpy change for the combustion of methane can be evaluated
for the path that decomposes the reactants into elements and then
recombines those elements to form the products.
A state function does not depend on the path you choose to go from the initial state to the
final conditions.
Hess’ Law
Allows us to measure the heat of reaction indirectly
by adding reactions together that have a known DHr
Hess's law of heat summation
the law that states that if you add two or more
thermochemical equations to give a final equation, then
you also add the heats of reaction to give the final heat
of reaction
Hess’ Law Example
C + 1/2 O2  CO
DHr° = -110 kJ/mol
CO + 1/2 O2  CO2
DHr° = -283 kJ/mol
C + O2
 CO2
DHr° =
=
Algebra and Hess’ Law
Whatever is done to one side of an equation, must be
done to the other.
If a chemical equation is doubled, tripled, etc. . . then the
enthalpy of that reaction is doubled, tripled, etc. . .
If a chemical equation is reversed, then the enthalpy is
reversed by changing the sign from negative to positive
(or vice versa)
Hess’ Law Example 2
• Find the heat of formation for:
• C(graphite) + 2H2(g)  CH4(g)
ΔHf = _____ kJ
• C(graphite) + O2(g)  CO2(g)
• H2(g) + ½O2(g)  H2O(l)
• CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
ΔH = -393.5 kJ
ΔH = -285.8 kJ
ΔH = -890.3 kJ
Hess’ Law Practice
Find 4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g)
using the data below.
N2 (g) + O2 (g)  2 NO (g); DHf = 180.6 kJ
N2 (g) + 3 H2 (g)  2 NH3 (g); DHf = -91.8 kJ
2 H2 (g) + O2 (g)  2 H2O (g); DHf = -483.7 kJ
More Hess’ Law Practice
Calculate ΔH for the reaction:
CS2(l) + 2H2O(l)  CO2(g) + 2H2S(g)
Given:
H2S(g) + 3/2O2(g)  H2O(l) + SO2(g)
CS2(l) + 3O2(g)  CO2(g) + 2SO2(g)
ΔH = +50.0 kJ
ΔH = ? kJ
ΔH = -562.6 kJ
ΔH = -1075.2 kJ
More Hess’ Law Problems
Calculate ΔHf for C2H2(g)
2C(s) + H2(g)  C2H2(g)
Given the following thermochemical equations:
C2H2(g) + 5/2O2(g)  2CO2(g) + H2O(l) ΔH = -1299.5 kJ
C(s) + O2(g)  CO2(g)
ΔH = -393.5 kJ
H2(g) + 1/2O2(g)  H2O(l)
ΔH = -285.8 kJ
ΔH = +226.7 kJ