5.2 HESSS LAW

```TOPIC 5
ENERGETICS/THERMOCHEMISTRY
5.2
HESS’S LAW
By: Merinda Sautel
Alameda Int’l Jr/Sr High School
Lakewood, CO
msautel@jeffco.k12.co.us
ESSENTIAL IDEA
In chemical transformations energy
can neither be created nor destroyed
(the first law of thermodynamics).
NATURE OF SCIENCE (2.4)
Hypotheses – based on the conservation of
energy and atomic theory, scientists can
test the hypothesis that if the same
products are formed from the same initial
reactants then the energy change should be
the same regardless of the number of steps.
INTERNATIONALMINDEDNESS
Recycling of materials is often an
effective means of reducing the
environmental impact of production, but
varies in its efficiency in energy terms in
different countries.
THEORY OF KNOWLEDGE
Hess’s Law is an example of the
application of the Conservation of
Energy. What are the challenges and
limitations of applying general
principles to specific instances?
UNDERSTANDING/KEY IDEA
5.2.A
The enthalpy change for a
reaction that is carried out in a
series of steps is equal to the
sum of the enthalpy changes
for the individual steps.
APPLICATION/SKILLS
Be able to apply Hess’s law to
calculate enthalpy changes.
APPLICATION/SKILLS
Be able to determine the
enthalpy change of a reaction
that is the sum of multiple
reactions with known enthalpy
changes.
• Hess’s law states that the
enthalpy change for any
chemical reaction is
independent of the route,
provided the starting conditions
and final conditions, and
reactants and products, are the
same.
• Hess’s Law is designed so that
you will manipulate a series of
reactions (called step equations)
to reach the required reaction
which is most always given.
1. If you reverse a reaction,
reverse the sign of ΔH.
2. If you multiply or divide a step
equation by an integer, you must
also multiply or divide your
ΔH by the same factor.
• Hess’s law is a statement of the
Law of Conservation of Energy.
• It is used to measure enthalpy
changes which cannot be
measured directly in the
laboratory.
• The standard enthalpy of formation
(ΔHf&ordm; ) is defined as the change in
enthalpy that accompanies the
formation of one mole of a compound
from its elements with all substances
in their standard states.
• The degree symbol means that the
process was carried out under
standard conditions.
Example Problem
• Consider the standard enthalpy of
formation of methane.
C(s) + 2H2(g)
CH4(g)
• Notice that one mole of methane was
formed from its elements in their
standard states.
• This looks like a simple enough
equation where we could react carbon
with hydrogen to form methane and
determine the heat change.
• However, it is difficult to react
carbon with hydrogen and practically
impossible to do so experimentally.
• We can easily burn carbon, hydrogen
and methane to determine the
standard enthalpy of combustion
ΔHc&ordm; directly for each of the three
substances.
• This would give us our step
equations with enthalpy values that
we could manipulate to determine
the ΔH for the ΔHf&ordm; for methane.
• C(s) + O2(g)
CO2(g)
ΔHc&ordm; = -393 kJ mol-1
• H2(g) + 1/2 O2(g)
H2O(l)
ΔHc&ordm; = -286 kJ mol-1
• CH4(g) + 2O2(g)
CO2(g) + 2H2O(l) ΔHc&ordm; = -890 kJ mol-1
• The main equation again is
C(s) + 2H2(g)
CH4(g)
• I need CH4 on the right and 2 – H2’s on the left.
• I will have to reverse the 3rd equation so I change
the sign of ΔH.
• I will have to double the 2nd equation because I
need 2H2’s in the final equation so I will multiply
the ΔH by 2.
• C(s) + O2(g)
CO2(g)
ΔHc&ordm; = -393 kJ mol-1
• 2H2(g) + 2/2 O2(g)
2H2O(l)
ΔHc&ordm; = 2(-286) kJ mol-1
• CO2(g) + 2H2O(l) CH4(g) + 2O2(g) ΔHc&ordm; = +890 kJ mol-1
• Now cross out anything that is the same on both sides.
• C(s) + O2(g)
CO2(g)
ΔHc&ordm; = -393 kJ mol-1
• 2H2(g) + 2/2 O2(g)
2H2O(l)
ΔHc&ordm; = 2(-286) kJ mol-1
• CO2(g) + 2H2O(l) CH4(g) + 2O2(g) ΔHc&ordm; = +890 kJ mol-1
• Adding the equations will now give us the main equation.
•
C(s) + 2H2(g)
CH4(g)
• ΔHf&ordm; = +890 kJ mol-1 + 2(-286) kJ mol-1 + -393 kJ mol-1
= -75 kJ mol-1
APPLICATION/SKILLS
Be able to calculate the ΔH
reactions using ΔHf◦ data.
GUIDANCE
Enthalpy of formation data can
be found in the data booklet in
section 12.
GUIDANCE
Be able to use the following
equation:
ΔH reaction = Σ(ΔHf ◦products)
- Σ(ΔHf ◦reactants)
Calculating ΔHf
⁰
ΔHf⁰ = ∑nΔHf ⁰(products) - ∑nΔHf ⁰(reactants)
• You can calculate the standard heat of
reaction (ΔH⁰) for any reaction using the
above equation.
• You will be given values in a table, plug
them into the above equation and
multiply these values by the number of
moles in the equation. Then solve.
• The standard enthalpy of
formation (ΔHf⁰) for an element
is zero.
• There is no chemical change
and so no enthalpy change
when an element is formed from
itself.
Sample Problem
• Calculate the enthalpy change for
the following reaction given the
standard enthalpy changes of
formation.
• C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(l)
• ΔHf⁰ of C3H8(g) = -104 kJ mol-1
• ΔHf⁰ of CO2(g) = -394 kJ mol-1
• ΔHf⁰ of H2O(l) = -286 kJ mol-1
• C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(l)
• -104
0
3(-394)
4(-286)
•
• ΔHf⁰ = ∑nΔHf ⁰(products) - ∑nΔHf ⁰(reactants)
•
•
= [3(-394) + 4(-286)] – [(-104) + (0)]
= -2220 kJ mol-1
Citations
International Baccalaureate Organization. Chemistry
Guide, First assessment 2016. Updated 2015.
Brown, Catrin, and Mike Ford. Higher Level Chemistry.
2nd ed. N.p.: Pearson Baccalaureate, 2014. Print.
ISBN 978 1 447 95975 5
eBook 978 1 447 95976 2
Most of the information found in this power point
comes directly from this textbook.
The power point has been made to directly
complement the Higher Level Chemistry textbook by
Brown and Ford and is used for direct instructional
purposes only.
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