Transcript Energetics - Chemistry Resources for IB, AP, Alevel, GCSE

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Chemical reactions involves the breaking and the
making of bonds.
Energy is needed to break down a bond.
Energy is released when a bond is formed.
If more energy is released than is absorbed, the
reaction will be Exothermic.
If more energy is needed to break the bonds than
is given when new bonds are formed, the reaction
will be endothermic.
Energy
level
Activation
energy
Using a catalyst
might lower the
activation energy
Energy given
out by
reaction
Reaction progress
Enthalpy
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The energy contained in a chemical
bond that can be converted into heat
is known as enthalpy.
Enthalpy is given the symbol H.
Enthalpy can not be measured
directly, but we can measure the
enthalpy change in a reaction, ΔH
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A process is
endothermic
when H is
positive.
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A process is
endothermic
when H is
positive.
A process is
exothermic when
H is negative.
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EXOTHERMIC
– more energy
is given out
than is taken
in (burning,
respiration)
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ENDOTHERMIC –
energy is taken
in but not
necessarily given
out.
(photosynthesis)
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H changes sign when a process is reversed.
Therefore, a cyclic process has the value H = 0.
Same magnitude; different signs.
Chapter 6: Thermochemistry
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Using H
Values are measured experimentally
Negative values indicate exothermic reactions
Positive values indicate endothermic reactions
Changes sign when a process is reversed. Therefore, a
cyclic process has the value H = 0
For problem-solving, one can view heat
being absorbed in an endothermic reaction
as being like a reactant and heat being
evolved in an exothermic reaction as being
like a product
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We measure heat flow using calorimetry.
A calorimeter is a device used to make
this measurement.
A “coffee cup” calorimeter may be used
for measuring heat involving solutions.
A “bomb” calorimeter is used
to find heat of combustion; the
“bomb” contains oxygen and a
sample of the material to be
burned.
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Specific heat of a substance is the amount of
heat required to raise the temperature of one
gram by 1 o C or by 1 Kelvin.
Specific heat = C = q/mT
units of C: J g–1 oC–1 or J g–1 K–1
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EOS
Chapter 6: Thermochemistry
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The standard enthalpy of reaction (Ho) is the
enthalpy change for a reaction in which the
reactants in their standard states yield
products in their standard states
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The standard enthalpy of formation (Hof) of
a substance is the enthalpy change that
occurs in the formation of 1 mol of the
substance from its elements when both
products and reactants are in their standard
states
•Is the enthalpy change that occurs during
the complete combustion of one mole of a
substance.
•The enthalpy of combustion is defined in
terms one mole of reactants, where the
enthalpy of formation is defined in terms of
one mole of products.
•the symbol Ho C is used to represent
standard enthalpy change os combustion.
EOS
Chapter 6: Thermochemistry
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The heat of a reaction is constant,
regardless of the number of steps in the
process
Hoverall = S H’s of individual reactions
When it is necessary to reverse a chemical
equation, change the sign of H for that
reaction
When multiplying equation coefficients,
multiply values of H for that reaction
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H is well known for many
reactions, and it is inconvenient to
measure H for every reaction in
which we are interested.
However, we can estimate H
using H values that are published
and the properties of enthalpy.
The change in
enthalpy, H, is the
enthalpy of the
products minus the
enthalpy of the
reactants:
H = Hproducts − Hreactants
Hess’s law states
that “If a reaction is
carried out in a
series of steps, H
for the overall
reaction will be
equal to the sum of
the enthalpy
changes for the
individual steps.”
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O
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Imagine this as occurring
in 3 steps:
C3H8 (g)  3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
(l
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O
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The sum of these
equations is:
C3H8 (g)  3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
(l
We can use Hess’s law in this way:
H = S n Hf(products) - S m Hf(reactants)
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where n and m are the stoichiometric coefficients.
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O
H =
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=
[3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)]
[(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)]
(-2323.7 kJ) - (-103.85 kJ)
-2219.9 kJ
(l