Transcript Chapter 17 Reaction Energy & Kinetics

Chapter 16
Reaction Energy & Kinetics
16-1 Thermochemistry
Thermochemistry

The study of the
transfers of energy as
heat that accompany
chemical reactions
and physical changes
 This heat can be
measured in a
calorimeter
Units


Temperature units may be in Kelvin (K) or
degrees Celsius (°C)
Energy units are the joule (J) which is the
SI unit for energy. The kJ maybe used as
well.
Heat Transfer

Think of heat transfer as the difference
between two objects’ temperatures. The
energy will flow from areas of high
temperature to areas of low temperature.
Heat Capacity & Specific Heat
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Different materials can absorb energy
differently.
Specific Heat is the amount of energy
required to raise the temperature of one
gram of a substance by one degree
Celsius or Kelvin.
See page 533 Table 16-1
Heat Capacity & Specific Heat
Cp = q / (m x ∆T) or q = Cp x m x ∆T
Where:
Cp = specific heat
q = energy lost or gained
m = mass of sample
∆T = difference between initial & final
temperatures

Heat of Reaction
The quantity of heat released or absorbed
during a chemical rxn.
 When a rxn includes this information, we
call it a thermochemical equation:
2H2 (g) + O2 (g) → 2H2O (g) + 483.6 kJ
If you double the amount of reactants, you
double the amount of energy.
Sometimes fractional coefficients are used.
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Thermochemistry
Enthalpy – The heat energy of reaction
∆H
Enthalpy can only be
measured by a
change, not directly.
∆H is the amount of
energy absorbed or
lost by a rxn at
constant pressure
∆H will be
negative since
energy has left
the system
Reaction Pathways
The products
have less
energy than the
reactants. The
rxn released
energy (heat) =
exothermic
∆H will be
positive since
energy has been
added to the
system
Reaction Pathways
The products
have more
energy than the
reactants. The
rxn absorbed
energy (heat) =
endothermic
Tips

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Coefficients = moles
Physical states must be indicated as ΔH
changes for different states of matter
ΔH is directly proportional to the molar
coefficients, as they change, ΔH
changes
Temperature usually does not effect ΔH
Heat of Formation, ΔHf


The energy released or absorbed when
one mole of a substance is formed from
its elements (in terms of product)
Standard Heats of Formation, ΔHf0 are
values when the reactants and products
are at their standard states of matter, at
atmospheric pressure and room
temperature (298 K)
Heat of Formation, ΔHf
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Compounds with a high negative heat of
formation, largely exothermic, are very
stable and occur readily.
An element’s ΔHf0 = zero
Compounds with a positive, or small
negative heat of formation values are
unstable and will decompose into their
constituents.
Heat of Combustion, ΔHc


The energy released when one mole of
substance undergoes complete
combustion (in terms of reactant)
All reactants are in their standard state.
C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g)
ΔHc = -2219.2
Calculating Heats of Reactions: Hess’s Law
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The overall enthalpy change in a rxn is
equal to the sum of enthalpy changes for
the individual steps in the process.
Rules:
If a rxn is reversed, the sign on ΔH is
reversed
Multiply the coefficients, if necessary, on
rxns so the end rxn has the desired
quantities
These values
are only for the
formation of
the compound
Heat of Formation, ΔHf
NaOH + HC2H3O2  NaC2H3O2 + H2O
Na + O2 + H2  NaOH
∆Hf = - 469.15 kJ
O2 + H2 + C  HC2H3O2
∆Hf = - 484.5 kJ
Na + O2 + H2 + C  NaC2H3O2
O2 + H2  H2O
∆Hf = -285.83 kJ
∆Hf = 298.15
NaOH + HC2H3O2  NaC2H3O2 + H2O
These are not being
formed, rather they are
being used
Reverse these
signs
Na + O2 + H2  NaOH
∆Hf = - 469.15 kJ
O2 + H2 + C  HC2H3O2
∆Hf = - 484.5 kJ
Na + O2 + H2 + C  NaC2H3O2
O2 + H2  H2O
∆Hf = 298.15
∆Hf = - 285.83 kJ
Keep these the same since
these compounds are
being formed in the rxn
NaOH + HC2H3O2  NaC2H3O2 + H2O
NaOH  Na + ½O2 + ½H2
∆Hf = 469.15 kJ
HC2H3O2  O2 + 2H2 + 2C
∆Hf = 484.5 kJ
Na + O2 + 3/2H2 + 2C  NaC2H3O2 ∆Hf = 298.15 kJ
½O2 + H2  H2O
The value is
positive, so the
rxn is
endothermic
∆Hf = -285.83 kJ
∆H = 965.97 kJ
Hess’s Law
Add
them
Practice Problem!
Calculate ∆H for the following reaction:
2N2 (g) + 5O2 (g)  2N2O5 (g)
Using the following data:
½O2 (g) + H2 (g)  H2O (l)
∆Hf° = -285.8 kJ
N2O5 (g) + H2O (l)  2HNO3 (l)
∆H° = -76.6 kJ
3/2O2(g)+½H2(g)+ ½N2(g) HNO3 (l) ∆Hf° = -174.1 kJ
∆ H = 28.4 kJ
Chapter 16
Reaction Energy & Kinetics
16-2 Driving Force of
Reactions
Enthalpy & Reaction Tendency
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Most rxns tend to favor production of
more stable products from less stable
reactants.
That means exothermic rxns are usually
favored in nature.
However, some endothermic rxns do
occur spontaneously.
Entropy, S
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A measure of the degree of randomness
of the particles in a system; tendency
toward disorder.
There is a tendency for systems in nature
to favor the disordered state.
Solid
Liquid
Increase in Entropy
Gas
Entropy
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A positive value for ∆S means an increase
in entropy (disorder)
A negative value for ∆S means a decrease
in entropy (disorder)
Rxns tend to favor high entropy and low
enthalpy
Free Energy, G
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A combined enthalpy-entropy function
Rxns tend to favor a lower free energy
system
Free Energy cannot be measured, only
the change can be measured, ∆G
∆G0 = ∆H0 - T∆S0
@ Standard States
Free Energy
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If ∆G is positive, the reaction is
nonspontaneous and will not readily
occur.
If ∆G is negative, the reaction is
spontaneous and will occur.
Relationship between Enthalpy, Entropy & Free
Energy
∆H
∆S
∆G
-
+
-
Exothermic
Disordering
Exothermic
Ordering
+
+
Endothermic
Disordering
@ high temps
+
-
+
Endothermic
Ordering
Always
-
-
Always
-
@ low temps
-
Practice
Predict whether the value of ∆S will be greater
than, less than or equal to zero:
1) 3H2 (g) + N2 (g)  2NH3 (g) ∆ S < 0

2)
2Mg (s) + O2 (g)  2MgO (s) ∆ S < 0
3)
C6H12O6 (s) + 6O2 (g)  6CO2 (g)+ 6H2O (g)
4)
KNO3 (s)  K+ (aq) + NO3-(aq)
∆S>0
∆S>0
Practice
Calculate the value of ∆G0 for the reaction
below. Will the reaction be spontaneous at 298
K?
Cu2S (s) + S(s)  2CuS(s)
∆H0 = -26.7 kJ
∆S0 = -19.7 J/(mol∙K)
∆ G0 = -20.8 kJ/mol

Yes, it is spontaneous