ah chemistry unit 1 - cathkin high school€¦ · the total energy of the universe (system and...
TRANSCRIPT
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AH Chemistry – Unit 1
Reaction Feasibility
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THERMOCHEMISTRY
The study of changes in ENERGY which occur during chemical reactions
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First Law Of Thermodynamics
System
energy
surroundings
Energy can be converted from one Form to another while
the Total Energy remains the same
Energy can neither be created nor destroyed
The Total Energy of the Universe (System and
Surroundings) is constant
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Energy Pathway Diagrams
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Enthalpy ~ Exothermic
∆H = Hfinal
- Hinitial
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Enthalpy ~ Endothermic
Actual Enthalpy values, H,
cannot be measured.
Changes in Enthalpy, ∆H,
can be measured.∆H = H
final- H
initial
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Standard Enthalpies
reactants and products are in their standard states
measured at a specific temperature, usually 298K
measured at a pressure of one atmosphere, 105
Nm-2
concentrations of solutions are one mol l -1
designated using superscript ° , ∆H°
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Enthalpy of Formation
is the heat change when 1 mole of a compound is formed from its
elements under standard conditions
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Standard Enthalpies of Formation
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Named Enthalpies
Combustion
Ionisation
Neutralisation
Formation
Bond Forming
Vaporisation
Atomisation
Solution
Hydration
Bond Breaking
Fusion
Lattice Forming
Lattice Breaking
Solvation
Sublimation
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Named Enthalpies
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Using Enthalpies of Formation
The Enthalpy of Combustion of methane can be calculated using
Enthalpies of Formation
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Step1 : Reactants to Elements
This is the ∆H°fof methane reversed.
By definition, the ∆H°fof an element = zero
∆Hstep 1
= - (∆H°f methane
+ ∆H°f oxygen
) = -Σ∆H°f reactants
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Step2 : Elements to Products
∆Hstep 2
= (∆H°f CO2
+ 2 x ∆H°f water
) = Σ∆H°f products
This is the ∆H°f
of
carbon dioxide.
This is the 2 x ∆H°f
of
water.
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General Expression
∆Hreaction
= ∆Hstep 1
+ ∆Hstep 2
∆Hreaction
= Σ∆H°f products
-Σ∆H°f reactants
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Formulae
∆Hreaction
= ΣBond Breaking + ΣBond Forming
(reactants) (products)
∆Hreaction
= Σ∆H°f products
-Σ∆H°f reactants
∆H = Hfinal
- Hinitial
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Energy Rules!Some Processes involve no change in temperature but a major change in the
energy of particles is still occurring.
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EntropyThese changes in the energy of particles have an overall affect on the level
of disorder shown by a substance.
The disorder in a substance is known as its ENTROPY, S .
The Third Law of Thermodynamics provides a reference against
which Entropies can be measured.
“the Entropy of a perfect crystal at 0
K is zero”
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Entropy - State
Molecular Motion
Translation Rotation Vibration
no freedom to move
restricted freedom to move
total freedom to move
no freedom to rotate
some freedom to rotate
total freedom to rotate
free to vibrate
free to vibrate
free to vibrate
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Entropy - Temperature
Entropy(S)
Temperature
Entropy of
Fusion
Entropy of Vaporisation
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Entropy - Dissolving
Less Randomness More Randomness
Less Entropy More Entropy
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Entropy - Molecules
NO NO2
N2
O4
Fewer Vibrations
Less Entropy
More Vibrations
More Entropy
More Vibrations
More Entropy
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Entropy - Numbers
Less Randomness More Randomness
Less Entropy More Entropy
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Entropy - Mixtures
Less Randomness More Randomness
Less Entropy More Entropy
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Entropy Values
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Entropy Calculations
Similar to a previous formula:
∆So = ∑ Soproducts - ∑ So
reactants
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Entropy Changes, ∆S
• spontaneous endothermic reactions tend to have certain
characteristics in common
• the number of moles of product are greater than the number of
moles of reactant
• a large proportion of the products are either liquids or gases
• reactants are often solids or liquids
(NH4)2CO3 + 2CH3COOH 2NH4CH3COO + CO2 + H2O
The trend solids liquids gases is associated with an
increase in disorder .
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Entropy - The Answer?Is an Increase In Entropy the driving force behind a spontaneous chemical
reaction ?
Both ∆S = +ve & ∆S = -ve processes can be spontaneous
The direction of a spontaneous process will depend on temperature
A spontaneous process will depend on both ∆S and ∆H
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Entropy - The Answer?The ‘problem’ can be resolved if we take into account changes taking place
in the Surroundings.
The driving force behind a spontaneous process turns out to be an Overall
Increase In Entropy
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Entropy - The Answer?
Water freezing is a spontaneous process whenever there is an Overall
Increase In Entropy
Water freezing leads to a decrease in
entropy within the system.
Being Exothermic, however, leads to
an increase in entropy in the
surroundings
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Entropy - The Answer?
Water melting is a spontaneous process whenever there is an Overall
Increase In Entropy
Being Endothermic leads to a decrease
in entropy in the surroundings
There will have to be an increase in
entropy within the system.
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Measuring ∆Sosurr
Trying to Calculate the effect on the surroundings would appear, at first,
an impossible task.
Where do the surroundings start & finish?
What is the entropy of air? Glass? etc.
How many moles of ‘surroundings’ are there?
Fortunately it is much, much simpler than that.
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Measuring ∆Sosurr
Firstly the change in Entropy of the Surroundings is caused by the
Enthalpy change of the Surroundings, and…….
∆Hosurr = -∆Ho
syst
so ∆Sosurr ∝ -∆Ho
syst
Temperature has an inverse effect. For example, energy released into the
surroundings has less effect on the entropy of the surroundings, the
hotter the surroundings are.
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Measuring ∆Sosurr
In fact, it turns out that ..
It is the Overall Entropy Change that must be considered.
∆Sosurr = -∆Ho
systT
∆Sototal = ∆So
syst + ∆Sosurr
∆Sototal = ∆So
surr -∆Hosyst
T
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Measuring ∆Sototal
We are interested in the point at which the Total Entropy becomes a
positive value (ceases being a negative value). We can ‘solve’ for
∆Stotal= 0
0 = T∆Sosyst -∆Ho
syst
0 = ∆Sosurr -∆Ho
systT
Multiplying throughout by T gives us
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Measuring ∆Sototal
Remember that this is really the formula for ∆Stotal
∆Sototal = T∆So
syst -∆Hosyst
Armed with ∆S , ∆H and values for T we can calculate the overall change in
Entropy and a positive value would be necessary for a spontaneous reaction.
However, for reasons that are beyond this Topic, a term called the Gibbs
Free Energy, G, is preferred. A negative value for ∆G is equivalent to a
positive value for ∆S. This requires a slight adjustment in our final formula.
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Gibbs Free Energy ∆Go
∆Go = ∆Hosyst - T∆So
syst
The convenient thing about this expression is that it allows us to do
calculations using only values that can be directly measured or easily
calculated.
Strictly speaking, the Second Law of Thermodynamics states that Entropy
must increase for a Spontaneous Process.
In practice, the Second Law of Thermodynamics means that Gibbs Free
Energy must decrease for a Spontaneous Process.
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Gibbs Free Energy ∆Go
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Calculating ∆Go
∆Ho = ∑ ∆Hfo
products - ∑ ∆Hfo
reactants
∆So = ∑ Soproducts - ∑ So
reactants
∆Go = ∆Ho - T∆So
Fe2O3 + 3 CO ➝ 2 Fe + 3 CO2
T in Kelvin
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∆Go of Formation
∆Go = ∑ ∆Gfo
products - ∑ ∆Gfo
reactants
The ∆G of a reaction can be calculated from ∆Gf values.
By themselves, they give useful information about relative stabilities.
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Ellingham Diagrams
∆Go = ∆Ho - T∆So y = c + mx
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Reversible Reactions
∆Go = ∆Ho - T∆SoFor a Chemical Reaction
If ∆G is negative for one direction, it must be positive for the reverse.
This implies that only one reaction can proceed (spontaneously) under a
given set of conditions.
However, ∆S calculations are based on 100% Reactant & 100% Product.
In reality, mixtures exist, so larger ∆S values will be obtained than those
calculated.
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Reversible Reactions
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Equilibrium position
∆G for 100% Reac➝ 100% Prod is
positive but 100% Reac➝ mixture is
negative so forward reaction can take place.
∆G for 100% Prod ➝ 100% Reac and
100% Prod ➝ mixture is more negative so
backward reaction is favoured
Equilibrium lies over to the left but only slightly
since value of ∆G is positive but relatively small
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Equilibrium position
∆G for 100% Reac➝ 100% Prod is very
positive but 100% Reac➝ mixture is still
slightly negative so forward reaction can
take place.
∆G for 100% Prod ➝ 100% Reac and
100% Prod ➝ mixture is more negative so
backward reaction is favoured
Equilibrium lies well over to the left since value of
∆G is positive and relatively large
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Equilibrium position
∆G for 100% Reac➝ 100% Prod is very
negative but 100% Prod ➝ mixture is still
slightly negative so backward reaction can
take place.
∆G for 100% Reac➝ 100% Prod and
100% Reac➝ mixture is more negative so
forward reaction is favoured
Equilibrium lies well over to the right since value of
∆G is negative and relatively large
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Equilibrium position
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Equilibrium Position
∆Go = - RT ln K
There is a mathematical relationship:
More simply:
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Thermodynamic Limits
Thermodynamics can predict whether a
reaction is feasible or not.
Thermodynamics can predict the
conditions necessary for a reaction to
be feasible.
Thermodynamics can predict the
position of equilibrium
Thermodynamics cannot predict how
fast a reaction might be.