3.4a
An
Introduction to the Gibbs free energy expression
All advanced level
syllabuses require a knowledge of entropy concepts and simple
calculations and some specifications require a 'working knowledge' of
the Gibbs free energy (ΔG) expressions.
Before dealing with
free energy changes (ΔG) a quick recap on the calculations so far
before I explain what free energy changes are.
In order for a reaction
to be feasible the overall entropy change ΔS_{tot} must
be >0 and the overall free energy change ΔG_{sys} must be <0.
If ΔG_{sys} is ~0 it is likely to be a significant
equilibrium i.e. not significantly on the right or the left of an
equilibrium situation.
For a reaction the overall
entropy change is given by:
ΔS^{θ}_{sys}
= ΣS^{θ}_{products
} ΣS^{θ}_{reactants} and
ΔS^{θ}_{surr}
= ΔH^{θ}_{sys}/T
giving the overall entropy change
equation: ΔS^{θ}_{tot}
= ΔS^{θ}_{sys} + ΔS^{θ}_{surr}
This has been dealt with in section
3.3b
ΔS, Entropy changes and the
feasibility of a chemical change
Before getting into free
energy change calculations a little digression.
If the criteria for a feasible reaction is
ΔS_{tot}
>0 and ΔG<0 what is ΔG? and what is the connection between them?
PLEASE NOTE that you do not need to be able to derive the Gibbs
free energy equation, but, I think it is important you realise there is a
direct connection with the entropy expressions and that delta G
expressions have not been pulled out of a hat!
From above and
substituting: ΔS^{θ}_{tot}
= ΔS^{θ}_{sys} ΔH^{θ}_{sys}/T,
then multiplying through by
T gives
TΔS^{θ}_{tot}
= TΔS^{θ}_{sys} ΔH^{θ}_{sys},
then changing signs
throughout gives
TΔS^{θ}_{tot}
= TΔS^{θ}_{sys} + ΔH^{θ}_{sys}
TΔS^{θ}_{tot}
is called the Gibbs Free Energy change denoted by the letter G,
so giving the familiar
free energy equation:
ΔG^{θ}_{sys}
= ΔH^{θ}_{sys}
TΔS^{θ}_{sys}
A working knowledge of
which is required by some courses.

3.4b Calculating free
energy changes for chemical reactions applications
Using the free energy equation:
ΔG^{θ}_{system/reaction}
= ΔH^{θ}_{system/reaction}
TΔS^{θ}_{system/reaction}
The free energy can be thought of
as energy that is available to do useful work.
A reaction is feasible if
ΔG^{θ} < 0
(its the parallel to feasibility when ΔS^{θ}_{tot}
> 0)
Rearrangements of :
ΔG^{θ}
= ΔH^{θ}
TΔS^{θ}
ΔH^{θ} = ΔG^{θ}
+ TΔS^{θ}
ΔS^{θ}
= (ΔH^{θ}
ΔG^{θ}) / T
T
= (ΔH^{θ}
ΔG^{θ}) / ΔS^{θ}
Note: To calculate temperature point
of 'first' feasibility, at which
ΔG^{θ}
= 0,
you argue that
at this point: ΔH^{θ}_{sys}
TΔS^{θ}_{sys} = 0, so T = ΔH^{θ}_{sys}
/ ΔS^{θ}_{sys}
If you know the entropy change and enthalpy
change of the system, you can then calculate whether any
reaction is feasible at any given temperature by using the
equation
ΔG^{θ}
= ΔH^{θ}
TΔS^{θ}, though this does assume that
enthalpy and entropy changes do not change with temperature,
unfortunately this is not the case, BUT it does give a fair
indication of feasibility;
Units note: Remember T is in K,
ΔH and ΔG are usually given in kJmol^{1},
but S and ΔS are
usually in JK^{1}mol^{1}, so its
probably best to convert ΔS into kJ (that's what I usually
do).
In the case of cells the free energy
equation is:
ΔG^{θ}_{sys}
= neE^{θ}F (see
section
3.4c)
which gives the electrical
energy available.
The free
energy change of a feasible reaction, ΔG^{θ}_{sys},
must be <0, effectively the net cell p.d. must be positive.
If ΔG^{θ}_{sys}
is close to zero, then you are likely to be dealing with a
significant equilibrium
situation.
and remember however feasible a
reaction might be thermodynamically, it does not necessarily mean it will
happen spontaneously because of kinetic limitations.
For more details on this last point see
3.6 Kinetic stability versus thermodynamic instability
for a more detailed discussion and
"KINETICS"
section 6.
How can you calculate the temperature at
which the free energy changes for two different reactions are the same?
ΔG
= ΔH TΔS, therefore for two
reactions
ΔG_{1}
= ΔH_{1} TΔS_{1}
and
ΔG_{2} =
ΔH_{2}
TΔS_{2}
∴ ΔH_{1} TΔS_{1}
= ΔH_{2} TΔS_{2}
∴ TΔS_{2} TΔS_{1}
= ΔH_{2} ΔH_{1}
∴ T(ΔS_{2} TΔS_{1})
= ( ΔH_{2} ΔH_{1})
∴
T = ( ΔH_{2}
ΔH_{1}) / (ΔS_{2} ΔS_{1})
Introduction to the examples of
free energy calculations and the linked concepts
ΔG^{θ}_{system/reaction}
= ΔH^{θ}_{system/reaction}
TΔS^{θ}_{system/reaction}
ΔG
= ΔH
TΔS and a chemical reaction is truly
thermodynamically feasible if ΔG < 0
All the data (from standard data books) for a
question I've tabulated under the equation for convenience, but a handy way of
preceding the enthalpy, entropy and free energy calculations.
I've worked everything out from first
principles, so in an exam you might be given some data which I've worked out, but no
matter, its all good practice in following the conceptual and numerical argument
for free energy calculations.
The
are four possible scenarios depending whether the enthalpy and entropy changes
are positive and negative.
(1)
ΔH positive (endothermic), ΔS positive (entropy increases) 
ΔG_{(sys)}
=
free energy change 
ΔH_{(sys)}
enthalpy change 
TΔS_{(sys)}
temp (K) x entropy change 
Reaction maybe feasible at higher temperatures 
positive, endothermic 
ΔS
positive, entropy increases 
Not feasible at low temperature,
the TΔS term can't outweigh the positive ΔH term, BUT,
at high temperature the TΔS term becomes sufficiently negative to
outweigh the positive ΔH term and give a negative ΔG free energy
overall, so the higher the temperature the more feasible the reaction. 

(2)
ΔH positive (endothermic), ΔS negative (entropy decreases) 
ΔG_{(sys)}
= 
ΔH_{(sys)} 
TΔS_{(sys)} 
Reaction never feasible at any temperature 
positive, endothermic 
ΔS negative, entropy decreases 
Since = +, the TΔS
term will always be positive, and, since ΔH is also positive, ΔG is
always positive (can't be < 0), and so the reaction can NEVER be
feasible. 

(3)
ΔH negative (exothermic), ΔS positive (entropy increases) 
ΔG_{(sys)}
= 
ΔH_{(sys)} 
TΔS_{(sys)} 
Reaction always feasible at any temperature 
negative, exothermic 
ΔS positive, entropy increases 
Since both
ΔH and
TΔS terms are always negative,
ΔG is
always < 0, so the reaction is ALWAYS feasible. 

(4)
ΔH negative (exothermic), ΔS negative (entropy decreases) 
ΔG_{(sys)}
= 
ΔH_{(sys)} 
TΔS_{(sys)} 
Reaction maybe feasible at lower temperatures 
negative, exothermic 
ΔS negative, entropy decreases 
Since = +, at low
temperatures the TΔS term is not too positive to outweigh the
negative ΔH and the reaction is feasible, BUT at high
temperatures the TΔS term ( = +) becomes to positive ( = +)
and outweighs the negative ΔH, so ΔG becomes positive and the reaction
becomes less and less feasible. 
Examples of calculations (1) to (7) using the Gibbs
free energy equation explained above
Examples (1) to (4) match the 'concept'
analysis above, after that its whatever came into my head!
In the questions, the following definitions
apply (to reduce repetition!)
ΔH^{θ}_{f}
= standard enthalpy of formation of a compound at 298K, 1 atm./101kPa pressure
in kJmol^{1}
(for elements in their normal stable
standard state ΔH^{θ}_{f}
= 0).
ΔG^{θ}_{f}
= standard free energy of formation of a compound at 298K, 1 atm./101kPa
pressure in kJmol^{1}
(for elements in their normal stable
standard state ΔG^{θ}_{f}
= 0).
S^{θ}_{f}
= standard entropy energy content at 298 K, 1 atm./101kPa pressure in JK^{}1mol^{1}
(standard entropy can never be zero!)
In calculating the free energy change at
temperatures other than 298K it is assumed the enthalpy and entropy of the
species does not change, however, this is not true, so any such calculations are
an approximation, but the general trends in feasibility are quite correctly
deduced.
TOP OF PAGE
(1)
ΔH positive (endothermic), ΔS positive (entropy increases) 
ΔG
= 
ΔH 
TΔS 

positive, endothermic 
ΔS
positive, entropy increases 
Not feasible at low temperature,
the TΔS term can't outweigh the positive ΔH term, BUT,
at high temperature the TΔS term becomes sufficiently negative to
outweigh the positive ΔH term and give a negative ΔG free energy
overall, so the higher the temperature the more feasible the reaction. 
Example (1) The thermal decomposition of
lithium carbonate
Thermodynamic data 
Li_{2}CO_{3}(s) 
====> 
Li_{2}O(s) 
+ 
CO_{2}(g) 
ΔH^{θ}_{formation}
(kJmol^{1}) 
1216 

596 

394 
ΔG^{θ}_{formation}
(kJmol^{1}) 
1133 

560 

395 
S^{θ}_{species}
(JK^{1}mol^{1}) 
90.4 

39.2 

214 
To calculate the enthalpy change
ΔH^{θ}_{sys}
= ΣΔH^{θ}_{f(products)
} ΣΔH^{θ}_{f(reactants)}
ΔH^{θ}_{sys}
= (596 394) (1216) = +226 kJmol^{1}
To calculate the free energy change
ΔG^{θ}_{sys}
= ΣΔG^{θ}_{f(products
} ΣΔG^{θ}_{f(reactants)}
ΔG^{θ}_{sys}
= (560 395) (1133) = +178kJmol^{1}
To calculate the entropy change
ΔS^{θ}_{sys}
= ΣS^{θ}_{products
} ΣS^{θ}_{reactants}
ΔS^{θ}_{sys}
= (39.2 + 214) (90.4) = +162.8 JK^{1}mol^{1}
(0.1628 kJK^{1}mol^{1})
Substituting into the Gibbs free energy change equation
ΔG^{θ}_{sys}
= ΔH^{θ}_{sys}
TΔS^{θ}_{sys}
ΔG^{θ}_{sys}
= +226 0.1628T kJmol^{1}
To calculate the temperature at which the
decomposition of lithium carbonate becomes feasible
When ΔG^{θ}_{sys}
= 0, you have reached the point of first feasibility
Therefore, at this point: ΔH^{θ}_{sys}
TΔS^{θ}_{sys}
= 0
So
T = ΔH^{θ}_{sys} / ΔS^{θ}_{sys} =
226 / 0.1628 = 1388 K (1115^{o}C)
and
the decomposition will occur at any temperature above 1388 K, below 1388 it
becomes less and less feasible, negligible product (theoretically!).
Suppose you given a series of free energies at various temperatures for a given
reaction?
How do you calculate the enthalpy change and entropy change?
ΔG^{θ}_{sys}
= ΔH^{θ}_{sys}
TΔS^{θ}_{sys}
y =
mx + c, is the classic algebraic equation for a linear graph connecting two
variables x and y, slope = m, c = a constant.
This
can be applied to the Gibbs free energy equation
y
=
ΔG^{θ}_{sys}, gradient = m =
ΔS^{θ}_{sys},
x = T(K), constant = c = ΔH^{θ}_{sys}
So,
by plotting a graph of
ΔG^{θ}_{sys}
versus T, you can derive ΔH^{θ}_{sys} and TΔS^{θ}_{sys}
I actually used the free energy equation to
generate the perfect data for the graph below, so that we are now working as if
the starting point was
ΔG data and not ΔH, S or ΔS
data.
You
can see from the graph, the point of reaction feasibility, when
ΔG is =<0, is around 1400K or
higher temperatures.
With
the help of Excel software and applying the y = mx + c logic:
From
the gradient: m =
ΔS
= 0.1628,
ΔS = 0.163 kJK^{1}mol^{1}
or +163JK^{1}mol^{1}
From
the constant c,
ΔH = +226 kJmol^{1}
(2)
ΔH positive (endothermic), ΔS positive (entropy increases) 
ΔG
= 
ΔH 
TΔS 

positive, endothermic 
ΔS negative, entropy decreases 
Since = +, the TΔS
term will always be positive, and, since ΔH is also positive, ΔG is
always positive (can't be < 0), and so the reaction can NEVER be
feasible. 
Example (2) The 'synthesis' of ethene from
its elements
Thermodynamic data 
C(s) 
+ 
2H_{2}(g) 
====> 
C_{2}H_{4}(g) 
ΔH^{θ}_{formation}
(kJmol^{1}) 
0 

0 

+52.3 
ΔG^{θ}_{formation}
(kJmol^{1}) 
0 

0 

+68.1 
S^{θ}_{species}
(JK^{1}mol^{1}) 
5.7 

2 x 131 

219 
To calculate the enthalpy change
ΔH^{θ}_{sys}
= ΣΔH^{θ}_{f(products)
} ΣΔH^{θ}_{f(reactants)}
ΔH^{θ}_{sys}
= +52.3 (0 + 0) = +52.3 kJmol^{1}
To calculate the free energy change
ΔG^{θ}_{sys}
= ΣΔG^{θ}_{f(products
} ΣΔG^{θ}_{f(reactants)}
ΔG^{θ}_{sys}
= +68.1 (0 + 0) = +68.1 kJmol^{1}
To calculate the entropy change
ΔS^{θ}_{sys}
= ΣS^{θ}_{products
} ΣS^{θ}_{reactants}
ΔS^{θ}_{sys}
= (219) {(5.7 + (2 x 131)} = 48.7 JK^{1}mol^{1}
(0.0487 JK^{1}mol^{1})
Substituting into the Gibbs free energy
equation for other
temperatures
ΔG^{θ}_{sys}
= ΔH^{θ}_{sys}
TΔS^{θ}_{sys}
ΔG^{θ}_{sys}
= +52.3 (0.0487T)
= 52.3 + 0.0487 kJmol^{1}
Both terms on the right are always
positive, so ΔG can never be
negative, so the reaction is never feasible at any temperature!
Suppose you given a series of free energies at various temperatures for a given
reaction?
How do you calculate the enthalpy change and entropy change?
The arguments are given in detail in Example
(1) above, so I'm just starting from the given graph of free energy versus
temperature from now on.
With
the help of Excel software and applying the y = mx + c logic:
From
the gradient: m =
= ΔS
= 0.0487,
ΔS = 0.0.0487 kJK^{1}mol^{1}
or 48.7JK^{1}mol^{1}
From
the constant c,
ΔH = +52.3 kJmol^{1}
TOP OF PAGE
(3)
ΔH negative (exothermic), ΔS positive (entropy increases) 
ΔG
= 
ΔH 
TΔS 

negative, exothermic 
ΔS positive, entropy increases 
Since both
ΔH and
TΔS terms are always negative,
ΔG is
always < 0, so the reaction is ALWAYS feasible. 
Example (3) The decomposition of hydrazoic
acid (hydrogen azide, a colourless volatile explosive liquid)
I could only find thermodynamic data on the
gaseous state.
Thermodynamic data 
HN_{3}(g) 
====> 
^{1}/_{2}H_{2}(g) 
+ 
^{3}/_{2}N_{2}(g) 
ΔH^{θ}_{formation}
(kJmol^{1}) 
+294 

0 

0 
ΔG^{θ}_{formation}
(kJmol^{1}) 
+328 

0 

0 
S^{θ}_{species}
(JK^{1}mol^{1}) 
237 

131/2 

^{3}/_{2}192 
To calculate the enthalpy change
ΔH^{θ}_{sys}
= ΣΔH^{θ}_{f(products)
} ΣΔH^{θ}_{f(reactants)}
ΔH^{θ}_{sys}
= (0 + 0) (+294) = 294 kJmol^{1}
To calculate the free energy change
ΔG^{θ}_{sys}
= ΣΔG^{θ}_{f(products
} ΣΔG^{θ}_{f(reactants)}
ΔG^{θ}_{sys}
= (0 + 0) (+328) = +328 kJmol^{1}
To calculate the entropy change
ΔS^{θ}_{sys}
= ΣS^{θ}_{products
} ΣS^{θ}_{reactants}
ΔS^{θ}_{sys}
= (131/2 + ^{3}/_{2}192) (237) = +116.5 JK^{1}mol^{1}
(+0.1165 kJK^{1}mol^{1})
Substituting into the Gibbs free energy equation for other
temperatures
ΔG^{θ}_{sys}
= ΔH^{θ}_{sys}
TΔS^{θ}_{sys}
ΔG^{θ}_{sys}
= 294
0.1165T
Both terms are always negative, therefore
ΔG^{θ}_{sys}
is always negative and the reaction feasible at any temperature!
Suppose you given a series of free energies at various temperatures for a given
reaction?
How do you calculate the enthalpy change and entropy change?
The arguments are given in detail in Example
(1) above, so I'm just starting from the given graph of free energy change
versus temperature.
Using
Excel software and applying the y = mx + c logic:
From
the gradient: m =
= ΔS
= 0.117,
ΔS = 0.117 kJK^{1}mol^{1}
or +117JK^{1}mol^{1}
From
the constant c,
ΔH = +52.3 kJmol^{1}
(4)
ΔH positive (endothermic), ΔS positive (entropy increases) 
ΔG
= 
ΔH 
TΔS 

negative, exothermic 
ΔS negative, entropy decreases 
Since = +, at low
temperatures the TΔS term is not too positive to outweigh the
negative ΔH and the reaction is feasible, BUT at high
temperatures the TΔS term ( = +) becomes to positive ( = +)
and outweighs the negative ΔH, so ΔG becomes positive and the reaction
becomes less and less feasible. 
Example (4) The synthesis of ammonia
Thermodynamic data 
^{1}/_{2}N_{2}(s) 
+ 
^{3}/_{2}H_{2}(g) 
====> 
NH_{3}(g) 
ΔH^{θ}_{formation}
(kJmol^{1}) 
0 

0 

46.2 
ΔG^{θ}_{formation}
(kJmol^{1}) 
0 

0 

16.6 
S^{θ}_{species}
(JK^{1}mol^{1}) 
^{1}/_{2}192 

^{3}/_{2}131 

193 
To calculate the enthalpy change
ΔH^{θ}_{sys}
= ΣΔH^{θ}_{f(products)
} ΣΔH^{θ}_{f(reactants)}
ΔH^{θ}_{sys}
= (46.2) (0 + 0) = 46.2 kJmol^{1}
To calculate the free energy change
ΔG^{θ}_{sys}
= ΣΔG^{θ}_{f(products
} ΣΔG^{θ}_{f(reactants)}
ΔG^{θ}_{sys}
= (16.6) (0 + 0) = 16.6 kJmol^{1}
To calculate the entropy change
ΔS^{θ}_{sys}
= ΣS^{θ}_{products
} ΣS^{θ}_{reactants}
ΔS^{θ}_{sys}
= (193) (192/2 + /_{2}131) = 99.5 JK^{1}mol^{1}
(0.0995 kJK^{1}mol^{1})
Substituting into the Gibbs free energy equation for other
temperatures
ΔG^{θ}_{sys}
= ΔH^{θ}_{sys}
TΔS^{θ}_{sys}
ΔG^{θ}_{sys}
= 46.2
(0.0995T) = 46.2 + 0.0995T
Which
means it is not favoured by high temperatures, no matter how fast you want the
reaction to take place!
For further discussion see
Applying Le Chatelier's Principle to Industrial
Processes
The Haber Synthesis of ammonia
Suppose you given a series of free energies at various temperatures for a given
reaction?
How do you calculate the enthalpy change and entropy change?
The arguments are given in detail in Example
(1) above, so I'm just starting from the given graph of free energy change
versus temperature.
You
can see from the graph, the point of reaction feasibility, when
ΔG is =<0, is around 460K or
lower temperatures.
Using
Excel software and applying the y = mx + c logic:
From
the gradient: m =
= ΔS
= 0.0994,
ΔS = 0.994 kJK^{1}mol^{1}
or 99.4JK^{1}mol^{1}
From
the constant c,
ΔH = +46.2 kJmol^{1}
I did find a set of values for the
equilibrium constant K_{p}, for the equation
N_{2}(g) + 3H_{2}(g)
2NH_{3}(g)
K_{p} = p_{NH3}^{2}/p_{N2}p_{H2}^{3}
Although I've done the thermodynamic
calculations based on the formation of 1 mole of ammonia, you can plainly see
the dramatic decrease in the equilibrium constant.
In reality, a fast low yield of ammonia
is acceptable at a moderate temperature, with the help of high pressure and
a catalyst, see ....
Applying Le Chatelier's Principle to Industrial
Processes
The Haber Synthesis of ammonia
TOP OF PAGE
Example (5) The thermal decomposition of
ammonium chloride (a sublimation, solid ===> gases)
Thermodynamic data 
NH_{4}Cl(s) 
====> 
NH_{3}(g) 
+ 
HCl(g) 
ΔH^{θ}_{formation}
(kJmol^{1}) 
315 

46.2 

92.3 
ΔG^{θ}_{formation}
(kJmol^{1}) 
204 

16.6 

95.3 
S^{θ}_{species}
(JK^{1}mol^{1}) 
94.6 

39.2 

187 
To calculate the enthalpy change
ΔH^{θ}_{sys}
= ΣΔH^{θ}_{f(products)
} ΣΔH^{θ}_{f(reactants)}
ΔH^{θ}_{sys}
= (46.2 92.3) (315) = +176.5 kJmol^{1}
To calculate the free energy change
ΔG^{θ}_{sys}
= ΣΔG^{θ}_{f(products
} ΣΔG^{θ}_{f(reactants)}
ΔG^{θ}_{sys}
= (16.6 95.3) (204) = +92.1 kJmol^{1}
To calculate the entropy change
ΔS^{θ}_{sys}
= ΣS^{θ}_{products
} ΣS^{θ}_{reactants}
ΔS^{θ}_{sys}
= (39.2 + 187) 94.6 = +131.6 JK^{1}mol^{1}
(0.3294 kJK^{1}mol^{1})
Substituting into the Gibbs free energy equation for other
temperatures
ΔG^{θ}_{sys}
= ΔH^{θ}_{sys}
TΔS^{θ}_{sys}
ΔG^{θ}_{sys}
= 176.5
0.329T
To calculate the temperature at which the
decomposition of ammonium chloride becomes feasible
When ΔG^{θ}_{sys}
= 0, you have reached the point of first feasibility
Therefore, at this point: ΔH^{θ}_{sys}
TΔS^{θ}_{sys} = 0
So
T = ΔH^{θ}_{sys} / ΔS^{θ}_{sys} =
176.5 / 0.3294 = 539 K (263^{o}C)
so,
the decomposition will occur at any temperature above 539 K, below 539 it
becomes less and less feasible.
You
can see from the graph, the point of reaction feasibility, when
ΔG is =<0, is around 540K or
higher temperatures.
You see this when heat ammonium chloride, it
readily decomposes on heating it in a test tube with a bunsen burner, but when
the gases reach the upper cooler part of the test tube, the solid crystal
reforms.
Although this is a very endothermic reaction
and the lattice enthalpy is 620 kJmol1, indicating a strong ionic bonding, it
is the formation of gases and the accompanying large increase in entropy that
drives the reaction forward at higher temperatures.
Its more feasible at lower temperatures than
lithium carbonate and calcium carbonate because of the bigger entropy change 
two moles of gas formed compared to one. Numerically the negative TΔS
term is about twice as big.
Example (6) Cracking an alkane hydrocarbon
to produce ethene
Suppose you wanted to 'ideally' convert
butane into (i) two molecules of ethene for polymers and (iii) one molecule of
hydrogen for fuel or hydrogenation of vegetable oil.
Calculate whether or not the reaction is
feasible and what sort of minimum temperature is theoretically needed.
Thermodynamic data 
C_{4}H_{10}(g) 
====> 
2C_{2}H_{4}(g) 
+ 
H_{2}(g) 
ΔH^{θ}_{formation}
(kJmol^{1}) 
125 

2 x +52.3 

0 
ΔG^{θ}_{formation}
(kJmol^{1}) 
15.7 

2 x 68.1 

0 
S^{θ}_{species}
(JK^{1}mol^{1}) 
310 

2 x 219 

131 
Enthalpy change
ΔH^{θ}_{sys}
= ΣΔH^{θ}_{f(products)
} ΣΔH^{θ}_{f(reactants)}
ΔH^{θ}_{sys}
= {(2 x 52.3) + 0} (125) = +229.6 kJmol^{1}
Free energy change
ΔG^{θ}_{sys}
= ΣΔG^{θ}_{f(products
} ΣΔG^{θ}_{f(reactants)}
ΔG^{θ}_{sys}
= {(2 x 68.1) + 0} (15.7) = +151.9 kJmol^{1}
Entropy change
ΔS^{θ}_{sys}
= ΣS^{θ}_{products
} ΣS^{θ}_{reactants}
ΔS^{θ}_{sys}
= {(2 x 219) + 131} (310) = +259 JK^{1}mol^{1}
(0.259 kJK^{1}mol^{1})
Gibbs free energy change equation for other
temperatures
ΔG^{θ}_{sys}
= +229.6
0.0259T
To calculate the temperature at which this
specific cracking reaction becomes feasible
When ΔG^{θ}_{sys}
= 0, you have reached the point of first feasibility
Therefore, at this point: ΔH^{θ}_{sys}
TΔS^{θ}_{sys} = 0
So,
T = ΔH^{θ}_{sys} / ΔS^{θ}_{sys} =
229.6 / 0.259 = 886 K (613^{o}C)
therefore, the decomposition will occur at any temperature above 886 K, below
886 it becomes less and less feasible.
However, you might need a very special catalyst to give these useful products
and nothing else!
You
can see from the graph, the point of reaction feasibility, when
ΔG is =<0, is around 900K or
higher temperatures.
(7) The thermal decomposition of calcium
carbonate (limestone ==> lime)
Thermodynamic data 
CaCO_{3}(s) 
====> 
CaO(s) 
+ 
CO_{2}(g) 
ΔH^{θ}_{formation}
(kJmol^{1}) 
1207 

635 

394 
ΔG^{θ}_{formation}
(kJmol^{1}) 
1129 

604 

395 
S^{θ}_{species}
(JK^{1}mol^{1}) 
92.9 

40 

214 
Enthalpy change
ΔH^{θ}_{sys}
= ΣΔH^{θ}_{f(products)
} ΣΔH^{θ}_{f(reactants)}
ΔH^{θ}_{sys}
= (635 394) (1207) = +178 kJmol^{1}
Free energy change
ΔG^{θ}_{sys}
= ΣΔG^{θ}_{f(products
} ΣΔG^{θ}_{f(reactants)}
ΔG^{θ}_{sys}
= (604 395) (1129) = +130 kJmol^{1}
Entropy change
ΔS^{θ}_{sys}
= ΣS^{θ}_{products
} ΣS^{θ}_{reactants}
ΔS^{θ}_{sys}
= (40 + 214) 92.9 = +161.1 JK^{1}mol^{1}
(0.1611 kJK^{1}mol^{1})
Gibbs free energy change equation for other
temperatures
ΔG^{θ}_{sys}
= ΔH^{θ}_{sys}
TΔS^{θ}_{sys}
ΔG^{θ}_{sys}
= 178 0.1611T
(its numerically similar to the free energy equation for the thermal
decomposition of lithium carbonate)
To calculate the temperature at which the
decomposition of calcium carbonate becomes feasible
When ΔG^{θ}_{sys}
= 0, you have reached the point of first feasibility
Therefore, at this point: ΔH^{θ}_{sys}
TΔS^{θ}_{sys} = 0
So
T = ΔH^{θ}_{sys} / ΔS^{θ}_{sys} =
178 / 0.0.1611 = 1105 K (263^{o}C)
so,
the decomposition will occur at any temperature above 1105 K (832^{o}C), below
1105K it
becomes less and less feasible.
You
can see from the graph, the point of reaction feasibility, when
ΔG is =<0, is around 1100K or
higher temperatures.
Although this is a very endothermic reaction,
and strong ionic bonding in calcium carbonate, it
is the formation of carbon dioxide and the accompanying large increase in entropy that
drives the reaction forward at higher temperatures.
Note: I've done the same calculation on the
entropy calculations page, which gave a value of
1112K.
The graph for calcium carbonate would be
similar to that for the thermal decomposition of lithium carbonate in example
(1).
TOP OF PAGE
3.4c Electrochemical Cell Emf's,
Feasibility and Gibb's Free Energy Change
Relating E^{θ}_{cell(sys)} to the direction of overall chemical change
and feasibility of reaction. If you
calculate a ve cell voltage for a given reaction, that is not the way
cell reaction goes! please reverse the equation and recalculate!
The E_{batterycell}
must be >0.00V
for the cell, and any other redox reaction, to be theoretically
feasible.
The free
energy change must be negative <0 for the cell reaction to be
feasible (matching the E_{cell} rule of >0V for
feasibility).
ΔG^{θ}_{cell(sys)} =
nE^{θ}F J
(ΔG not needed by all A2 advanced level syllabuses)
n =
number of moles of electrons transferred in the reaction per mol of
reactants involved,
E^{θ} is the
Emf for the overall reaction in volts,
F = the Faraday
constant (96500 C mol^{1}).
Ex 3.4c1 for the
zinccopper Daniel cell producing +1.10V
2 electrons
transferred: M^{2+}_{(aq)} + 2e^{} <=> M_{(s)}
full cell redox reaction: Cu^{2+}_{(aq)}
+ Zn_{(s)} ==> Cu_{(s)} + Zn^{2+}_{(aq)}
ΔG^{θ}_{cell}
= 2 x 1.10 x 96500 = 212300 Jmol^{1} or 212.3 kJmol^{1}
A very negative
free energy value so very feasible!
If work the
calculation for the reverse reaction (Cu + Zn^{2+} ==>) you get
a very positive free energy value, so definitely NOT a feasible chemical
change.
For calculating an equilibrium constant for a redox reaction see section
3.5b
3.4d Free
Energy Changes and the Extraction of Metals
In order for a metal
extraction to be thermodynamically feasible from an oxide the decrease
in Gibbs free energy on oxide formation must be numerically less than
the free energy released by the formation of carbon monoxide/carbon
dioxide if carbon/carbon monoxide is used in its extraction.
This is
conveniently explored via an Ellingham diagram like the one shown below.
An Ellingham diagram shows the variation of the free energy change with temperature
for particular reactions reaction. The more negative the free
energy change the more feasible it is.
A metal can be theoretically extracted using
carbon, if the free energy change in forming carbon monoxide or
carbon dioxide, is more negative than the free energy change in
forming the oxide. 
You can theoretically extract
magnesium (Mg) from magnesium oxide (MgO) using carbon at a temperature
of over 2300K, but technologically not very convenient since most
furnace materials would melt at this temperature!
Aluminium would require an even higher
temperature of well over 2500K, so not surprisingly, these highly reactive
metals are extracted by electrolysis.
However, iron can be
extracted from iron oxides using carbon at temperatures above 1300K
which of course is what happens in a blast furnace. Zinc can be
extracted above 1100K.
Note that many metal mineral ores are
sulfides and carbonates which are often roasted to give the oxide which
is then reduced by some means.
For
more on extraction of metals and Ellingham free energy diagrams see
section 3.4d
A
summary of the maths of free energy, entropy and feasibility

In order for a reaction
to feasible the overall entropy change ...

For a reaction the overall
entropy change is given by:

Total entropy change =
entropy change of system + entropy change of surroundings

ΔS^{θ}_{tot}
= ΔS^{θ}_{sys} + ΔS^{θ}_{surr}

ΔS^{θ}_{sys}
= = ΣS^{θ}_{products
} ΣS^{θ}_{reactants}

and ΔS^{θ}_{surr}
= ΔH^{θ}_{sysreaction}/T(K)

Note the connection
between ΔS and ΔG.

From above and
substituting ...

ΔS^{θ}_{tot}
= ΔS^{θ}_{sys} ΔH^{θ}_{sys}/T

multiplying through by
T gives

TΔS^{θ}_{tot}
= TΔS^{θ}_{sys} ΔH^{θ}_{sys}

changing signs
throughout gives

TΔS^{θ}_{tot}
= TΔS^{θ}_{sys} + ΔH^{θ}_{sys}

TΔS^{θ}_{tot}
is called the Gibbs Free Energy change symbol G,

so giving the familiar
free energy equation ...

ΔG^{θ}_{sys}
= ΔH^{θ}_{sys}
TΔS^{θ}_{sys}

The free energy can be
thought of as heat energy that is available to do work.

OR in the case of
cells, where ΔG^{θ}_{sys}
= neE^{θ}F,

this gives the
electrical energy available to do useful work.

PLEASE NOTE you do
not have to derive the Gibbs Free Energy equation but some syllabuses
require you to be able to use when supplied with enthalpy and entropy
data.

The free
energy change of the reaction, ΔG^{θ}
must be <=0 to be feasible.

If ΔS^{θ}_{tot}
or ΔG^{θ}_{sys}
is close to zero, then you are likely to be dealing with an equilibrium
situation.
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which reaction becomes feasible A Level chemistry exams,
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reaction becomes feasible A Level chemistry Explaining the
Gibbs free energy equation for determining the feasibility
of a reaction temperature. How to work out if a reaction is
feasible at a particular temperature using enthalpy and
entropy changes. How the free energy equation applies to any
chemical changes including the extraction of metals (e.g.
feasibility of carbon reduction of metal oxide). How
calculating free energy can tell you the direction of
chemical change in an electrical cell 'battery' of two
halfcells. How do you calculate the minimum temperature for
a reaction to become feasible is described using enthalpy
and entropy change data. Explaining how to obtain enthalpy
change and entropy change from a graph of free energy versus
temperature. Calculating the enthalpy, entropy and free
energy changes for the thermal decomposition of lithium
carbonate, calculating the lowest temperature the reaction
is feasible, the enthalpy, entropy and free energy change in
synthesising of ethene from carbon and hydrogen, when does
the synthesis of ammonia become feasible using the Gibbs
free energy equation and enthalpy, entropy and free energy
data, using the enthalpy, entropy and Gibbs free energy
change equation to calculate the lowest decomposition
temperature for ammonium chloride, using enthalpy, entropy
and free energy data to calculate the lowest thermal
decomposition temperature of calcium carbonate (limestone)
to make calcium oxide (lime), what is the free energy change
in cracking alkanes to alkenes (cracking butane to ethene)
















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