Enthalpy
The enthalpy of formation of a compound is a change in energy that
occurs when it is formed from its element.
CH4 + 2O2 --> CO2 + 2H2O
, 
H = -890.2 kJ/mol
4Fe(s) + 3O2(g) --> 2Fe2O3(s)
, H = -1647 kJ/mol
N2O4(g) --> 2NO2(g)
, H = 57.24 kJ/mol
The enthalpies of formation may be either positive or negative.
A positive enthalpy of formation indicates that energy has to be
provided in order for the reaction to proceed. A negative
enthalpy of formation indicates that energy is evolved.
What the enthalpy of formation tells us is the change of enthalpy
that would occur if a reaction pathway could be found that
takes one of the elements to the compound of interest.
Hess's law: The overall reaction enthalpy is the sum of
reaction enthalpies of each step into which the reaction may
formally be divided.
Entropy
The entropy, S, of a system is a measure of its randomness or,
its disorder.
Second law of thermodynamics: The entropy of the universe
increases in the course of every natural change.
Consider the oxidation of iron (25oC).:
4Fe(s) + 3O2(g) --> 2Fe2O3(s)
S =
2(87.37J/(K mol)) - 4(27.29J/(K mol)) - 3(205.2J/(K mol)) =
-550.12J/(K mol)
(H = -1647kJ/mol)
The negative enthalpy in this reaction (exothermic reaction),
causes the
S(surroundings) to be
positive:
S(surroundings) =
-H/T
S(surroundings) =
-(-1647 kJ/mol)/(298.15 K) = +5524 J/(K mol)
This increase in entropy arises because the reaction releases
energy into the surroundings.
In contrast to oxidation of iron, the reaction:
N2O4(g) --> 2NO2(g)
is an endothermic reaction
(H = +57.2 kJ/mol):
S =
2(240.1J/(K mol)) - 304.38J/(K mol) = 175.82J/(K mol)
S(surroundings) =
-(57.24 kJ/mol)/(298.15 K) = -192 J/(K mol)
All endothermic reactions decrease the entropy of the surroundings.
The sum of the total entropy changes
S(total) =
-H/T +
S(surroundings)
is a measure of the total change of energy in the universe.
For every 4 mol of Fe converted to 2 mol Fe2O3
there is an overall increase in the entropy of the universe:
S(total) = +5524J/(K mol)
+ (-550.12J/(K mol)) = +4973.88 J/(K mol)
This large positive quantity indicates that the reaction is spontaneous
and a thermodynamic reason why steel corrodes.
The dissociation of N2O4 indicates that the
total entropy change:
S(total) = -192J/(K mol)
+ (175.82J/(K mol)) = -16.18 J/(K mol)
Because this is a small, negative value, we can conclude that this
reaction does not go to completion.
Gibbs Free Energy.
When a reversible chemical change occurs, the difference between
H and
TS
represents the amount of available energy released or
absorbed as a result of the chemical change.It is expressed by:
-TS(total) =
H -
TS
The new quantity -TS(total)
is now given a new symbol and a new name. It is called the
Gibbs function of reaction or
Gibbs free
energy and denoted
G. The last equation
therefore becomes:
G =
H -
TS
This energy represents
the driving force of the reaction.
When G is negative the
reaction will be a spontaneous one;
when G is positive the reverse of the reaction as written
will be the spontaneous one;
when G is zero equilibrium exists.
The magnitude of G is a
measure of the extent to which the
reaction will go to completion. Thus, a knowledge of
G values enables us
to predict the course of a reaction. For example, we can
predict that the reaction:
Cl2 (g) + 2I-(aq) --> 2Cl-(aq) + I2(g)
G = -38 kcal
will be a spontaneous one since
G is negative. Had
G been positive, the
spontaneous reaction would have been the reverse of the
reaction as written. -
State symbols
Tab. Thermochemistry (state symbols)
State symbols Example Interpretation (g) H2(g) Gas (l) H2O(l) Liquid (s) Cu(s) Solid (aq) Cu+2(aq) Aqua
Related topics:
Signs and symbols for Molecular Calculator
Signs and symbols for Balance
Example
Ammonium nitrate ,NH4NO3, decomposes explosively:
NH4NO3(s) > N2O(g) + H2O(g)
a) Calculate 
H
for the balanced reaction.
b) If 36 grams of H2O are formed from the reaction, how much
heat was released?
Solution:
a) Insert and calculate: NH4NO3(s) > N2O(g) + H2O(g)
b) Arg. H2O by 36g, insert and calculate:
NH4NO3(s) > N2O(g) + 36gH2O(g)
Here's an example of an equation using all the state symbols (g),(l),(s) and (aq)
CaCO3(s) + H2SO4(aq) > Ca+2(aq) + CO2(g) + SO4-2(aq) + H2O(l)
Software for Chemistry - Home
Second law of thermodynamics: The entropy of the universe increases in the course of every natural change.
Consider the oxidation of iron (25oC).:
4Fe(s) + 3O2(g) --> 2Fe2O3(s)
S =
2(87.37J/(K mol)) - 4(27.29J/(K mol)) - 3(205.2J/(K mol)) =
-550.12J/(K mol)
(H = -1647kJ/mol)The negative enthalpy in this reaction (exothermic reaction), causes the
S(surroundings) to be
positive:S(surroundings) =
-H/T S(surroundings) =
-(-1647 kJ/mol)/(298.15 K) = +5524 J/(K mol) This increase in entropy arises because the reaction releases energy into the surroundings.
In contrast to oxidation of iron, the reaction: N2O4(g) --> 2NO2(g)
is an endothermic reaction (
H = +57.2 kJ/mol):S =
2(240.1J/(K mol)) - 304.38J/(K mol) = 175.82J/(K mol)S(surroundings) =
-(57.24 kJ/mol)/(298.15 K) = -192 J/(K mol) All endothermic reactions decrease the entropy of the surroundings.
The sum of the total entropy changes
S(total) =
-H/T +
S(surroundings)
is a measure of the total change of energy in the universe.For every 4 mol of Fe converted to 2 mol Fe2O3 there is an overall increase in the entropy of the universe:
S(total) = +5524J/(K mol)
+ (-550.12J/(K mol)) = +4973.88 J/(K mol)This large positive quantity indicates that the reaction is spontaneous and a thermodynamic reason why steel corrodes.
The dissociation of N2O4 indicates that the total entropy change:
S(total) = -192J/(K mol)
+ (175.82J/(K mol)) = -16.18 J/(K mol)Because this is a small, negative value, we can conclude that this reaction does not go to completion.
Gibbs Free Energy.
When a reversible chemical change occurs, the difference between
H and
TS
represents the amount of available energy released or
absorbed as a result of the chemical change.It is expressed by:
-TS(total) =
H -
TS
The new quantity -TS(total)
is now given a new symbol and a new name. It is called the
Gibbs function of reaction or
Gibbs free
energy and denoted
G. The last equation
therefore becomes:
G =
H -
TS
This energy represents
the driving force of the reaction.
When G is negative the
reaction will be a spontaneous one;
when G is positive the reverse of the reaction as written
will be the spontaneous one;
when G is zero equilibrium exists.
The magnitude of G is a
measure of the extent to which the
reaction will go to completion. Thus, a knowledge of
G values enables us
to predict the course of a reaction. For example, we can
predict that the reaction:
Cl2 (g) + 2I-(aq) --> 2Cl-(aq) + I2(g)
G = -38 kcal
will be a spontaneous one since
G is negative. Had
G been positive, the
spontaneous reaction would have been the reverse of the
reaction as written. -
State symbols
Tab. Thermochemistry (state symbols)
State symbols Example Interpretation (g) H2(g) Gas (l) H2O(l) Liquid (s) Cu(s) Solid (aq) Cu+2(aq) Aqua
Related topics:
Signs and symbols for Molecular Calculator
Signs and symbols for Balance
Example
Ammonium nitrate ,NH4NO3, decomposes explosively:
NH4NO3(s) > N2O(g) + H2O(g)
a) Calculate H
for the balanced reaction.
b) If 36 grams of H2O are formed from the reaction, how much
heat was released?
Solution:
a) Insert and calculate: NH4NO3(s) > N2O(g) + H2O(g)
b) Arg. H2O by 36g, insert and calculate:
NH4NO3(s) > N2O(g) + 36gH2O(g)
Here's an example of an equation using all the state symbols (g),(l),(s) and (aq)
CaCO3(s) + H2SO4(aq) > Ca+2(aq) + CO2(g) + SO4-2(aq) + H2O(l)
Software for Chemistry - Home
| State symbols | Example | Interpretation |
|---|---|---|
| (g) | H2(g) | Gas |
| (l) | H2O(l) | Liquid |
| (s) | Cu(s) | Solid |
| (aq) | Cu+2(aq) | Aqua |
Related topics:
Signs and symbols for Molecular Calculator
Signs and symbols for Balance