Thermochemistry
Chemistry
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Introduction
Every chemical change and every physical change involves an energy
change. These energy changes that accompany the transformation of
matter help us understand better the nature of chemical and physical
changes.
First law of thermodynamics : Energy can be neither created nor
destroyed.
Thermodynamic Quantities
Chemical thermodynamics is
the study of the energy effects accompanying chemical and physical
changes. Three of these thermodynamic quantities are: enthalpy,
H, entropy, S, and Gibbs free energy,
G.
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 |
CHEMIX
School - Thermochemistry
Related topics:
Signs and symbols for
Molecular
Weight Calculator
Signs and symbols for
Balancing Chemical Equations
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)
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