|The Third Law of
|Predicting S for Physical and Chemical Changes|
|It is often a relatively simple matter to
predict whether a particular change in a reaction will cause the energy of
the reactants to become more spread out (have greater entropy) or less spread
out (have lesser entropy). One of the things to look for is the state
of the reactants and products. Gases have more entropy than liquids which
have more than solids. During chemical reactions, the freedom of movement
of the atoms often changes because of changes in the complexity of the molecules.
For example, consider the reaction
|Among the reactants are six atoms combined
into two molecules of NO2. Among the products, these same six
atoms are confined to one molecule. In the reaction vessel, dividing the
energy of the six atoms between two molecules allows their energy to be
spread out more in their free movement than when the atoms are combined into
just one molecule. Therefore, we can conclude that as this reaction occurs,
there is an entropy decrease.
|Two general rules for predicting entropy changes.
|1. Look at the states first. (gases
|2. If both states are the
same then look at the number of moles of reactants and products and decide
if there has been an increase in the number of moles or a decrease.
|The entropy of a pure crystal
is zero at absolute zero.
|The entropy of a substance (i.e; the extent
of its dispersal of energy at a given temperature) varies with the temperature
of the substance. The lower the temperature, the lower the entropy.
|For example, at a pressure of 1 atm and a
temperature above 100oC, water exists as a
|Yet even in the crystalline form, the water
molecules still have some entropy. There is enough thermal energy left to
cause them to vibrate within the general area of their lattice sites. Thus
at any particular instance, we would find the molecules near, but probably
not exactly at, their lattice positions. If we cool the solid further, we
decrease the thermal energy and the molecules spend less time away from their
lattice positions. The amount of energy and its spreading out decreasese and
the entropy decreases also. Finally, at absolute zero, the ice will be in
a state of absolutely minimal energy of molecular movement and its entropy
will be zero. This leads us to the statement of the Third Law of Thermodynamics.
At absolute zero, the entropy of a pure crystal is also zero.
| i.e. S = 0
at T = 0K
|Because we know the point at which entropy
has a value of zero, it is possible by measurement and calculation to determine
the actual amount of entropy that a substance possesses at temperatures
above 0 K. If the entropy of one mole of a substance is determined at a temperature
of 298 K (25oC) and a pressure of 1 atm, we call it the standard
entropy, So. In your databook there is a listing of standard entropies
next to the standard enthalpies.
********* IMPORTANT *********
Entropy has units of J/K degree J/K not kJ/K!!!!!
|Once we know the entropies of a variety of
substances we can calculate the standard entropy change, ΔSo, for
chemical reactions in much the same way that we calculated Ho.
= (sum of So of products) - (sum of So of reactants)
|The values of ΔSfo
have not been calculated for you. If you need them, then you must calculate
then from the values of So.
|Urea (from urine) hydrolyzes slowly in the
presence of water to produce ammonia and carbon dioxide.
|What is the standard entropy change, in J/K,
for this reaction when 1 mole of urea reacts with water?
= [CO2(g) + (2)NH3(g)] - [CO(NH2)2(aq)
= [213.6 J/K + (2)192.5 J/K] - [173.8 J/K + 96.96 J/K]
= (598.6 J/K)-(243.8 J/K)
= 354.8 J/K