Nernst Heat Theorem
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Describe Nernst Heat Theorrem.
How to find a quantitative relationship between ∆G and ∆H
in a chemical reaction?
Nernst Heat Theorem:
In the earlier development of chemical thermodynamics, the
difficult task was to ascertain a quantitative relationship between ∆G and ∆H
in a chemical reaction and to find out ∆G from thermal data. i.e., from ∆H. To
deal with this problem a preliminary discussion will be given which will deal
with this problem ultimately leading to the Nernst heat theorem and third law
of thermodynamics.
1. Joule and Thomson’s
concept: According to this concept, the two terms ∆G and ∆H are
identical. To support their statement they gave well known example of Daniel
cell where ∆G and ∆H had been found to be the same.
2. Berthelot’s concept:
Berthelot came forward to give his famous concept as:
“When
heat is given out in a reaction, the free energy of the system decreases.”
Although this concept described relation
between ∆G and ∆H in a qualitative manner and was found to be true in a number
of examples such as condensed systems at ordinary temperatures, this system got
failed in a number of other cases.
3. Gibb’s Helmholtz concept: For
the first time Gibb’s and Helmholtz deduced a quantitative relation between ∆G
and ∆H for a chemical reaction when it is carried out at constant pressure and
called it Gibb’s Helmholtz equation.
∆G = ∆H +T[δ(∆G)/δT]P eq……(1)
Case I :
When δ(∆G)/δT = 0 eq (1) becomes
∆G = ∆H
An example of this case is found in Daniel
cell where ∆G is found to be the same as ∆H.
Case
II
: When ∆G ≈ 0 eq (1) becomes as follows
∆H = -T[δ(∆G)/δT]V
From the above equation it follows that ∆H
must acquire a large value. An example of this is the fusion of solids at their
melting point where ∆G is almost zero and ∆H is very large.
Case
III
: When T =0K eq (1) yields,
∆G = ∆H
This condition has little practical
importance because it is impossible to carry out a process at the absolute
zero.
Except in the above three cases an inequality
exists between ∆G and ∆H in all the processes
Equation (1) therefore allows the
calculation of ∆H from ∆G. But the reverse seems too to be practicable. Thus the
main limitation of Gibb’s Helmholtz equation is that it does not allow the
calculation of ∆G from ∆H.
4. Richard’s concept: In
1902 Richard measure the EMF of cells at low temperature and concluded that δ(∆G)/δT
gets decreased gradually with the lowering of temperature. In other words ∆G
and ∆H approach each other more closely at extremely low temperatures i.e.
LimT→0 ∆G
=LimT→0 ∆H
5. Nernst Heat Theorem: In
1906 Nernst relied on data given by Richard and made a bold statement for a
process in condensed system that the value of δ(∆G)/δT approaches zero
asymptotically as the absolute zero is approached. In other words ∆G and ∆H curves
meet each other at a short region above absolute zero and run coinciding with
each other up to absolute zero. This behavior is shown by full lines not dotted
lines. The dotted line curve reveals that the two values ∆G and ∆H not only
become equal to each other at absolute zero but their approach to each other
becomes rapid and not gradual.
Mathematically it can be written as
LimT→0[δ(∆G)/δT]P
= LimT→0[δ(∆H)/δT]P
= 0 eq……(2)
Nernst heat theorem can also be explained as “There is no
change in entropy of the system when the process occurs in the vicinity of absolute
zero”. This statement can be proved using equation
dG = VdP – SdT
At constant pressure we have
[δG/δT]P = -S
For a process G and S can be replaced by ∆G and ∆S
respectively, hence
[δ(∆G)/δT]P = -∆S
According to eq (2) left hand side of the equation will
become zero at 0 K, hence
LimT→0
∆S = 0
LimT→0 (S2 – S1) =0
This
implies that when T =0, then S1 = S2.
Nernst heat theorem can be used for the study of variation
in molar specific heat at constant pressure for the process occurring in the neighborhood
of absolute zero. The molar heat capacity at constant pressure can be defined
as
CP = (δH/δT)P
For the change in H and Cp, the above equation
can be written as
∆CP =[δ(∆H)/δT]P
In the vicinity of absolute zero, the above expression can
be written as
LimT→0(∆CP)
= LimT→0[δ(∆H)/δT]P
According to Nernst heat theorem the right hand side of the
equation is equal to zero, therefore
LimT→0(∆CP)
= 0
LimT→0(CP2-CP1) = 0
LimT→0CP2 = LimT→0CP1
According to above equation the molar heat capacity at constant
pressure during any transformation at absolute zero remains unchanged.
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