Group 14
CHAPTER EIGHT
REACTIONS INVOLVING
GASES
8.1 Introduction
As result of no interactive forces among the particles of perfect gases,
no heat effects occur when two (or more) of such gases are mixed
together.
On the other hand, when H2 and O2 gases are mixed in presence of a
suitable catalyst, the heat released is of considerable magnitude.
The thermodynamics of such a system can be treated in either of two
ways: either the mixture can be regarded as being a highly non-ideal
gas mixture of H2 and O2, the thermodynamic equilibrium state of
which at given T and P can be described in terms of the fugacities of H2
and O2; or the H2 and O2 have reacted with one another to some
extent to give use to the physical appearance of H2O. The latter, is by
far the more convenient one.
8.2 Reaction Equilibrium in Gas Mixture to form
Reaction's Products
Consider the process:
A(g) + B(g) = 2C(g) ,
and by starting with 1 mole of A and 1 mole of B;
and as 1 atom of A reacts with 1 atom of B to
produce 2 atoms of C, stociometric conditions, and
assuming that
the extent of the process at time t is y then at
time t:
n(A) = n(B) = 1-y
where n(A) and n(B) are the number of moles
of A and B, respectively, at time t.
Consider that the process occurs at constant T and P, thus at
any moment during the process, the integral free energy of
the system is given by:
G' = n(A) G(A) + n(B) G(B) + n(C) G (C)
Let us consider the following setting:
A
B
C
n at t = 0
1
1
0
n at t = t
n
n
2(1-n)
mole fraction (x)
n/2 n/2
(1-n)
extent of the process
since:
(1-n) (1-n)
--
G(i) = Go(i)+ RTlnx(i)P
total
2
2
--
--
thus, if the pressure=1 atm, the free energy of the system
can be given by:
G' = n(Go(A)+ RT
(A)) + n (Go(B)+RT
+2(1-n) (Go(B (C)+RT
(B))
= nGo(A) + nGo(B) + 2(1-n) Go(C) +2n RT
ln (n/2)+2(1-n)RT ln (1-n)
= n Go A + n Go(B) + 2Go(C) – 2n Go(C) + 2n RT
ln (n/2)+2(1-n)RT ln (1-n)
therefore:
G' – (Go(A) + Go(B)) =(1-n)(2Go(C) - Go(A) - Go(A)) + 2RT(n
ln(n/2) + (1-n) ln (1-n)
or: ΔG = (1-n) ΔGo(R) + 2RT(n ln (n/2) + (1-n) ln (1-n))
The right-hand side of the previous equation has two
terms, the first, (1-n) ΔG0(R), represents the standard
free energy change due to the chemical reaction, and
the second term 2RT [ n ln n/2 + (1-n) ln (1-n) ]
represents the free energy change due to mixing of
the gases.
Figure (8.1) shows the variation of the contribution to
the free energy change due to the chemical reaction
(line II), the contribution to the free energy change
due to gas mixing (curve III) and the sum of these two
contribution ( curve I ) with the extent of the reaction:
A(g) + B(g) = 2C(g)
for which ΔG0(R) = -1000 cal at 500K
Figure (8.1): the variation of the contribution to the free
energy change for the reaction A+B=2C for
which ΔG0(R)= -1000 cal at 500K.
•The equilibrium point of the previous reaction occurs at the
minimum point of curve 1, point S, which can be obtained at the
condition of (∂ΔG/∂n)T,P= 0 ; this means that:
-ΔGo(R) + 2RT(1+ln(n/2)-1-ln(1-n))=0
i.e.
ΔGo(R) = -RT ln (1-n)2/(n/2)2
or
ΔGo(R) = -RT ln [X(C) /x (A) x (B)]
if P is constant, thus:
ΔGo(R) = -RT ln [P(C)2 /P(A) P(B)]
in general as:
G' = n(A) G(A) + n(B) G(B) + 2(1-n(A)) G (C)
and by differentiation and setting (∂G'/∂n)T,P= 0 ; we
have:
at equilibrium, G(A)+G(B)-2G(C)=0
i.e. the reaction equilibrium occurs at
Go(A) + RT ln a(A) + Go(B) +RT ln a(B)= 2 Go(C) + 2RT ln a(C)
where:
a(i) =P(i) / P0(i)
ΔGo(R) = -RT ln (a2(C) /a(A) a(B))eq
= -RT ln (P2(C)/P(A) P(B))eq
= -RT ln (P2(C)/P(A) P(B))
and hence:
ΔGo(R) = -RT ln Kp
thus;
Kp
= ( P (C)2/P(A) P(B) )
where Kp is known as the equilibrium constant.
thus;
8.4 The Effect of Temperature on the Equilibrium Constant
(van't Hoff equation)
•The Gibbs – Helmholtz equation states that:
 (  G /T )
0

  H
  (1 / T )  p
0
As
ΔGo = -RT lnKP
then
(∂ ln K/∂(1/T))P= -ΔHo/R
•Thus, for endothermic reaction, the equilibrium constant increases with
increasing temperature, i.e, the equilibrium shift in that direction which involves
absorption of heat, i.e. in the forward direction.
•Conversely, for exothermic reaction, the equilibrium constant
decreases with increasing temperature, i.e. the equilibrium shift in
that direction which involves the absorption of heat, i.e. in the
backward direction.
It is clear now that the variation of Kp with temperature obeys
the le-Chatelier's principle, i.e if heat is added to a system,
the reaction equilibrium is displaced in that direction which
involves absorption of heat
van't Hoff equation shows also that if ΔHo is independent of
temperature, then In Kp varies linearly with (1/T) which agrees
with the experimentally obtained Arrhenius equation