Page 40 - Chemical equilibria Volume 4

16 Chemical Equilibria
Thus, by using the symmetry of the characteristic matrix:
Q = T ∑ ν k k S + ∑ ν μ k [1.64]
P
k
k k
However, we know that we have:
μ = G k = H k − T k S [1.65]
k
For the heat of reaction we find:

∑
Q = ν k H k = Δ r H [1.66]
P
k
Thus, the heat of transformation, at constant temperature and pressure, is
equal to the enthalpy associated with the reaction.
NOTE 1.3.– For a long time, chemists counted the heat released by an
exothermic reaction positively. With this old convention, relation [1.66] was
transformed into:

Q =− Δ H
P
r
1.5.2. Heat of transformation at constant volume and
temperature

Let us look again at expression [1.58]. By choosing the set of variables
(T, V, ξ), we obtain:
S ∂ S ∂ S ∂
δ QT dT T+ dV T+ dξ − T d S [1.67]
=
∂ T ∂ V ξ ∂ i

At constant volume and temperature we have:

S ∂
δ Q = T dξ − T d S [1.68]
V
ξ ∂ i