Thermochemistry deals with determining quantities of heat produced or absorbed by a reaction both by measurement and by calculation. It rests on Lavoisier and Laplace's law and on Hess's law:
Lavoisier and Laplace's law: The energy change accompanying any reaction is equal and opposite to energy change accompanying the reverse process
Hess' law: The energy change accompanying any reaction is the same whether the process occurs in one or in several steps
These statements preceded the first law of thermodynamics (1845) and helped in its formulation.
Entropy changes ΔS and Thermodynamic Equilibrium – Solved Examples
Entropy Changes ΔS and Thermodynamic Equilibrium – Solved Examples
In a previous post entitled “Entropy, Free Energy and Thermodynamic Equilibrium” the Boltzmann definition of entropy was given and how entropy changes ΔS are associated with chemical processes was discussed. Below, some examples are given regarding entropy changes ΔS and chemical reactions.
Choose the reaction expected to have the greatest increase in entropy:
a) N2(g) + O2(g) ———› 2 NO(g)
b) H2O (l) ———› H2O (g)
c) 2 XeO3(s) ———› 2 Xe(g) + 3 O2(g)
d) C(s) + O2(g) ———› CO2 (g)
The reaction with the greatest increase in the moles of gas will have the greatest increase in entropy. Answer (c) is correct. In general, when a reaction involves gaseous molecules and the number of gaseous products (or moles of gaseous products) is greater than the number of molecules of gaseous reactants (or moles of gaseous reactants) the entropy change ΔS increases (ΔS > 0).
More gaseous molecules means more possible configurations and therefore a greater probability to occur.
Predict the sign of ΔS° for the oxidation of SO2 in air:
2 SO2(g) + O2(g) ———› 2 SO3(g)
Three molecules of gaseous reactants produce 2 molecules of gaseous products. The number of gaseous products is less than the number of the reactants and therefore ΔS < 0