If we consider a reaction occurring at constant pressure and temperature , we know that enthalpy represent the net heat released or absorbed by the system and internal energy accounts for both heat and expansion work. ( Further reading )
suppose we perform a reaction H2(g) + O2(g) —> H2O(g) at constant temperature 25°C and pressure 1 atm
• Bond energy of H2 ≈ 436 kj/mol
• Bond energy of O2 ≈ 498 kj/mol
• Bond energy of O-H bond in H2O(g) ≈ 464 kj /mol
• ∆H = Energy required to break the bonds of H2 and O2 - Energy released in formation of water molecule -
= (436 + 498/2) - (2 × 464) = -243 kj
• Expansion work = Wexp = ∆(ng)RT ≈ +1.2 kj/mol
• ∆U = ∆H - ∆PV = ∆H - (∆ng)RT = -241.8 kj
Data is taken from : Question out of syllabus
As we can see here enthalpy is evaluated just by calculating difference between bond energies and kinetic energy of reactants and products are not included i.e. enthalpy is not related to kinetic energy.
Now I want to know that system absorbed some expansion work energy but in what form does it store it within itself i.e. kinetic energy or potential energy.
In the book Chemical Physics Electrons and Excitations By Sven Larsson · 2012 page 146 Mr Sven writes that we the term pv ( which accounts for expansion work) takes care of kinetic energy i.e. expansion work is equal to change in kinetic energy during reaction.
But problem is that the expression of kinetic energy and expansion work although looks like similar but have a difference of factor 3/2 as kinetic energy change at constant temperature ( According to kinetic gas theory ) is equal to 3/2(∆ng)RT but expansion work is equal to (∆ng)RT.
Could someone prove that change in kinetic energy in a reaction ( occurring at constant pressure and temperature) is equal to Expansion work ?