# Kinetic energy change in reaction

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

Calculation

• ∆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 ?

• Why do not you use enthalpies of formation instead of bond energies? Commented Aug 6, 2023 at 10:41
• Standard enthalpy of formation of H2 = 0. Standard enthalpy of formation of O2 = 0. Standand enthalpy of formation of H2O = −241.818 Commented Aug 6, 2023 at 10:56
• 3/2nRT is ideal gas thermal energy(=translational kinetic energy in this case) only for monoatomic gases or for very low T of polyatomic ones. For low enough T it is 5/2nRT for biatomic and linear triatomic ones, 6/2nRT for other triatomic ones, due 2 or 3 axis of rotation. For higher T, there are being progressively added energies of vibration modes, up to nRT each. Commented Aug 6, 2023 at 11:04
• that is enough for rotation,not for vibration. H2 and O2 5/2RT, H2O 6/2RT. (All not exactly,but approximately, you can search for tables or charts of molar heat capacities = f(T) ) // See also en.wikipedia.org/wiki/Molar_heat_capacity // Expansion work is equal to the difference between system heat capacities at constant pressure and constant volume. Commented Aug 6, 2023 at 11:16
• Select wisely what you ask about, as the last resort choice. Frequent asking is demanding on resources of others and some users could get tired and annoyed by it. Most such things have been already described by many ways on many places, few keywords and mouse clicks away. Bring rather just particular details from such resources you do not understand even after failed effort. Commented Aug 6, 2023 at 11:21