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.

enter image description here

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 ?

  • $\begingroup$ Why do not you use enthalpies of formation instead of bond energies? $\endgroup$
    – Poutnik
    Aug 6 at 10:41
  • $\begingroup$ Standard enthalpy of formation of H2 = 0. Standard enthalpy of formation of O2 = 0. Standand enthalpy of formation of H2O = −241.818 $\endgroup$ Aug 6 at 10:56
  • $\begingroup$ 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. $\endgroup$
    – Poutnik
    Aug 6 at 11:04
  • $\begingroup$ But reaction described here is at low temperature 25°C $\endgroup$ Aug 6 at 11:08
  • $\begingroup$ 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. $\endgroup$
    – Poutnik
    Aug 6 at 11:16

1 Answer 1


While doing some more research and collecting data , I found ( Quantum Bits and Quantum Secrets page 128 ) that average kinetic energy of a particle at room temperature is equal to 4 × 10 ^ (-21) joule so now if we further calculate energy of one of gas at room temperature that will be equal to 4 × 6.023 × 100 j ≈ 2.4 kj.

enter image description here

Net kinetic energy change of reaction ≈ Change in no of moles × energy of 1 mole ≈ (1/2) × 2.4 ≈ 1.2 kj

To my wonder this is exactly same as volume work done by surrounding to system.

Comment if there is anything wrong!

  • 1
    $\begingroup$ Taking 1mol of He, being heated isobarically by 1 K, the provided heat is 5/2R J = 3/2 R J to increase He thermal energy + R J of heat is converted to volume work. $\endgroup$
    – Poutnik
    Aug 6 at 12:10
  • $\begingroup$ @Poutnik last question, according to kinetic theory of gas kinetic energy of a molecule is 3/2kt whereas as I described above that kinetic energy of a atom at room temperature is kt , which is correct ? $\endgroup$ Aug 6 at 12:55
  • 1
    $\begingroup$ Neither is correct. // The mean translational kinetic energy of gaseous molecules or atoms is 3/2kT. But in case of molecules, it is not the only form of their kinetic energy. They have another 2/2kT (linear ones) or 3/2kT of rotational energy (except of very low T where they progressively stop rotate due rotation quantization) and at higher T also progressively (due vibration quantization) 2/2kT for each vibration. At very high T comes also excitation and ionization. Generally, in classical thermodynamics, each degree of freedom is covered by 1/2kT of energy contribution. $\endgroup$
    – Poutnik
    Aug 6 at 13:10

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