0
$\begingroup$

In the reaction $$\ce{C_{(s)} + O2_{(g)} -> CO2_{(g)}} $$ one mole gaseous reactant gives one mole gaseous product, so there is no significant change in entropy. It can be verified by line of $\ce{C}$ gives $\ce{CO2}$ in Ellingham diagram being almost horizontal.

But oxygen has a degree of freedom 5 and at higher temperature $\ce{CO2}$ has 9. So there should be an increase in extent of random motion, leading to significant positive entropy change. Why is it not the case ? Does degree of freedom not affect entropy change of a reaction greatly ? Why so ?

$\endgroup$
5
  • $\begingroup$ What about the change in chemical potential in going from C and O2 to CO2? $\endgroup$ Jun 13 at 12:02
  • $\begingroup$ Yes number of dof is important, and the entropy will be similar but not identical for both gasses. Using the Sakur-Tetrode eqn. for translational entropy, this depends on $\ln(mass)$ and rotational entropy on $\ln(\text{moment of inertia})$ so small changes between O2 and CO2. Vibrational contribution from oxygen is negligible at room temperature and larger but still small for CO2, this by population of vibration from Boltzmann eqn. Btw, try not to think of energy as random motion but as number of positions a molecule can take or the number of energy levels it can possibly be in. $\endgroup$
    – porphyrin
    Jun 13 at 12:19
  • $\begingroup$ @porphyrin thanks, although I don't know the equations you have mentioned . So the effect is small in this reaction... but in some other case, can there be a significant effect of dof ? $\endgroup$
    – Sagnik
    Jun 13 at 16:06
  • $\begingroup$ And you meant entropy , not energy , in last sentence, right ? $\endgroup$
    – Sagnik
    Jun 13 at 16:08
  • $\begingroup$ yes my typo it should be 'try not to think of entropy as random motion'. You will find the equations in most phys. chem. textbooks, usually in sections dealing with statistical mechanics. You will also know that the heat capacity of O2 and CO2 differ and that $S(T)=S_0+\int_0^T C_p/T dT$ so that way also you expect the entropy to differ. To use this eqn. you have to know how $C_P$ varies with temperature . $\endgroup$
    – porphyrin
    Jun 13 at 17:49

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.