0
$\begingroup$

I have tried heating $\ce{PbCO3}$ at $\pu{400 ^\circ C}$ temperature to form $\ce{Pb3O4}$. After sometime of heating, only a layer of $\ce{Pb3O4}$ is made and the reaction does not complete. The substance stays white even after hours of heating.

$\endgroup$
2
  • 2
    $\begingroup$ 1. More heat. 2. If a crust forms, it prevents oxygen getting in. Agitate and crush. $\endgroup$ Feb 27 at 23:15
  • 1
    $\begingroup$ Crust means partial oxidation. Yes, more heat and more reaction time. $\endgroup$ Feb 28 at 3:03

1 Answer 1

2
$\begingroup$

According to your post, you have tried heating $\ce{PbCO3}$ at temperature $\pu{400 ^\circ C}$ to form Red Lead, $\ce{Pb3O4}$. You haven't give how long you have heat, but mentioned the substance $(\ce{PbCO3})$ stays white even after hours of heating. I think your starting material is wrong in this instance (it should be beneficial if you have given us what literature procedure you have been following).

According to studies done in Ref.1, your starting material should be Basic Lead White $\ce{\left(2PbCO3·Pb(OH)2\right)}$ instead of $\ce{PbCO3}$ itself. Accordingly,

The thermal decomposition of basic lead(II) carbonate $\ce{2PbCO3·Pb(OH)2}$ in static air under atmospheric conditions has been studied for the first time using a combination of thermoanalytical, X-ray diffraction and Raman spectroscopic techniques. New intermediate compounds of the type $\ce{4PbCO3·3PbO}$ and $\ce{PbCO3}$ have been observed in addition to previously reported ones of the type $\ce{2PbCO3·PbO}$, $\ce{PbCO3·PbO}$ and $\ce{PbCO3·2PbO}$. Post-decomposition studies have identified the presence of the tetragonal-to-orthorhombic phase transition in $\ce{PbO}$. Results are also presented for the decomposition processes in nitrogen and oxygen atmospheres. Oxidative decomposition studies have shown that $\ce{Pb3O4}$ forms in oxygen and in flowing air.

According to the work described in here and here, it is clear that the starting material for this conversion is better to be basic lead(II) carbonate, $\ce{2PbCO3·Pb(OH)2}$ and temperature should be around $\pu{475 ^\circ C}$ (the heating should be for longer time period).


References:

  1. Dan A. Ciomartan, Robin J. H. Clark, Lachlan J. McDonald, and Marianne Odlyha, "Studies on the thermal decomposition of basic lead(II) carbonate by Fourier-transform Raman spectroscopy, X-ray diffraction and thermal analysis," J. Chem. Soc., Dalton Trans. 1996, 3639-3645 (DOI: https://doi.org/10.1039/DT9960003639).
$\endgroup$
2
  • $\begingroup$ the substance I have used is PbCO3 because when I overheat it, it turns to a dark substance (PbO2). and I have tested all of the reaction period such as 30mins ,1 hour . after a layer of Pb3O4 is made it turns to yellow when I continue heating $\endgroup$
    – Jak
    Feb 28 at 13:35
  • 1
    $\begingroup$ You should have included these observations in your question at first place. that is important to reader to give you their thoughts. It is still not too late to do so. $\endgroup$ Feb 29 at 15:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.