0
$\begingroup$

The thermal decomposition of siderite equation: $\ce{FeCO3 -> FeO + CO2}$ looks correctly balanced to me, you can notice $\ce{1 Fe , 1 C , 3 O}$ on the Left side and the same number of molecules on the other side.

However my text book writes it as $\ce{2FeCO3 -> 2FeO + 2CO2}$ and I wanted to know why mine is not correct.

$\endgroup$
  • 3
    $\begingroup$ Textbook's version is wrong, as written, because you have to write the smallest possible coefficients, therefore 4A -> 4B+4C is also incorrect for something which should be A -> B+C. $\endgroup$ – M. Farooq Sep 29 '19 at 13:42
  • $\begingroup$ @M.Farooq So FeCO3→FeO+CO2 is the correct answer ? $\endgroup$ – AmirWG Sep 29 '19 at 13:45
  • 1
    $\begingroup$ If the products are carbon dioxide and FeO, then your version is correct. There is not need of an additional 2. It must be a typo in the book. $\endgroup$ – M. Farooq Sep 29 '19 at 13:48
  • 1
    $\begingroup$ @M.Farooq thanks for your time , that helped me a lot. $\endgroup$ – AmirWG Sep 29 '19 at 13:56
  • 1
    $\begingroup$ Please do not use MathJax formatting on the title. $\endgroup$ – Mathew Mahindaratne Sep 29 '19 at 14:21
4
$\begingroup$

Well, just like any other reactions, it depends upon the the heating temperature and the reaction environment. Contrary to what the reaction tells, the decomposition products of the reaction is actually iron(II,III) oxide, $\ce{Fe2O3}$ if it is heated in oxygen environment$\ce{^{[1]}}$ and $\ce{Fe3O4}$ if it is heated in $\ce{CO2}$ environment$\ce{^{[2]}}$. $\ce{FeO}$ also formed during the reaction but for a small period of time. Hence it is an intermediate product.

  1. Figure 1 shows the thermal decomposition of iron carbonate at temperature at 600°C. It is shown from one DTA curve in Fig. 1 one exothermic peak at 495°C accompanied with a weight loss of about 32% (theoretical value 31.1%). This step is due to the decomposition of iron carbonate to $\ce{Fe2O3}$.

    X-ray diffraction patterns of the decomposition products of iron carbonate were determined for the solid calcined at 550 and 800°C Figure 3 (a) shows X-ray diffractogram for the sample dried at room temperature corresponding to iron carbonate. The diffractogram of the sample calcined at 550°C consists of very poorly crystallized α-Fe2O3 phase. .Heating of ferrous carbonateat 800°C led to the formation of well crystalline $\ce{α-Fe2O3}$

  2. In air, the X-ray pattern of $\ce{FeCO3}$ was replaced by that of $\ce{Fe3O4}$ as the sample was heated. The change was essentially complete at 500°C. No further changes were noted. If, however, the furnace atmosphere was changed from air to $\ce{CO2}$ or helium above 1000°C. the sample transformed slowly to $\ce{Fe3O4}$. Restoring the air atmosphere again resulted in the pattern of $\ce{Fe2O3}$. When $\ce{FeCO3}$ was decomposed in a $\ce{CO2}$ atmosphere, the first decomposition product observed was $\ce{Fe3O4}$.[...]the brief appearance of $\ce{FeO}$ was suspected in the decomposition. By oscillating the Geiger counter through a very small arc encompassing the strongest line of FeO, the presence of this oxide was indicated. The line appeared for only a very short time and was at no time very strong. In the inert atmosphere, the products of decomposition were $\ce{FeO}$ and $\ce{Fe3O4}$. The $\ce{Fe3O4}$ in the sample could not be converted to $\ce{FeO}$ even by heating to the highest temperature available with the furnace (about 1150°C. in helium).

So, the reaction indeed forms $\ce{FeO}$ but it gets oxidized depending on the environment the reaction is proceeded.

$$\ce{FeCO3 -> FeO + CO2 }$$

$$\ce{4FeO + O2 -> 2Fe2O3}$$

$$\ce{3FeO + CO2 -> Fe3O4 + CO}$$

References

  1. A.A. El-Bellihi, "Kinetics of Thermal Decomposition of Iron Carbonate", Egypt. J. Chem. 53, No. 6, pp.871- 884 (2010) DOI: https://ejchem.journals.ekb.eg/article_1268_82097e00ce1c4e475f17cfec61edb6d2.pdf
  2. H. E. KISSINGER, H. F. McMURDIE, and B. E. SIMPSON,Thermal Decomposition of Manganous and Ferrous Carbonates,DOI: https://ceramics.onlinelibrary.wiley.com/doi/pdf/10.1111/j.1151-2916.1956.tb15639.x
| improve this answer | |
$\endgroup$
2
$\begingroup$

The thermal decomposition of siderite is interesting aspect. First, the answer to your question is your textbook has mistakenly put a common factor (of 2) to balance the chemical equation. It is not wrong but traditionally we do not put a common factor on equations, unless we intended to add or subtract two of more equations. Said that, I have a doubt that this equation is correct at all, based on literature. Although this must have been too much for the OP, I'd like to make my point.

One such publication reveals the following (Ref.1):

Abstract: In order to provide a better theoretical foundation for utilisation of Xinjiang siderite resources in China, its thermal decomposition behaviour was studied in neutral and oxidising atmospheres by employing thermodynamics analysis, chemical titration, thermogravimetric, and X-ray diffraction means. Isothermal experiments were conducted to investigate the thermal decomposition kinetics of siderite lump in a weakly oxidising atmosphere at $\pu{500–850 ^\circ C}$. The results reveal that siderite has self-magnetisation characteristics under controlled conditions, and the phase evolution process and final products of decomposition depend temperature and atmosphere. The phase transformation process in weak oxidising atmosphere follows the steps as: $\ce{FeCO3 -> Fe3O4 -> \gamma-Fe2O3}$ at $\pu{550 ^\circ C}$, and $\ce{FeCO3 -> FeO + Fe3O4 -> Fe3O4 -> \gamma-Fe2O3 -> \alpha-Fe2O3}$ at $\pu{800 ^\circ C}$. In inert atmosphere, the decomposition pathway is $\ce{FeCO3 -> Fe3O4}$ below $\pu{733 ^\circ C}$ and $\ce{FeCO3 -> FeO + Fe3O4}$ above $\pu{733 ^\circ C}$. The molar ratio of $\ce{FeO/Fe3O4}$ increases with temperature. The decomposition kinetics of siderite lump in oxygen-deficient atmosphere is consistent with chemical reaction control in the temperature range $\pu{500–700 ^\circ C}$ and nucleation and growth mechanism in the $\pu{750–850 ^\circ C}$. The corresponding activation energies are $53.73$ and $\pu{38.15 KJ mol−1}$, respectively.

It is clear from this reference that decomposition products of siderite ($\ce{FeCO3}$) are dependent on the temperature. Yet, the authors have not obtained iron(II) oxide ($\ce{FeO}$) as a sole product. Another recent publication (Ref.2) claimed $\ce{Fe2O3}$ as the final product when $\ce{FeCO3}$ is heated above $\pu{400 ^\circ C}$, based on its thermal decomposition studies in air by DTA-TGA, XRD, SEM, and Mossbauer measurements. The authors have observed a weight loss of about $32 \%$ at the single exothermic peak at $\pu{495 ^\circ C}$. They claimed the theoretical weight loss is $31.1 \%$, thus decomposition should be consistent with the equation: $$\ce{2FeCO3 -> Fe2O3 + CO + CO2}$$

This finding is also consistent with an old Science publication claiming $\ce{FeCO3}$ decomposed to magnetite ($\ce{Fe3O4}$) and graphite when heating in equilibration with graphite at $\pu{455–465 ^\circ C}$ (Ref.3).

All these evidence directed to doubt the simple equation given in OP's textbook.

References:

  1. Y. H. Luo, D. Q. Zhu, J. Pan, X. L. Zhou, “Thermal decomposition behaviour and kinetics of Xinjiang siderite ore,” Mineral Processing and Extractive Metallurgy - Transactions of the Institutions of Mining and Metallurgy: Section C 2016, 125(1), 17-25 (https://doi.org/10.1080/03719553.2015.1118213).
  2. A. A. El-Bellihi, “Kinetics of Thermal Decomposition of Iron Carbonate,” Egypt. J. Chem. 2010, 53(6), 871-884 (DOI: 10.21608/EJCHEM.2010.1268).
  3. Bevan M. French1, P. E. Rosenberg, “Siderite ($\ce{FeCO3}$): Thermal Decomposition in Equilibrium with Graphite,” Science 1965, 147(3663), 1283-1284 (DOI: 10.1126/science.147.3663.1283).
| improve this answer | |
$\endgroup$
1
$\begingroup$

Both equations are balanced, both equations are correct. Using other coefficients such as $0.5, \sqrt{2}$ or ${e^3}$ would also be correct. Only if the ratio of coefficients were different (i.e. not $1:1:1$) it would be wrong.

However, it is commonly preferable to write the smallest possible integers, so $1,1,1$ is preferred over $2,2,2$.

| improve this answer | |
$\endgroup$

Your Answer

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

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