Can sulphur react with calcium carbonate to form sulphur trioxide and calcium carbide, as the energy of 3(S=O) is greater than ((C=O) +2(C-O))?

EDIT: i forgot to say that this is in an oxygen-free environment.

  • $\begingroup$ There is not enough carbon atoms per calcium atom to form CaC2. $\endgroup$ – Francis L. Jul 19 '19 at 22:13
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    $\begingroup$ Room temperature, certainly no reaction. However at temperature extremes > 1200 C, it might be worth checking in the presence of oxygen. $\endgroup$ – M. Farooq Jul 20 '19 at 0:54
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    $\begingroup$ Given the presence of oxygen, things will start to happen way below 1200 °C. Without it, you may have to go that high. But there is no way you'll end up with SO3 and/or CaC2. $\endgroup$ – Ivan Neretin Jul 20 '19 at 7:23

If you mix CaCO3 + S and heat the mix, the first thing to occur will be that sulfur melts. Second, CaCO3 loses CO2 and becomes CaO. Third, sulfur boils off (or has already boiled off).

A thought that perhaps CaS + O2 would form can be eliminated on the basis that roasting metal sulfides in air produces SO2 + the metal oxide. So the metal sulfide is unlikely.

If any reaction had a possibility, it might be 3 CO2 + S --> 3 CO + SO3, but the heats of formation favor 3 CO2 + S by a large margin. The heat of formation of SO3 is about the same as CO2, and the heat of formation of SO2 is even less.

So the final products will be CaO + sulfur + CO2.

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  • $\begingroup$ But if the final products are CaO, CO2 and S, then why wouldn't sulphur reduce calcium oxide to calcium as the bond energy of 3(Ca-O) is far less than that of 3(S=O). (3CaCO3 + S ->Ca + SO3 + 3CO2) $\endgroup$ – klj 1 Jul 21 '19 at 13:50
  • $\begingroup$ If you wind up with CaO and S (all the CO2 is gaseous and gone), It might be possible to arrange conditions where an excess of sulfur takes the oxygen from CaO and gives CaS + SO2. If you can remove all the SO2 produced, any equilibrium would leave you with CaS. Individual bond energies are not as significant as heats of formation of a compound. Bonds are only parts of a compound. $\endgroup$ – James Gaidis Jul 21 '19 at 14:45

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