What is the heat of combustion of 1 kg carbon dioxide w/ magnesium?

In summary, the conversation is about finding the heat of combustion of one kilogram of carbon dioxide with magnesium. The equation 2 Mg(s) + CO2 ---> 2 MgO(s) + C(s) yields a standard enthalpy of -810.1 kJ/mol, with the standard enthalpy of formation of C(s) being zero joules/mol. It is important to distinguish between the standard molar reaction enthalpy, ΔrH°, and the standard molar enthalpy of formation, ΔfH°, which is the heat change accompanying the formation of one mole of compound from the elements at standard state. The standard molar enthalpy of formation for MgO(s) is -601
  • #1
TL;DR Summary
I am trying to find out the heat of combustion of one kilogram of carbon dioxide with magnesium. I am looking for results for stochiometric conditions of these two reactants.
I am trying to find out the heat of combustion of one kilogram of carbon dioxide with magnesium. I am looking for results for stochiometric conditions of these two reactants.

I have found the following equation online

2 Mg(s) + CO2 ---> 2 MgO(s) + C(s) yields a standard enthalpy of --810.1 kJ/mol.

I would think that the C(s) is in its standard state and thus has a standard enthalpy of formation of zero joules/mol. Thus, I would tend to believe that the standard enthalpy of formation of 2 MgO(s) from the precursor magnesium and carbon dioxide is --810.1 kJ/mol.
 
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  • #2
You must clearly distinguish between the standard molar reaction enthalpy, ΔrH°, and the standard molar enthalpy of formation (or standard molar heat of formation), ΔfH°.

ΔrH° = -810.1 kJ/mol is the standard molar reaction enthalpy for the reaction 2Mg(s) + CO2 ---> 2MgO(s) + C(s) .


The standard molar enthalpy of formation of a compound is the heat change accompanying the formation of one mole of compound from the elements at standard state.

For the standard molar enthalpy of formation of MgO(s) one has ΔfH° = -601.7 kJ/mol for the formation reaction Mg(s) + 1/2O2(g) ---> MgO(s).
 
  • #3
Thanks for your reply Lord Jestocost. Now I understand the reaction.
 

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