Carbon monoxide reacts with oxygen to form
carbon dioxide by the following reaction:
2 CO(g) + O2(g) → 2 CO2(g)
∆H for this reaction is −566 kJ/mol rxn.
What is the ∆H for CO(g)?
To find the ∆H for CO(g), we can use the fact that the reaction given is the formation of carbon dioxide from carbon monoxide and oxygen. The reaction equation is:
2 CO(g) + O2(g) → 2 CO2(g)
The ∆H for this reaction is given as -566 kJ/mol rxn, which represents the enthalpy change for the formation of two moles of carbon dioxide.
We want to find the ∆H for the formation of one mole of CO(g), so we need to divide the given ∆H by the stoichiometric coefficient of CO. In this case, the stoichiometric coefficient is 2, since there are two moles of CO in the reaction equation.
Therefore, the ∆H for CO(g) would be:
∆H for CO(g) = (-566 kJ/mol rxn) / 2
∆H for CO(g) = -283 kJ/mol
So, the ∆H for the formation of carbon monoxide (CO(g)) is -283 kJ/mol.
To determine the ΔH for CO(g), we can use the stoichiometry of the reaction and the given ΔH value.
The balanced equation for the reaction is:
2 CO(g) + O2(g) → 2 CO2(g)
From the equation, we can see that 2 moles of CO(g) react to form 2 moles of CO2(g). Therefore, the enthalpy change for the reaction is for two moles of CO(g).
Given that ΔH for the reaction is -566 kJ/mol rxn, we can write:
ΔH = -566 kJ/mol rxn
To find the enthalpy change for CO(g), we divide the ΔH by the stoichiometric coefficient of CO(g), which is 2:
ΔH for CO(g) = (-566 kJ/mol rxn) / 2 = -283 kJ/mol
Therefore, the ΔH for CO(g) is -283 kJ/mol.
See this post.
http://www.jiskha.com/display.cgi?id=1350618193