Given $2 C O (g) + O_{2} (g) \to 2 C O_{2} (g)$ $2CO(g) + O_2(g) rarr 2CO_2(g)$ $DeltaH^@=-564kJmol^-1$ ...what is the enthalpy associated with the formation of a mass of $44.0*g$ of carbon dioxide by these means?

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Given $2 C O (g) + O_{2} (g) \to 2 C O_{2} (g)$ $2CO(g) + O_2(g) rarr 2CO_2(g)$ $DeltaH^@=-564kJmol^-1$ ...what is the enthalpy associated with the formation of a mass of $44.0*g$ of carbon dioxide by these means?

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Answer 1 · 2017-01-09T17:52:57

$DeltaH^@=-282*kJ$

Explanation:

$"2CO +O"_2rarr"2CO"_2$ $;DeltaH^@=-564*kJ*mol^-1$

$DeltaH^@$ is always quoted per mole of reaction as written. That is synthesis of 2 moles of carbon dioxide from 2 moles of carbon monoxide and stoichiometric dioxygen has an associated enthalpy change of $-564*kJ$ . Because we combusted 1 mole of carbon monoxide, rather than 2 moles as specified in the equation, the enthalpy calculation is modified and HALVED.........

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Given $2 C O (g) + O_{2} (g) \to 2 C O_{2} (g)$ $2CO(g) + O_2(g) rarr 2CO_2(g)$ $DeltaH^@=-564kJmol^-1$ ...what is the enthalpy associated with the formation of a mass of $44.0*g$ of carbon dioxide by these means?

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Given 2CO(g) + O_2(g) rarr 2CO_2(g)2CO(g)+O2(g)→2CO2(g)2CO(g) + O_2(g) rarr 2CO_2(g) DeltaH^@=-564*kJ*mol^-1DeltaH^@=-564*kJ*mol^-1...what is the enthalpy associated with the formation of a mass of 44.0*g44.0*g of carbon dioxide by these means?

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Explanation:

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Given $2 C O (g) + O_{2} (g) \to 2 C O_{2} (g)$ $2CO(g) + O_2(g) rarr 2CO_2(g)$ $DeltaH^@=-564kJmol^-1$ ...what is the enthalpy associated with the formation of a mass of $44.0*g$ of carbon dioxide by these means?