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Question: The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283 \[kJmo{l^{ - 1}}\] re...

The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283 kJmol−1kJmo{l^{ - 1}} respectively. The enthalpy of formation of carbon monoxide per mole is:
A.-676.9 kJmol−1kJmo{l^{ - 1}}
B.676.5 kJmol−1kJmo{l^{ - 1}}
C.110.5 kJmol−1kJmo{l^{ - 1}}
D.-110.5 kJmol−1kJmo{l^{ - 1}}

Explanation

Solution

Enthalpy of formation can be understood as the amount of energy required to form a given compound while the enthalpy of combustion is the amount of energy released upon combustion of a compound

Complete Step-by-Step answer:
From the given data, we can infer the following:
Enthalpy of combustion of carbon = -393.5 kJmol−1kJmo{l^{ - 1}}
Enthalpy of combustion of carbon monoxide = - 283 kJmol−1kJmo{l^{ - 1}}
The corresponding chemical reaction for the combustion of carbon and carbon monoxide can be given as:
C(s)+O2 (g)→CO2  (g){C_{(s)}} + {O_{2\,(g)}} \to C{O_{2\,\,(g)}}
CO(g)+12O2 (g)→CO2  (g)C{O_{(g)}} + \dfrac{1}{2}{O_{2\,(g)}} \to C{O_{2\,\,(g)}}
Now the quantity we are required to find is the enthalpy of formation of carbon monoxide. The chemical reaction for the same can be given as:
C(s)+12O2 (g)→CO (g){C_{(s)}} + \dfrac{1}{2}{O_{2\,(g)}} \to C{O_{\,(g)}}
From Hess's law we can say that the total change in the enthalpy for any particular reaction is equivalent to the change of enthalpy of formation from the reactants to the products. This can be represented mathematically as:
ΔHreaction=(ΔHproducts)formation−(ΔHreactan⁡ts)formation\Delta {H_{reaction}} = {(\Delta {H_{products}})_{formation}} - {(\Delta {H_{reac\tan ts}})_{formation}}
or,
ΔHreaction=(ΔHreactan⁡t)combustion−(ΔHproduct)combustion\Delta {H_{reaction}} = {(\Delta {H_{reac\tan t}})_{combustion}} - {(\Delta {H_{product}})_{combustion}}
Hence, the total enthalpy change that occurs for the formation of the product, is mathematically equivalent to the enthalpy of combustion product subtracted from the enthalpy of combustion of reactant.
(ΔHproduct)formation=(ΔHreactan⁡t)combustion−(ΔHproduct)combustion{(\Delta {H_{product}})_{formation}} = {(\Delta {H_{reac\tan t}})_{combustion}} - {(\Delta {H_{product}})_{combustion}}
Hence, we can calculate the enthalpy of formation of carbon monoxide as:
(ΔH(CO))formation=(ΔH(C))combustion−(ΔH(CO))combustion{(\Delta H(CO))_{formation}} = {(\Delta H(C))_{combustion}} - {(\Delta H(CO))_{combustion}}
= -393.5 – (- 283) = -110.5 kJ

Hence, Option D is the correct option

Note: “Enthalpy change of formation” can be either positive or negative, since a reaction to form 1 mole of a substance can be either exothermic or endothermic. “Combustion” is always an exothermic process. Due to this, enthalpy change of combustion must always be positive. Also, in endothermic reactions, the change in enthalpy is always negative.