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Question: The enthalpies of formation of \(CO\) and \(C{O_2}\) are \( - 110.5\,kJ\,mo{l^{ - 1}}\) and\( - 393....

The enthalpies of formation of COCO and CO2C{O_2} are −110.5 kJ mol−1 - 110.5\,kJ\,mo{l^{ - 1}} and−393.5 kJ mol−1 - 393.5\,kJ\,mo{l^{ - 1}} respectively. The enthalpy of combustion of carbon monoxide is:
(A) −110.5 kJ mol−1 - 110.5\,kJ\,mo{l^{ - 1}}
(B) −283.0 kJ mol−1 - 283.0\,kJ\,mo{l^{ - 1}}
(C) 504 kJ mol−1504\,kJ\,mo{l^{ - 1}}
(D) 221.2 kJ mol−1221.2\,kJ\,mo{l^{ - 1}}

Explanation

Solution

According to the question, first we have to define the enthalpies of given compounds and then denote the enthalpies of both the compound symbolically. Lastly, calculate the final enthalpy of combustion of Carbon Monoxide by adding the enthalpies of both the given compounds.

Complete step-by-step solution: Enthalpy of any compound is a thermodynamic quantity equal to the internal energy of a system plus the product of its volume and pressure, “enthalpy is the amount of energy in a system capable of doing mechanical work” heat content, total heat, H.
And, the enthalpy of combustion of any compound is the addition of enthalpies of formation of two compounds. Enthalpy changes of combustion reactions are used in industrial heating and in rocket fuels and in domestic fuels.
Enthalpy of formation of COCO = −110.5 kJ mol−1 - 110.5\,kJ\,mo{l^{ - 1}}
Enthalpy of formation of CO2C{O_2} = −393.5 kJ mol−1 - 393.5\,kJ\,mo{l^{ - 1}}
C+O2→CO2C + {O_2} \to C{O_2}
∴ ΔHc=−393.5 kJ mol−1\therefore \,\Delta {H_c} = - 393.5\,kJ\,mo{l^{ - 1}}
Here, ΔHc\Delta {H_c} is the enthalpy of the CO2C{O_2} .
Now,
Formation of Carbon Monoxide:
C+12O2→COC + \dfrac{1}{2}{O_2} \to CO
∴ Hc′=−110.5 kJ mol−1\therefore \,H{'_c} = - 110.5\,kJ\,mo{l^{ - 1}}
Here, Hc′H{'_c} is the enthalpy of COCO .
So, the enthalpy of combustion of Carbon Monoxide is as below:
ΔH=ΔHc+ΔHc′\Delta H = \Delta {H_c} + \Delta H{'_c}
⇒ΔH=−393.5−(−110.5)\Rightarrow \Delta H = -393.5-(-110.5)
⇒ΔH=−393.5+110.5\Rightarrow \Delta H = -393.5+110.5
∴ΔH=−283.0 kJ mol−1\therefore \Delta H = - 283.0\,kJ\,mo{l^{ - 1}}
Here, ΔH\Delta H = Enthalpy of combustion of Carbon Monoxide.
ΔHc\Delta {H_c}= Enthalpy of the CO2C{O_2} .
ΔHc′\Delta H{'_c} = Enthalpy of COCO .

Hence, the correct option is B. −283.0 kJ mol−1 - 283.0\,kJ\,mo{l^{ - 1}}.

Note: The formation reactions for most organic compounds are hypothetical. For example, carbon and hydrogen won't legitimately respond to shape methane , so the standard enthalpy of development can't be estimated straightforwardly. Anyway the standard enthalpy of burning is promptly quantifiable utilizing bomb calorimetry.