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Question: The enthalpy of combustion of carbon and carbon monoxide are \[ - 390kJ\]and \[ - 298kJ\] respective...

The enthalpy of combustion of carbon and carbon monoxide are −390kJ - 390kJand −298kJ - 298kJ respectively. The enthalpy of formation CO in kJ is-
A.668kJ668kJ
B.112kJ112kJ
C.−112kJ - 112kJ
D.−668kJ - 668kJ

Explanation

Solution

Enthalpy is the measure of energy in an observed thermodynamic system. Basically it is the total heat present in a system equalising to the system’s internal energy and the product of volume present and the pressure.

Complete answer:
The standard enthalpy of combustion is defined because the enthalpy change when one mole of a compound is totally burnt in oxygen with all the reactants and products in their standard state under standard conditions i.e 298K298K and 11 bar pressure.
The standard enthalpy of formation is defined because the enthalpy change when one mole of a compound is made from its elements in their most stable state of aggregation i.e stable state of aggregation at temperature: 298.15K298.15K, pressure: 11 atm
Now, solving the above question,
I. C+O2=CO2C + {{ }}{O_2}{{ }} = {{ }}C{O_2}
ΔH1=−390\Delta {H_1} = {{ }} - 390
II. CO+12O2=CO2CO{{ }} + {{ }}\dfrac{1}{2}{{ }}{O_2}{{ }} = {{ }}C{O_2}
ΔH2=−298\Delta {H_2}{{ }} = {{ }} - {{ }}298
Now , by subtracting (I) and (II) we will get,
III. C+12O2=COC{{ }} + {{ }}\dfrac{1}{2}{{ }}{O_2} = {{ }}CO
ΔH=ΔH1−ΔH2=−112\Delta H{{ }} = {{ }}\Delta {H_1}{{ }} - {{ }}\Delta {H_2}{{ }} = {{ }} - 112
Here, we can see that,
Equation III represents the formation of 1mole1mole of CO
Thus we have the enthalpy of formation of CO is 112kJ112kJ.

Hence,Option B is the correct choice.

Additional Information:
One must keep in mind that the quality enthalpy change of combustion, is that the energy released as heat when a substance undergoes complete combustion at degree Centigrade and pressure whereas the quality enthalpy of formation, is that the change of enthalpy (so it might be + ive or - ive) when exactly 1 mole of a compound is made , provided that the reactants and products are at standard temperature and pressure.

Note:
We know that combustion is always an exothermic process. And because of this, enthalpy change of combustion must always be positive while on the other hand, 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. Sometimes we may come to notice that these two might have equivalent values but they are not the same thing.