Question

What is the enthalpy of formation of carbon monoxide in $ KJ/mol $ ?
 $ {{C}_{(s)}}+{{O}_{2(g)}}\to C{{O}_{2(g)}} $ , $ \Delta H{}^\circ =-393kJ $
 $ 2C{{O}_{(s)}}+{{O}_{2(g)}}\to 2C{{O}_{2(g)}} $ , $ \Delta H{}^\circ =-588kJ $

Answer 1

Hint :Here, the reactions we are given shows the combustion reaction of carbon and carbon monoxide. The enthalpy of formation is the change in enthalpy when the carbon monoxide will be formed $ 1 $ mole from its constituent elements in their natural state. Hence, by changing or comparing the given equation, we can get the equation for formation of carbon monoxide.

Complete Step By Step Answer:
Let us understand and note the given data here.
Here, both the given reactions can be explained as a reaction of a substance with oxygen. As the heat is released in these reactions, they can also be explained as combustion of substance with oxygen.
Hence, this proves the definition of enthalpy of combustion which is defined as the energy released by a substance when it completely burns in presence of oxygen.
Hence, the above reactions are reactions of combustion of carbon and carbon monoxide. There reactions are exothermic reactions.
Hence consider the reactions as,
 $ {{C}_{(s)}}+{{O}_{2(g)}}\to C{{O}_{2(g)}} $ …… $ (1) $
 $ \therefore \Delta {{H}_{1}}{}^\circ =-393kJ $
 $ 2C{{O}_{(s)}}+{{O}_{2(g)}}\to 2C{{O}_{2(g)}} $ …… $ (2) $
 $ \therefore \Delta {{H}_{2}}{}^\circ =-588kJ $
Now, to get one mole of product for the second reaction, we can divide the complete reaction by $ 2 $
 $ C{{O}_{(s)}}+\dfrac{1}{2}{{O}_{2(g)}}\to C{{O}_{2(g)}} $ …… $ (3) $
Now, by Hess’ Law, if we divide the whole chemical reaction is halved, the enthalpy of reaction is also halved
 $ \therefore \Delta {{H}_{3}}{}^\circ =-294kJ $
Now, let us reverse the above reaction, so that the product becomes reactant and vice-versa.
 $ C{{O}_{2(g)}}\to C{{O}_{(s)}}+\dfrac{1}{2}{{O}_{2(g)}} $ …… $ (4) $
By Hess’ Law, we understand that if the reaction is inverted, the sign of the enthalpy of the reaction is changed.
 $ \therefore \Delta {{H}_{4}}{}^\circ =294kJ $
Now, let us combine the first and the fourth reaction, from which we get the final reaction as
 $ {{C}_{(s)}}+\dfrac{1}{2}{{O}_{2(g)}}\to C{{O}_{(g)}} $
Now, in this reaction we can see that mole carbon monoxide is formed from its constituent particles in their natural states. Hence, the above reaction is the reaction for formation of Carbon Monoxide.
By Hess’s Law, if reactions are combined algebraically, their enthalpies are also combined in the exact same way.
Hence, the enthalpy of formation of carbon monoxide is given as,
 $ \therefore \Delta {{H}_{f}}{}^\circ =\Delta {{H}_{1}}{}^\circ +\Delta {{H}_{4}}{}^\circ $
Substituting the values,
 $ \therefore \Delta {{H}_{f}}{}^\circ =-393kJ+294kJ $
 $ \therefore \Delta {{H}_{f}}{}^\circ =-99kJ $
This is the enthalpy of formation of Carbon Monoxide.

Note :
Here, the point should always be remembered that the change in enthalpy is called the enthalpy of formation only when the product formed is $ 1 $ mole. Also, some reactions that are shown for the calculation may not be practically possible. For example, carbon monoxide is formed by the reaction of carbon and carbon dioxide. But to obtain the enthalpy of formation, we have to consider its constituent elements i.e. one atom of carbon and one atom of oxygen.