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Question: Find the value of standard enthalpy of the formation for the given reaction, \[{C_{(s)}} + \dfrac{...

Find the value of standard enthalpy of the formation for the given reaction,
C(s)+12O2→CO(g){C_{(s)}} + \dfrac{1}{2}{O_2} \to C{O_{(g)}} ,  ΔH=? \,\Delta H = ?\,

Explanation

Solution

We can see here in the reaction, formation of a compound is happening, Hence, the enthalpy here is that of formation. So, this can be calculated using Hess’s law of constant heat summation. The Hess’s law is due to the state function of enthalpy, which enables one to quantify the total shift in enthalpy by simply summarising the changes for each step of the way in which the product is formed. All steps have to continue at the same temperature and it is important to balance out the calculations for the individual steps.

Complete step by step answer:
Let us first understand what is the standard enthalpy of formation  ΔH \,\Delta H\,;
The standard enthalpy of a reaction is the enthalpy change accompanying the reaction when all the substances are in their standard states.
Now, since there is a compound being formed, the enthalpy will be calculated for the formation of the compound.
Standard enthalpy of formation of a compound is defined as enthalpy change that accompanies a reaction in which one mole of pure compound in its standard state is formed from its elements also in their standard states.
The enthalpy of formation can be calculated from Hess’s law:
Hess’s law states that the changes in the enthalpy for a reaction is the same whether the reaction takes place in one or a series of steps.
To solve the given equation let us consider the following equations:

C(S)+O2(g)→CO2(g)  {C_{(S)}} + {O_{2(g)}} \to C{O_{2(g)}} \\\ \\\
 ΔH=−393.5kJ \,\Delta H = - 393.5kJ\,
Also,  2CO(g)+O2(g)→2CO2(g) ΔH=−566kJ \,2CO(g) + {O_2}(g) \to 2C{O_2}(g)\,\Delta H = - 566kJ\,
As our required equation has  CO \,CO\,in the right- hand side, we reverse the equation and also divide it by  2 \,2\,, as the equation get reversed we have to change the sign of  ΔH \,\Delta H\,as well and divide that also by  2 \,2\,, Hence, we get;
CO(g)+12O2→CO2(g)C{O_{(g)}} + \dfrac{1}{2}{O_2} \to C{O_{2(g)}}  ΔH=+283.0kJ  \,\Delta H = + 283.0kJ\,\,
Now, adding both the equations, we get,
C(s)+12O2→CO(g){C_{(s)}} + \dfrac{1}{2}{O_2} \to C{O_{(g)}}
Here,  ΔH=−393.5+283 \,\Delta H = - 393.5 + 283\,
=−110.5kJ= - 110.5kJ

ADDITIONAL INFORMATION:
Hess’s law is a direct consequence of the fact that enthalpy is a state function and so the enthalpy change depends only on the initial and the final states of the system and not on the path by which the reaction takes place. The importance of this law is that it treats the thermochemical equations mathematically. The thermochemical equations can be added, subtracted or multiplied by a number of factors like ordinary algebraic equations.

Note:
Enthalpy of formation of carbon monoxide cannot be determined experimentally. Therefore, it is calculated by Hess’s law. It is one of the advantages of this law. One consequence of our observation of Hess 's Law is that the net heat produced or consumed during a reaction is independent of the direction connecting the reactant to the product, this declaration is again subject to our constraint that all reactions must occur under constant pressure conditions in the alternative path.