Calculate the standard enthalpy of formation of NOCl(g) at 25 ºC, knowing that the standard enthalpy of formation of NO(g) at that temperature is 90.25 kJ / mol, and that the standard molar enthalpy of the reaction 2 NOCl(g) → 2 NO(g) + Cl2(g) is 75.5 kJ / mol at the same temperature.

The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:

Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants

In this case, you have: 2 NOCl(g) → 2 NO(g) + Cl₂(g)

So, ΔH=[tex]2*H_{NO} +H_{Cl_{2} }-2*H_{NOCl}[/tex]

Knowing:

ΔH= 75.5 kJ/mol
[tex]H_{NO}[/tex]= 90.25 kJ/mol
[tex]H_{Cl_{2} }[/tex]= 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound the chlorine Cl₂)
[tex]H_{NOCl}[/tex]=?

Replacing:

75.5 kJ/mol=2* 90.25 kJ/mol + 0 - [tex]H_{NOCl}[/tex]

Solving

-[tex]H_{NOCl}[/tex]=75.5 kJ/mol - 2*90.25 kJ/mol

-[tex]H_{NOCl}[/tex]=-105 kJ/mol

[tex]H_{NOCl}[/tex]=105 kJ/mol

The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol