Answer:
-184.6 KJ/mole
Explanation:
H2(g) + Cl2(g) ------------> 2HCl(aq)
Δ H0 rxn = Δ H0f products - Δ H0f reactants
= 2×-92.3 - (0 + 0)
= -184.6KJ/mole
H2(g) + Cl2(g) ------------> 2HCl(aq) Δ H0rxn = -184.6KJ/mole
Given :
Hydrogen gas,[tex]H_2[/tex] , reacts explosively with gaseous chlorine, [tex]Cl_2[/tex], to form hydrogen chloride, HCl(g).
The standard enthalpy of formation of HCl(g) is −92.3 kJ/mol.
[tex]H_2(g)+Cl_2(g) - >2HCl(g)[/tex]
To Find :
The enthalpy change for the reaction of 2 mole of [tex]H_2(g)[/tex] with 2 mole of [tex]Cl_2(g)[/tex] if both the reactants and products are at standard state conditions .
Solution :
We know ,
[tex]\Delta H^o_{rxn}=\Delta H^{o}f_{products}-\Delta H^of_{reactants}\\\\\Delta H^o_{rxn}=2\times (-92.3)-(0+0)\\\\\Delta H^o_{rxn}=-184.6\ kJ/mol[/tex]
( Since , [tex]H_2[/tex] and [tex]Cl_2[/tex] are their most elementary form . So their enthalpy of formation is 0 kJ )
Therefore , the enthalpy change for the reaction of 2 mole of [tex]H_2(g)[/tex] with 2 mole of [tex]Cl_2(g)[/tex] if both the reactants and products are at standard state conditions is -184.6 kJ/mol .
Hence , this is the required solution .
