Hydrogen gas, H2, reacts explosively with gaseous chlorine, Cl2, to form hydrogen chloride, HCl(g). What is the enthalpy change for the reaction of 2 mole of H2(g) with 2 mole of Cl2(g) if both the reactants and products are at standard state conditions? The standard enthalpy of formation of HCl(g) is −92.3 kJ/mol. H2(g)+Cl2(g)→2HCl(g)

Given :

Hydrogen gas, , reacts explosively with gaseous chlorine, , to form hydrogen chloride, HCl(g).

The standard enthalpy of formation of HCl(g) is −92.3 kJ/mol.

To Find :

The enthalpy change for the reaction of 2 mole of with 2 mole of if both the reactants and products are at standard state conditions .

Solution :

We know ,

( Since , and are their most elementary form . So their enthalpy of formation is 0 kJ )

Therefore , the enthalpy change for the reaction of 2 mole of with 2 mole of if both the reactants and products are at standard state conditions is -184.6 kJ/mol .

Hence , this is the required solution .

the correct answer is option d

explanation:

i was wrong its d because they dont use cm when calculating square roots