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 .

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c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current