Calculate the enthalpy change for the reaction 2 hcl → h2 + cl2 1. +184 kj/mol of rxn 2. +428 kj/mol of rxn 3. 0 kj/mol of rxn 4. −184 kj/mol of rxn 5. −244 kj/mol of rxn

Below is the list of enthalpies of formation of the compounds involved in the chemical reaction.

HCl: -92.3 kJ/mol
H2: 0 
Cl2: 0

Both H2 and Cl2 do not have enthaply of formation because they are pure substances. 

Since, in the reaction there are 2 moles of HCl,
   Enthalphy of formation = enthalpy of products - enthalphy of reactant/s
                = 0 - (-92.3)(2) = +184.6 kJ

Hence, the answer is the first choice. 

Ibelieve the answer is d

Idon’t really understand the question i’m sorry i can’t