Final answer:
The enthalpy change for the formation of 1 mole of HCl(g) is -92.3 kJ/mol. For the reaction of 1 mole of H2(g) with 1 mole of Cl2(g), the enthalpy change is -184.6 kJ, indicating an exothermic reaction where more energy is released in forming product bonds than is consumed in breaking reactant bonds.
Step-by-step explanation:
The enthalpy change for the reaction of 1 mole of H2(g) with 1 mole of Cl2(g) to form HCl(g) can be found using the standard enthalpy of formation.
The enthalpy change, ΔH, for the formation of 1 mole of HCl(g) from its elements in their standard states is -92.3 kJ/mol.
Since the balanced chemical equation shows the formation of 2 moles of HCl(g), the total enthalpy change would be twice this value, meaning ΔH for the reaction is -184.6 kJ.
The process of forming HCl(g) involves breaking the bonds of H-H and Cl-Cl and forming new H-Cl bonds. The bond energy for H-H is 436 kJ/mol, and for Cl-Cl, it is 243 kJ/mol. The energy released when forming two moles of H-Cl bonds is 2 x 432 kJ = 864 kJ. This reaction is exothermic, as it releases more energy in forming products than is required to break the bonds of the reactants.