A 0.5 mol sample of `H_(2)(g)` reacts with a 0.5 mol sample of `Cl_(2)(g)` to form 1 mol of HCl(g). The decrease in enthalpy for the reaction is 93 kJ. Draw an enthalpy diagram for this reaction.

Explanation

`(1)/(2)H_(2)(g)+(1)/(2)Cl_(2)(g)toHCl (g),DeltaH=-93kJ*mol^(-1)`
The reaction is associated with decrease in enthalpy. So, it is ann exothermic reaction. In such reactiton, the total enthalpy of the product(s) `(sumH_(P))` is less than of the reactant(s) `(sumH_(R))`. therefore, in the enthalpy diagram for the reaction, `sumH_(P)` lies below `sumH_(R)`.
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