Answer:
A. (1/2) N₂(g) + O₂ (g) → NO₂(g)
Explanation:
H°rxn = ∑n*H°f, products - ∑n*H°f, reactants (where n is the number of moles)
H°f is the enthalpy of formation at standard conditions, and it's 0 by the compounds formed by a single element at its atmospheric phase. So, to H°rxn = H°f, products, the number of moles of the product must be 1 and the H°f, reactants must be 0.
Both conditions are only satisfied in letter A.
In letter B, the reactants H and Br are not in the molecule formula, so their H°f is not 0. The mase happens in letter C for O, and in letter E for H. In letter D, 2 moles of the product are being formed.
The reaction in which ΔH°rxn is equal to the heat of formation of the product is 1/2N2 (g) + O2 (g) → NO2(g)
The standard enthalpy change for a reaction is obtained as the difference between the enthalpy change of the reactants and the products. It is a state function hence it depends on the initial and final states of the system.
Note that the standard enthalpy change for a reaction is calculated from the formula;
ΔH°rxn = [n∑(ΔH°products - n∑(ΔH°reactants]
For the reaction; 1/2N2 (g) + O2 (g) → NO2(g), the ΔH°reactants = zero because they are in standard state therefore, ΔH°rxn is equal to the heat of formation of the product.
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