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mr Goodwill
1 month ago
10

For which of the following reactions is ΔHrxn equal to ΔHf of the product? You do not need to look up any values to answer this

question. Select all that apply. Group of answer choices 1/2 O2(g) + H2O(g) LaTeX: \longrightarrow ⟶ H2O2(g) Na+(g) + F-(g) LaTeX: \longrightarrow ⟶ NaF(s) K(g) + 1/2 Cl2(g) LaTeX: \longrightarrow ⟶ KCl(s) O2(g) + 2 N2(g) LaTeX: \longrightarrow ⟶ 2 N2O(g) None of the above
Chemistry
1 answer:
Alekssandra [3K]1 month ago
7 0

Answer:

In all listed reactions, ΔH°rxn does not correspond to the ΔH°f of the resulting product.

Explanation:

The standard enthalpy of formation (ΔH°f) signifies the enthalpy change that occurs when 1 mole of a product is created from its basic elements in their standard states.

1/2 O₂(g) + H₂O(g) ⟶ H₂O₂(g)

ΔH°rxn does not equal ΔH°f of the product, since H₂O(g) is a compound rather than an element.

Na⁺(g) + F⁻(g) ⟶ NaF(s)

ΔH°rxn is not the same as ΔH°f of the product because Na and F are not in their standard states (Na(s); F₂(g)).

K(g) + 1/2 Cl₂(g) ⟶ KCl(s)

ΔH°rxn is not equal to ΔH°f of the product due to K being outside its standard state (K(s)).

O₂(g) + 2 N₂(g) ⟶ 2 N₂O(g)

ΔH°rxn does not match ΔH°f of the product as 2 moles of N₂O are produced.

In none of the above cases does ΔHrxn match ΔHf of the product.

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