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

Question

Chemistry

Adalyn Hill

Today, 11:59

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

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Morgan · Answer 1

In none of the reactions ΔH°rxn equal to ΔH°f of the product.

Explanation:

The standard enthalpy of formation (ΔH°f) is the enthalpy change when 1 mole of a product is formed from its constituent elements in the standard states.

1/2 O₂ (g) + H₂O (g) ⟶ H₂O₂ (g)

ΔH°rxn is NOT equal to ΔH°f of the product because H₂O (g) is not an element but a compound.

Na⁺ (g) + F⁻ (g) ⟶ NaF (s)

ΔH°rxn is NOT equal to 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 because K is not in its standard state (K (s)).

O₂ (g) + 2 N₂ (g) ⟶ 2 N₂O (g)

ΔH°rxn is NOT equal to ΔH°f of the product because 2 moles of N₂O are formed.

In none of the above ΔHrxn equal to ΔHf of the product.