Respuesta :
Answer:
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.
The study of chemicals and bonds is called chemistry. There are different types of elements and these are metal and nonmetals.
The correct answer to the question is mentioned below.
In none of the reactions ΔH°rxn equal to ΔH°f of the product.
What is standard enthalpy?
- 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.
[tex]\frac{1}{2} O_2(g) + H_2O(g) ---> H_2O_2g)[/tex]
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)).
[tex]K(g) + \frac{1}{2} Cl_2(g) ----> KCl(s)[/tex]
ΔH°rxn is NOT equal to ΔH°f of the product because K is not in its standard state (K(s)).
[tex]O_2(g) + 2 N_2(g) ----> 2 N_2O(g)[/tex]
ΔH°rxn is NOT equal to ΔH°f of the product because 2 moles of N₂O are formed.
In none of the above ΔHrxn is equal to ΔHf of the product.
For more information about the enthalpy, refer to the link:-
https://brainly.in/question/9531558
