Be sure to answer all parts. δh o f of hydrogen chloride [hcl(g)] is −92.3 kj/mol. given the following data, determine the identity of the two missing products and calculate the δh o rxn for equation 3. [hint: start by writing the chemical equation that corresponds to δh o f for hcl(g).] n2(g) + 4h2(g) + cl2(g) → 2nh4cl(s) δh o rxn = −630.78 kj/mol (1) n2(g) + 3h2(g) → 2nh3(g) δh o rxn = −296.4 kj/mol (2) nh4cl(s) → (include states of matter.) δh o rxn = kj/mol

NH₄Cl(s) → NH₃(g) + HCl(g)

ΔH°rxn 74,89 kJ/mol

Explanation:

The change in enthalpy of formation (ΔHf) is defined as the change in enthalpy in the formation of a substance from its constituent elements. For HCl(g):

(1) ¹/₂H₂(g) + ¹/₂ Cl₂ → HCl(g) ΔH = -92,3 kJ/mol

It is possible to sum ΔH of different reactions to obtain ΔH of a global reaction (Hess's law).

For the reactions:

(2) N₂(g) + 4H₂(g) + Cl₂(g) → 2NH₄Cl(s) ΔH°rxn = −630.78 kJ/mol

(3) N₂(g) + 3H₂(g) → 2NH₃(g) ΔH°rxn = −296.4 kJ/mol

The sum of -(2) + (3) gives:

-(2) 2NH₄Cl(s) → N₂(g) + 4H₂(g) + Cl₂(g) ΔH°rxn = +630.78 kJ/mol

(3) N₂(g) + 3H₂(g) → 2NH₃(g) ΔH°rxn = −296.4 kJ/mol

-(2) + (3) 2NH₄Cl(s) → 2NH₃(g) + H₂(g) + Cl₂(g)

ΔH°rxn = +630.78 kJ/mol −296.4 kJ/mol = +334,38 kJ/mol

Now, the sum of -(2) + (3) + 2×(1)

-(2) + (3) 2NH₄Cl(s) → 2NH₃(g) + H₂(g) + Cl₂(g) ΔH°rxn = +334,38 kJ/mol

2×(1)  H₂(g) + Cl₂(g)→ 2HCl(g) ΔH = 2×-92,3 kJ/mol

-(2) + (3) + 2×(1) 2NH₄Cl(s) → 2NH₃(g) + 2HCl(g)

ΔH°rxn = +334,38 kJ/mol + 2×-92,3 kJ/mol = 149,78 kJ/mol

The reaction of:

NH₄Cl(s) → NH₃(g) + HCl(g)

Has ΔH°rxn = 149,78kJ/mol / 2 = 74,89 kJ/mol

I hope it helps!

the solar system used to have much dust particles option no d

i think so it’s will be b yes i believe its