Question

the values are given in the table below

Substance ΔH°f (kJ/mol)
SO2 (g) −297
SO3 (g) −396
SO2Cl2 (g) −364
H2SO4 (ℓ) −814
H2O (ℓ) −286

The value of ΔH° for the following reaction is −62.1 kJ. What is the value of ΔH°f for HCl(g)?

SO2Cl2(g) + 2 H2O(ℓ)→ H2SO4(ℓ) + 2 HCl(g)

Answer 1

Answer:

ΔH°f,HCl = - 92.05 KJ/mol

Explanation:

• SO2Cl2(g) + 2 H2O(l) → H2SO4(l) + 2 HCl(g)

• ΔH°rxn = ∑ νi.ΔH°f,i

∴ ΔH°rxn = 62.1 KJ/mol

⇒ - 62.1 KJ/mol = 2 ΔH°f,HCl(g) + ΔH°f,H2SO4(l) - 2 ΔH°f,H2O(l) - ΔH°f,SO2Cl2(g)

⇒ 2 ΔH°f,HCl = - 62.1 KJ/mol + 2 ΔH°f,H2O(l) + ΔH°f,SO2Cl2(g) - ΔH°f,H2SO4(l)

⇒ 2 ΔH°f,HCl = - 62.1 KJ/mol + 2 (-286 KJ/mol) - 364 KJ/mol - (- 814 KJ/mol)

⇒ 2 ΔH°f,HCl = - 62.1 KJ/mol - 572 KJ/mol - 364 KJ/mol + 814 KJ/mol

⇒ 2 ΔH°Δf,HCl(g) = - 62.1 KJ/mol - 122 KJ/mol

⇒ 2 ΔH°f,HCl(g) = - 184.1 KJ/mol

⇒ ΔH°f,HCl(g) = ( - 184.1 KJ/mol ) / 2 = - 92.05 KJ/mol