Question

The enthalpy changes at the following reactions at 27˚C are

Na(s) + 1/2Cl₂ (g) → NaCl(s)

∆_rH = -411 kJmo

H₂(g) + S(s) + 2O₂ (g) → H₂SO₄ (l)

∆_rH = -811 kJ/mol

2Na(s) + S(s) + 2O₂ (g) → Na₂ SO₄(s)

∆_rH = -1382 kJ/mol

1/2H₂ (g) + 1/2Cl₂ (g) →HCl(g)

∆_rH = -92 kJ/mol; R = 8.3 J/K-mol 

From these data, the heat change of reaction at constant volume (in kJ/mol) at 27C for the purpose

2NaCl (s) +H₂SO₄ (l) ⇋Na₂SO₄ (s) +2HCl (g) is

(A) 67 

(B) 62.02

(C) 71.98 

(D) None

Answer 1

Correct option (B) 62.02 

Explanation:

∆H_reac. = ∆H_f + Na₂ SO₄ + 2∆H HCl 

–2 ∆H_f NaCl – ∆H_f H₂SO₄

= –1382 – 2 × 92 + 2 × 441 + 811

∆V = ∆H – ngRT 

= 62.02

∆H = –67