Thermodynamics ...

Calculate heat of the following reaction at constant pressure,
F₂O(g) + H₂O(g) → O₂ (g) + 2HF(g)
The heats of formation of F₂O (g), H₂O(g) and HF (g) are 5.5 kcal, –57.8kcal and 64.2 kcal respectively.

Calculate the heat of formation of benzene from the following data, assuming no resonance. Bond energies :
C – C = 83 kcal = 140 kcal C – H = 99 kcal
Heat of atomisation of C = 170 .9 kcal
Heat of atomisation of H = 52.1 kcal

Calculate ∆H at 358 K for the reaction
Fe₂O₃(s) + 3H₂(g) → 2Fe (s) + 3H₂O(l)
Given that, ∆H298= – 33.29 kJ mole^–1 and Cp for Fe₂O₃ (s), Fe (s), H₂O (l) and H₂ (g) are
103.8, 25.1, 75.3 and 28.8 J/K mole.

Ka for CH3COOH at 25ºC is 1.754 × 10^–5 . At 50ºC, Ka is 1.633 × 10^–5 What will be value of ∆Sº for the ionisation of CH₃COOH?

C₂H₆ (g) + 3.5 O₂ (g) → 2CO₂ (g) + 3H₂O (g)
∆S_vap (H₂O, l) = x₁ cal K^–1 (boiling point + T₁)
∆Hf (H₂O, l) = x₂
∆Hf (CO₂) = x₃
∆Hf (C₂H₆) = x₄
Hence DH for the reaction is –

C (s) + O₂ (g) → CO₂ (g); ∆H = –94.3 kcal/mol
CO (g) + O₂(g) → CO₂ (g); ∆H = – 67.4 kcal/mol
O₂(g) → 2O (g); ∆H = 117.4 kcal/mol
CO (g) → C (g) + O(g) ; ∆H = 230.6 kcal/mol
Calculate ∆H for C (s) → C (g) in kcal/mol.

The difference between ∆H and ∆E on a molar basis for the combustion of n–octane at 25°C would be : 25ºC

What is the work done against the atmosphere when 25 grams of water vaporizes at 373 K against a constant external pressure of 1 atm ? Assume that steam obeys perfect gas laws. Given that the molar enthalpy of vaporization is 9.72 kcal/mole, what is the change of internal energy in the above process ?

In the reaction CS₂ (l) + 3O₂ (g) → CO₂ (g) + 2SO₂ (g) ∆H = –265 kcal The enthalpies of formation of CO₂ and SO₂ are both negative and are in the ratio 4 : 3. The enthalpy of formation of CS₂ is + 26 kcal/mol. Calculate the enthalpy of formation of SO₂.

The bond dissociation energy of gaseous H₂, Cl₂ and HCl are 104, 58 and 103 kcal mol^{– 1} respectively. The enthalpy of formation for HCl gas will be