The standard enthalpy of vaporization ${∆}_{\mathrm{vap}}{\mathrm{H}}^{\mathrm{o}}$ for water at 100 ^oC is 40.66 kJ mol^–1. The internal energy of vaporization of water at 100 ^oC (in kJ mol^–1) is: 
(Assume water vapour behaves like an ideal gas.)

Calculate the H–H bond dissociation energy from the following reaction:

4H(g) → 2H₂(g)  ΔH = −869.6 kJ

1. -869.6 kJ
2. +434.8kJ
3. +217.4kJ
4. -434.8 kJ

Consider the following processes:

                                       ∆H (kJ/mol)
½ A → B                            + 150
3B → 2C + D                      –125
E + A → 2D                        +350

For B + D → E + 2C, ∆H will be:

1. 325 kJ/mol 

2. 525 kJ/mol

3. –175 kJ.mol 

4. –325 kJ/mol

The enthalpy changes for the following reactions are given:

(i) Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g); ΔH₁ = −26.88 kJ

(ii) FeO(s) + CO(g) → Fe(s) + CO₂(g); ΔH₂ = −16.5 kJ

Using Hess’s law, calculate the enthalpy change (ΔH) for the following reaction:

Fe₂O₃(s) + CO(g) → 2FeO(s) + CO₂(g)

1. -43.3 kJ

2. -10.3 kJ

3. +6.2 kJ

4. +10.3 kJ