Question

The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, CO2(g) and H2O (1) are produced and 3267.0 kJ of heat is liberated. Calculate the standard enthalpy of formation, ΔfHΘof benzene. Standard enthalpies of formation of CO2(g) and H2O(l) are -393.5 kJ mol1and 285.83kJ mol1respectively.

Answer 1

The formation reaction of benezene is given by :
6C (graphite) + 3H₂ g → C₆H₆ (l) ;

Δ_fH^Θ = ? … —————————————–(i)

The enthalpy of combustion of 1 mol of benzene is :

C₆H₆ (l) + 15/2 O₂ → 6CO₂ (g) + 3H₂O (l);
Δ_cH^Θ = -3267kJ mol⁻¹ … ——————————–(ii)

The enthalpy of formation of 1 mol of CO₂(g) :

C (graphite) + O₂(g) → CO₂ g ;
Δ_fH^Θ = -393.5kJ mol⁻¹ … ———————————————(iii)

The enthalpy of formation of 1 mol of H₂O(l) is :

H₂ (g) + 1/2 O₂ (g) → H₂O (l);

Δ_fH^Θ = -285.83kJ mol⁻¹ … ——————————————–(iv)

multiplying eqn. (iii) by 6 and eqn. (iv) by 3 we get:

6C (graphite) + 6O₂ g → 6CO₂ g ;
Δ_fH^Θ = -2361kJ mol⁻¹

3H₂ (g) + 3/2 O₂ (g) → 3H₂O (l);
Δ_fH^Θ = -857.49kJ mol⁻¹

Summing up the above two equations :

6C (graphite) + 3H₂ (g) + 15/2 O₂ g → 6CO₂ (g) + 3H₂O(l);

Δ_fH^Θ = −3218.49 kJ mol^-1 … ————————–(v )

Reversing equation (ii);

6CO₂ (g) + 3H₂O(l) → C₆H₆(l) + 15/2 O₂;

Δ_fH^Θ = −3267.0 kJ mol^-1 … ————————–(vi )

Adding equations (v) and (vi), we get

6C (graphite) + 3H₂ (g) → C₆H₆(l);

Δ_fH^Θ = 48.51 kJ mol^-1