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.

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

ΔfHΘ = ? … —————————————–(i)

The enthalpy of combustion of 1 mol of benzene is :

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

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

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

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

H2 (g) + 1/2 O2 (g) → H2O (l);

ΔfHΘ = -285.83kJ mol−1 … ——————————————–(iv)

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

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

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

Summing up the above two equations :

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

ΔfHΘ = −3218.49 kJ mol-1 … ————————–(v )

Reversing equation (ii);

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

ΔfHΘ = −3267.0 kJ mol-1 … ————————–(vi )

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

6C (graphite) + 3H2 (g) → C6H6(l);

ΔfHΘ = 48.51 kJ mol-1

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