Question

Calculate the standard enthalpy of formation of benzene give that the enthalpies of combustion of carbon hydrogen and benzene are 393.5kj/mol,-285.83kj/mol and 3267kj/mol respectively

Answer 1

The standard enthalpy of formation of benzene is ΔHf = 48.51 kJ mol-1

Explanation:

The formation reaction of benzene is given by:

6C (graphite) + 3H2 g → C6H6 (l)

ΔHf = ? ............(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 ;

ΔHf = -393.5kJ mol−1   ..............(iii)

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

H2 (g) + 1/2 O2 (g) → H2O (l);

ΔHf = -285.83 kJ mol−1 ............(iv)

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

6C (graphite) + 6O2 g → 6CO2 g ;

ΔHf = -2361kJ mol−1

3H2 (g) + 3/2 O2 (g) → 3H2O (l)

ΔHf = -857.49 kJ mol−1

Summing up the above two equations :

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

ΔHf = −3218.49 kJ mol-1  ..............(v )

Reversing equation (ii);

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

ΔHf = −3267.0 kJ mol-1  ..................(vi )

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

6C (graphite) + 3H2 (g) → C6H6(l);

ΔHf = 48.51 kJ mol-1

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