Question

Calculate the enthalpy of formation of liquid benzene (C₆H₆) given the enthalpies of combustion of carbon, hydrogen and benzene as -393.5 KJ, -285.83 KJ and -3267.0 KJ respectively.

Answer 1

Required equation

6C(s) + 3H₂(g) → C₆H₆ (l) Δ_fH = ?

Given Data : C(s) + O₂(g) → CO₂(g) Δ_cH = -393.5 KJ

H₂(g) + 1/2O₂(g) → H₂O(l) Δ_cH = -285.83 K

C₆H₆(l) + 15/2O₂(g) → 6CO₂ + 3H₂O Δ_cH = -3267.0 KJ

Answer 2

\(6C+ 3H_2 \to C_6H_6\)

\(\Delta H_{C_6H_6} = \Delta H_{C_6H_6} - 6\Delta H_C - 3\Delta H_{H_2}\)

\(= -3267 - (6 \times (-393.5)) - (3 \times (-285.83))\)

\(= -3267 + 2361 + 857.49\)

\(= -48.51\)