Question

Marks Calculate the enthalpy of combustion of benzene in which the water produced is a gas rather than a liquid. You may use the following information: 2 H2O(1)-»H20(g) AvapH° = 40.65 kJ mol-1 Answer

Answer 1

C₆H₆ (l)  +
15 2
O₂ (g) $\rightarrow$ 6 CO₂ (g) + 3 H₂O (g)

$\Delta$H_comb = [6* $\Delta$H_f (CO_2⁾ + 3*$\Delta$H^o_f H₂O (g) ] - [ $\Delta$H_f^o(C₆H₆) +
15 2
$\Delta$H(O₂)]

= [6*(-393.5) + 3*(-245.15) ] - [ 49.2+ 0]

= -2361 -735.45 - 49.2

= -3096.45 - 49.2

= -3145.65 KJ/mole

Note:

[ Enthalpy of Formation of H₂O (l) = -285.8 KJ/mole

and, H₂O (l) $\rightarrow$ H₂O (g)

then, Enthalpy of formation of H₂O (g) = Enthalpy of evaporation + Enthapy of formation of H₂O (l) = 40.65 +(-285.8) = - 245.15 KJ ]

Enthalpy of formation of CO₂ , H₂O (l) , C₆H₆ (l)  , O₂ (g) are taken from literature.