Question

A fuel contains 86.96% carbon and 13.04% h by weight. Calculate the higher heating value, lower heating value and enthalpy of formation of the fuel on a molar basis. The higher heating value is given as 44 kj/g. Latent heat of condensation of water is 44 kj/mol. The standard enthalpy of formation of co2 and h2o is -393509 j/mol and -241818 j/mol, respectively.

Answer 1

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Answer 2

Answer:

A fuel contains 86.96% C and 13.04% H by weight, that has the higher heating value of 44kJ/g, and the lower heating value of 41.13kJ/g

The enthalpy of formation of this fuel is  -43545.2 J/mol/C

Explanation:

• find fuel composition and formula weight:

        Weight%   Atomic weight   Ratio           normalizing with respect to C

C       86.96              12                $\frac{86.96}{12}=7.25$                  $\frac{7.25}{7.25} =1$

H      13.04                1                 $\frac{13.04}{1}=13.04$                $\frac{13.04}{7.25} =1.8$

  Thus the composition of fuel is CH₁.₈

   formula weight, $M=12\times1+1\times1.8=13.8$

• find lower heating value

The enthalpy of combustion of unit mass $\Delta \bar{h} _c_H$ is generally reported in terms of higher heating value (HHV)

                   $\Delta \bar{h} _c_H=-HHV=-44kJ/g$

The molar enthalpy of combustion $\Delta {h} _c_H$ is given by

             $\Delta{h} _c_H=M\times\Delta \bar{h} _c_H= 13.8\times-44000=-607200J/mol/K$

The equation for the combustion of this fuel is,

                   CH₁.₈ + 1.45 O₂  →  CO₂ + 0.9H₂O

Given that,

    latent heat of condensation of water,$\Delta {h} _v =44\:kJ/mol$

Then, the molar enthalpy of combustion corresponds to the lower heating value (LHV). is

     $\Delta{h} _c_L=\Delta {h} _c_H+0.9\times\Delta {h} _v=-607200+0.9\times44000=-567600J/mol/C$

lower heating value is given by

      $LHV=-\Delta \bar{h} _c_L=-\frac{\Delta{h} _c_L}{M}=-\frac{-567600}{13.8} =41130J/g$

• find the enthalpy of formation fuel

given that,

$\Delta {h} _f_C_O_{2} =-393509J/mol$

$\Delta {h} _f_H_{2}_O =-241818J/mol$

then the enthalpy of formation of this fuel is

       $\Delta {h} _c_f=\Delta {h} _f_C_O_{2} +0.9\times\Delta {h} _f_H_{2} _O- 1.45\times\Delta {h} _f_O_{2}-\Delta {h} _c_L$

                $=-393509+0.9\times-241818-1.45(0)-(-567600)$

                $=-43545.2J/mol/C$