Question 14.7: A mixture of benzene gas with 60 % excess air contained in a......

Given: (1) Combustion reactants: Air and benzene, (2) Excess air: 60%
Find out: heat transferred per kmol of benzene.
Properties: Molar mass of $C,  H_2$ and $O_2$ and air are: 12 kg/kmol, 2 kg/kmol, 32 kg/kmol and 29 kg/kmol
respectively.
Assumption: (1) Combustion is complete, (2) Air and combustion gases are ideal (3) Kinetic and potential energies are negligible (4) Combustion products are $H_2O,  CO_2$ and $N_2$ only.
Combustion reaction with no-excess air will be:

$C_6H_6 + (7.5  O_2 + 28.2  N_2) \longrightarrow 6  CO_2  +  3  H_2O (28.2  N_2)$

From the first law:
For constant volume,

$Q – W = \sum{N_R} (H^0_f + H – H_0 – pV)_R – \sum{N_P (H^0_f + H – H_0 – pV)_p}$

With W = 0,

$Q = \sum{N_R} (H^0_f + H – H_0 – pV)_R – \sum{N_P (H^0_f + H – H_0 – pV)_p}$

Assuming ideal gas condition:

$Q = \sum{N_R} (H^0_f + H – H_0 – RT)_R – \sum{N_P (H^0_f + H – H_0 – RT)_p}$

From the given table data:

$\sum{N_P} (H^0_f + H – H_0 – RT)_p = 6 \times (- 3,93,520 + 42,769 – 9,364 – 8.314 \times 1,000) + 3\times (-2,41,820 + 35,882 – 9,904 – 8.314 \times 1,000) + 28.2 \times (0 + 30,159 – 8,669 – 8.314 \times 1,000)$
= –24,81,766 kJ/kmol

$\sum{N_R} (H^0_f + H – H_0 – RT)_R = 1 \times (82,930 -8.314 \times 1,000) + 7.5 \times (- 8.314 \times 1,000) + 28.2 \times (- 8.314 \times 1,000)$
= –7,996.89 kJ/kmol.

So, Q = –24,81,766 – (–7,996.89) = –24,73,769.11 kJ/kmol.
Negative sign indicates heat outgoing.