Question 8.3: The oxidation of methane can take place according to either ...

The stoichiometric coefficients of the chemical reaction,

$CH_4 + 2 O_2 \xrightarrow{1} CO_2 + 2 H_2O$.

$2 CH_4 + 3 O_2 \xrightarrow{2} 2 CO + 4 H_2O$.

are $ν_{1,CH_4} = −1, ν_{1,O_2} = −2, ν_{1,CO_2} = 1, ν_{1,H_2O} = 2, ν_{2,CH_4} = −2, ν_{1,O_2} = −3 , ν_{2,CO} = 2$ and $ν_{2,H_2O} = 4$.

According to relation (8.11), the time evolution of the number of moles of a substance A taking part in the coupled reactions 1 and 2 is given by,

$N_A (t ) = N_A (0) + \sum\limits_{a=1}^{n}{\nu _{aA}\xi _a(t)}$      (8.11)

$N_A (t) = N_A (0) + ν_{1,A} \xi _1 (t) + ν_{2,A} \xi _2 (t)$.

The reactions stop at time $t_f$ when all the methane is burnt. Thus, according to relation (8.11), we write,

$N_{CH_4} (t_f) = N_{CH_4} (0) + ν_{1,CH_4} ξ_1 (t_f) + ν_{2,CH_4} ξ_2 (t_f) = 0$.

which implies that the initial number of moles of methane yields,

$N_{CH_4} (0) = ξ_1 (t_f) + 2 ξ_2 (t_f)$.

Initially, there is no water, i.e. $N_{H_2O} (0) = 0$. Thus, according to relation (8.11), we write,

Since $m_{H_2O} (t_f)$ = 12.6 g and $M_{H_2O}$ = 18 g, we obtain the following identity,

$N_{H_2O} (t_f) = \frac{m_{H_2O} (t_f)}{M_{H_2O}} =2 ξ_1 (t_f) + 4 ξ_2 (t_f)$ = 0.7 mol.

Initially, there is no carbon dioxide and monoxide, i.e. $N_{CO_2} (0) = 0$ and $N_{CO}(0) = 0$. Thus, according to relation (8.6), the time evolution of the carbon dioxide and monoxide are given by,

$N_A (t) = N_A (0) + ν_{aA} \xi _a (t)$.                   (8.6)

$N_{CO_2} (t_f) = N_{CO_2} (0) + ν_{1,CO_2} ξ_1 (t_f) = ξ_1 (t_f)$.

$N_{CO} (t_f) = N_{CO} (0) + ν_{2,CO} ξ_2 (t_f) = 2 ξ_2 (t_f)$.

The total mass $m(t_f)$ of products of their number of moles times their molar masses,

$m(t_f) = N_{CO_2} (t_f) M_{CO_2} + N_{CO} (t_f) M_{CO} + N_{H_2O} (t_f) M_{H_2O}$ = 24.8 g

which implies that,

$\frac{M_{CO_2}}{m(t_f)} N_{CO_2} (t_f) +\frac{M_{CO}}{m(t_f)} N_{CO} (t_f) + \frac{M_{H_2O}}{m(t_f)} N_{H_2O} (t_f)$ = 1 mol.

and can be recast as,

$\biggl(\frac {M_{CO_2}}{m(t_f)} +2 \frac {M_{H_2O}}{m(t_f)}\biggr) ξ_1 (t_f) + \biggl(2 \frac {M_{CO}}{m(t_f)} +4 \frac {M_{H_2O}}{m(t_f)}\biggr) ξ_2 (t_f)$= 1 mol.

Since $m(t_f) = 24.8 g, M_{H_2O} = 18 g, M_{CO}$ = 28 g and $M_{CO_2}$ = 48 g, we obtain the following identity,

$3.39 ξ_1 (t_f) + 5.26 ξ_2 (t_f)$ = 1 mol.

Solving the system of equations,

$2 ξ_1 (t_f) + 4 ξ_2 (t_f)$ = 0.7 mol.

$3.39 ξ_1 (t_f) + 5.26 ξ_2 (t_f)$= 1 mol.

we find that,

$ξ_1 (t_f)$ = 0.10 mol            and                $ξ_2 (t_f)$ = 0.12 mol.

Since $M_{CH_4}$ = 16 g, the initial mass of methane $m_{CH_4} (t_f)$ burned in this reaction is,

$m_{CH_4} (t_f) = N_{CH_4} (t_f) M_{CH_4} = \biggl( ξ_1 (t_f) + 2 ξ _2 (t_f)\biggr) M_{CH_4}$ = 5.4 g.