Question

Consider the reactions _1 2 N2(g) + _1 2 O2(g) → NO(g) _1 2 N2(g) + O2(g) → NO2(g) If these reactions come to equilibrium after combustion in an internal-combustion engine at 2000 K and 200 bar, estimate the mole fractions of NO and NO2 present for mole fractions of nitrogen and oxygen in the combustion products of 0.70 and 0.05.

Answer 1

Answer:

$w_{NO}=0.16$

$w_{NO2}=0.09$

Explanation:

Base of calculation: 100 mol

Initially in air :

79% N2

21% O2

Equilibrium:

$n_{N2}=79 mol - 0.5*n_{NO}- 0.5*n_{NO2}$

$n_{O2}=21 mol - 0.5*n_{NO}- n_{NO2}$

$n_{tot}=n_{N2}+ n_{NO2}+n_{NO}+n_{O2}$

$n_{tot}=100 mol - 0.5*n_{NO2}$

Knowing that:

$w_{N2}=\frac{n_{N2}}{n_{tot}}$

$w_{N2}=\frac{79 mol -0.5*n_{NO}- 0.5*n_{NO2}}{100 mol - 0.5*n_{NO2}}=0.7$

$79 mol - 0.5*n_{NO}- 0.5*n_{NO2}= 70 mol - 0.35*n_{NO2}$

$9 mol - 0.5*n_{NO}- 0.15*n_{NO2}=0$ (1)

$w_{O2}=\frac{n_{O2}}{n_{tot}}$

$w_{N2}=\frac{21 mol - 0.5*n_{NO}-*n_{NO2}}{100 mol -0.5*n_{NO2}}=0.05$

$21 mol - 0.5*n_{NO}- n_{NO2}= 5 mol - 0.025*n_{NO2}$

$16 mol - 0.5*n_{NO}- 0.975*n_{NO2}=0$  (2)

Solving 1 and 2:

$n_{NO}=15.45 mol$

$n_{NO2}=8.48 mol$

Fractions:

$w_{NO}=\frac{15.45}{100-0.5*8.48}=0.16$

$w_{NO2}=\frac{8.48}{100-0.5*8.48}=0.09$