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.

Explanation:

Base of calculation: 100 mol

Initially in air :

79% N2

21% O2

Equilibrium:

Knowing that:

(1)

 (2)

Solving 1 and 2:

Fractions:

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml