Question

The reaction

N₂(g) + O₂(g) ⇄ 2 NO(g)

contributes to air pollution whenever a fuel is burned in air at a high temperature, as in a gasoline engine. At 1500 K, K_c = 1.0×10^-3. Suppose a sample of air has [N₂] = 0.70 mol/L and [O₂] = 0.20 mol/L before any reaction occurs. Calculate the equilibrium concentrations of reactants and products after the mixture has been heated to 1500 K.

[N₂] = M

[O₂] = M

[NO] = M

Answer 1

FIRST CONSTRUCT THE ICE TABLE AND THEN SUBSTITUTE THE VALUES IN KC EXPRESSION AND YOU CAN GET THE X VALUE FROM THIS CALCULATE THE EQUILIBRIUM CONCENTRATIONS
         N2   +    O2   ------------------------> 2 NO

I       0.70       0.20                                     0

C      -X           -X                                      +2X

E    (0.70-X)   (0.20-X)                             2X

Kc = [NO]² / [N2][O2]

1.0*10^-3 = [2X]² / [0.70-X][0.20-X]

1.0*10^-3 * [0.70-X][0.20-X] = [2X]²

solving for X => 0.0058 M

at equilibrium [N2] = (0.70-X) =(0.70-0.0058) = 0.6942 M

                        [O2] = (0.40-X) = (0.40-0.0058) = 0.3942 M

                        [NO] = 2X = 2*0.0058 = 0.0116 M