Question

If one starts with pure NO₂(g) at a pressure of 0.422 atm, the total pressure inside the reaction vessel when 2NO₂(g) ? 2NO(g) + O₂(g) reaches equilibrium is 0.606 atm. Calculate the equilibrium partial pressure of NO₂. Round your answer to three significant figures.

Answer 1

2NO₂(g)   ?      2NO(g) + O₂(g)

0.422                     0            0                    initial

0.422-2x                2x             x                equilibrium

but total pressure is given

0.422-2x +2x + x = 0.606

x= 0.184

at equilibrium NO2 pressure = 0.422-2x = 0.054atm