Question

Consider the following equilibrated system: 2NO₂(g) 2NO(g) + O₂(g). If the K_p value is 0.997, find the equilibrium pressure of the O₂ gas if the NO₂ gas pressure is 0.520 atm and the P_NO is 0.300 atm at equilibrium.

I really need a good explaination toward the end when you actually solve for [02] because I get that far but don't know how to reaarange my equation.

Answer 1

Given chemical transformation,

2NO₂(g) -------------> 2NO(g) + O₂(g)

For this transformation the equilibrium constant is defined as,

Kp = [(P_NO(g))² x (P_O2(g) ] / (P_NO2(g))² ..........(1)

We have, Kp = 0.997, P_NO(g) = 0.300 atm, P_NO2(g) = 0.520 atm,   P_O2(g) = ?

Let us put all these values in eq.1 and we get,,

0.997 = [(0.300)² x (P_O2(g))] / (0.520)² .

0.997 x (0.520)² = (0.300)² x (P_O2(g))

0.997 x 0.2704 = 0.09 x (P_O2(g))

0.2696 = 0.09 x (P_O2(g))

P_O2(g) = 0.2696 / 0.09

P_O2(g) = 2.995 atm.

Hence at equilibrium pressure of O2 gas(i.e. equilibrium pressure) of O2 gas is 2.995 atm.