If a reaction vessel at this temperature initially contains 0.500 atm NO2, what will be the partial pressure of NO2 in the vessel when equilibrium has been attained?

Question

The equilibrium constant K_p for the thermal decomposition of NO₂ is 6.5 × 10^–6 at 450°C.

2NO2​(g)2NO(g)+O2​(g)

A. If a reaction vessel at this temperature initially contains 0.500 atm NO₂, what will be the partial pressure of NO₂ in the vessel when equilibrium has been attained?

B. What will be the partial pressure of NO in the vessel when equilibrium has been attained?

C. What will be the partial pressure of O₂ in the vessel when equilibrium has been attained?

Answer 1

2NO₂​(g) <===> 2NO(g) + O₂​(g)

0.500..................0..............0............Initial

-2x....................+2x...........+x...........Change

0.5-2x................2x.............x............Equilibrium

Kp = 6.5x10^-6 = (NO)²(O₂) / (NO₂)²

6.5x10^-6 = (2x)²(x) / 0.5-2x)² (if we assume x is small and ignore it in denominator, we can avoid quadratic)

6.5x10^-6 = (2x)²(x) / 0.5)²

6.5x10^-6 = 4x³ / 0.25

4x³ = 1.63x10^-6

x = 3.44x10^-4 (note: this is only about 0.06% of 0.500 so above assumption was valid)

Partial pressure NO₂ @ equilibrium = 0.5 - 2x = 0.500 - 6.88x10^-4 = 0.499 atm

Partial pressure NO @ equilibrium = 2x = 6.88x10^-4 atm

Partial pressure O₂ @ equilibrium = x = 3.44x10^-4 atm