Question

Nitrogen dioxide decomposes according to the reaction

2 NO₂(g) ⇌ 2 NO(g) + O₂(g)

where K_p = 4.48 × 10⁻¹³ at a certain temperature. If 0.70 atm of NO₂ is added to a container and allowed to come to equilibrium, what are the equilibrium partial pressures of NO(g) and O₂(g)?

I can't seem to figure this one out, I follow the steps to do the problem and I keep getting the answer 0 no matter how I enter it in the calculator and I used 2 different calculators. Can someone help me?

Answer 1

Equilibrium constant Kp for a given reaction is the ratio of product of partial pressures of gaseous products to product of partial pressures of gaseous reactants at equilibrium raised to power their stoichiometric coefficients.

So for given reaction

Let us make an ICE table for given reaction

  

Initial pressure (atm). 0.70 0 0

Change -2y +2y +y

Equilibrium pressure (atm) 0.70-2y 2y y

So

As the value of Kp is very small so we can ignore 2y in the denominator as 2y<<<<<0.70

2.20x10^-13=4y³

y³=2.20x10^-13/4=0.55 x 10^-13

So y=3.8x10^-5 atm

Thus equilibrium partial pressure of O₂==y=3.8x10^-5 atm

Equilibrium partial pressure of NO==2y=2x3.8x10^-5 atm=7.6x10^-5 atm