Question

The equilibrium constant (kp) for the formation of the air pollutant nitric oxide (no) in an automobile engine at 537°c is 4.5 × 10−11. n2(g) + o2(g) ⇌ 2no(g) (a) calculate the partial pressure of no under these conditions if the partial pressures of nitrogen and oxygen are 3.00 and 0.012 atm, respectively.

Answer 1

Hey there!:

Kp = 4.5*10⁻¹¹

Given the reaction:

N2 (g) + O2(g) ⇌ 2 NO (g)

Clearly :

Kp = P ( NO )² / P (N2) * P (O2 )

4.5*10⁻¹¹ = P ( NO )² / ( 3.00 ) * ( 0.012 )

P (NO)² = 3.00 * 0.012 * 4.5*10⁻¹¹

P (NO)² = 1.62*10⁻¹² atm

P (NO)² = √ 1.62*10⁻¹²

P (NO)² = 1.273*10⁻⁶ atm

Hope that helps!

Answer 2

1.3 × 10⁻⁶ atm

Step-by-step explanation:

Let's consider the following reaction at equilibrium.

N₂(g) + O₂(g) ⇄ 2 NO(g)

The equilibrium constant (Kp) is:

`Kp=(pNO^(2))/(pN_(2).pO_(2)) = 4.5 * 10^(-11)`

Given pN₂ = 3.00 atm and pO₂ = 0.012 atm, the partial pressure of NO is:

pNO² = Kp × pN₂ × pO₂ = 4.5 × 10⁻¹¹ × 3.00 × 0.012

pNO = 1.3 × 10⁻⁶ atm