Question

For the reaction 2 NO₂(g) ⇌ 2 NO(g) + O₂(g) Kp = 1.11 × 10⁻⁵ at 200 °C. A 2.50 L vessel at 200 °C is filled with NO₂(g) at an initial pressure of 4.00 atm and allowed to come to equilibrium. What will be the pressure (in atm) of NO(g) at equilibrium?

Answer 1

The equilibrium pressure of NO is 2.14 atm.

What is equilibrium constant?

The equilibrium constant is a numerical value that shows the extent to which reactants are converted into products. We have the reaction  2NO₂(g) ⇌ 2 NO(g) + O₂(g).

We have to set up the ICE table

From PV = nRT

P = nRT/V = 4 * 0.082 * 473/2.5 = 62 atm

     

          2NO₂(g) ⇌ 2 NO(g) + O₂(g)

I          62                 0            0

C         -x                +2x           +x  

E        62 - x           2x               x

Kp = pNO^2 * pO2/pNO2^2

1.11 × 10⁻⁵ = (2x)^2 (x)/(62 - x)^2

1.11 × 10⁻⁵(62 - x)^2 = 4x^3

1.11 × 10⁻⁵(3844 - 124 + x^2) = 4x^3

0.04 - 0.0014x +  1.11 × 10⁻⁵x^2 = 4x^3

4x^3 -  1.11 × 10⁻⁵x^2 +  0.0014x - 0.04 = 0

x = 0.107  atm

Since;

pNO = 2x = 2(0.107) = 2.14 atm

The equilibrium pressure of NO is 2.14 atm.

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