Question

Nitric oxide is formed in automobile exhaust when nitrogen and oxygen in air react at high temperatures.N2(g) + O2(g) ⇌ 2NO(g)The equilibrium constant Kp for the reaction is 0.14 at 1200 °C. If a container is charged with 0.175 atm of nitrogen and 0.127 atm of oxygen and the mixture is allowed to reach equilibrium, what will be the equilibrium partial pressure of oxygen? Report your answer to three significant figures.

Answer 1

Answer:

0.104 atm

Explanation:

The equilibrium constant based on the partial pressure (Kp) depends only on the gas substances. For a generic reaction aA(g) + bB(g) ⇄ cC(g) + dD(g), it can be calculated by:

$Kp = \frac{pC^cxpD^d}{pA^axpB^b}$, where pX represents the partial pressure of X. So, for the reaction given, let's do a equilibrium table:

N₂(g) + O₂(g) ⇄ 2NO(g)

0.175     0.127       0             Initial

-x            -x          +2x           Reacts (stoichiometry is 1:1:2)

0.175-x     0.127-x   2x         Equilibrium

$Kp = \frac{(2x)^2}{(0.175-x)*(0.127-x)}$

$0.14 = \frac{4x^{2} }{0.022225 - 0.175x -0.127x +x^2}$

$0.14 = \frac{4x^2}{0.022225 - 0.302x + x^2}$

4x² = 0.0031115 - 0.04228x + 0.14x²

3.86x² + 0.04228x - 0.0031115 = 0

Using Bhaskara:

Δ = (0.04228)² - 4x3.86x(-0.0031115)

Δ = 0.04983

x = (-0.04228 +/ √0.04983)/(2*3.86)

x must be positive, so let's calculate only for the positive:

x = 0.0234 atm

So, the equilibrium partial pressure of oxygen is

0.127 - 0.0234 = 0.104 atm