Question

Consider the reaction below: NO3(g) + NO(g)  2 NO2(g) If the concentration equilibrium constant for this reaction is Kc = 3.36 at reaction conditions, what will the equilibrium concentration of NO(g) be if a sealed reaction vessel is filled with gases at initial concentrations of 0.0 M of NO3(g), 1.8 M of NO(g), and 2.8 M of NO2(g)?

Answer 1

[NO] = 2.75 M

Step-by-step explanation:

We determine the equilibrium:

2NO₂(g) ⇄ NO₃ (g) + NO(g)

We analyse the situations:

Initial: 2.8 moles - 1.8 moles

React: x x/2 1.8 + x/2

In the reaction x amount has reacted, so by stoichiometry we add x/2 to the NO₃ (initially we had nothing) and x/2 to NO (initially we had 1.8)

Eq: 2.8-x x/2 1.8+x/2

Let's make the expression for Kc

Kc = [NO₃] . [NO] / [NO₂]² → (x/2) (1.8+x/2) / (2.8-x)²

3.36 = (x/2) (1.8+x/2) / 7.84 - 5.6x + x²

3.36 (7.84 - 5.6x + x²) = (x/2) (1.8+x/2)

26.34 - 18.816x + 3.36x² = 0.9x + x²/4

26.34 - 18.816x + 3.36x² = 0.9x + x²/4 → (3.6x +x²) /4

4 (26.34 - 18.816x + 3.36x²) = 3.6x + x²

105.37 - 75.264x + 13.44x² - 3.6x + x² = 0

105.37 - 78.864x +12.44x² = 0 → A quadratic function

a = 12.44; b = -78.864 ; c = 105.37

(-b +- √(b² - 4ac)) / (2a)

x₁ = 4.42

x₂ = 1.91 → We consider this value, because x₂ give us negative concentration

[NO] = 1.8 + x/2 → 1.8 + 1.91 /2 = 2.75 moles