At 2000 ∘c the equilibrium constant for the reaction 2no (g) ⇌n2 (g) + o2 (g) is kc=2.4*103. part a if the initial concentration of no is 0.171 m, what is the equilibrium concentration of no?

[NO] = 1.72 x 10⁻³ M.

Explanation:

For the reaction:

2NO (g) ⇌ N₂ (g) + O₂ (g),

Kc = [N₂][O₂] / [NO]².

At initial time: [NO] = 0.171 M, [N₂] = [O₂] = 0.0 M. At equilibrium: [NO] = 0.171 M - 2x, [N₂] = [O₂] = x M.

∵ Kc = [N₂][O₂] / [NO]².

∴ 2400 = x² / (0.171 - 2x) ².

Taking the aquare root for both sides:

√ (2400) = x / (0.171 - 2x)

48.99 = x / (0.171 - 2x)

48.99 (0.171 - 2x) = x

8.377 - 97.98 x = x

8.377 = 98.98 x.

∴ x = 8.464 x 10⁻².

∴ [NO] = 0.171 - 2 (8.464 x 10⁻²) = 1.72 x 10⁻³ M.

∴ [N₂] = [O₂] = x = 8.464 x 10⁻² M.