Consider the following equation: N₂O₄(g) ⇌ 2 NO₂(g) Kc = 5.8 × 10^-3
If the initial concentration of N₂O₄(g) = 0.040 M and the initial concentration of NO₂(g) is 0 M, what is the equilibrium concentration of N₂O₄(g) ?

Question

Quizplus · Accepted Answer

Using the equation for Kc, we can set up an ICE table: N₂O₄(g) ⇌ 2 NO₂(g) I 0.040 M 0 M C -x +2x E 0.040-x 2x Now we can plug in the values for equilibrium concentrations into the Kc expression and solve for x: Kc = [NO₂]2 / [N₂O₄] 5.8 × 10^-3 = (2x)2 / (0.040-x) Solving for x gives us: x = 3.3 × 10^-2 Thus, the equilibrium concentration of N₂O₄ is: 0.040 - x = 0.040 - 3.3 × 10^-2 = 3.7 × 10^-2, which corresponds to choice C.

Consider the Following Equation: N2O4(g) ⇌ 2 NO2(g) Kc =

Consider the Following Equation: N₂O₄(g) ⇌ 2 NO₂(g) Kc =