Answer to Question #333134 in Chemistry for SEVEN

Question

For the following reaction at 373 K, Kc = 0.50. If the initial concentration of N2O4 is 0.152 M, what is the equilibrium concentration of NO2?

N2O4 (g) ⇆ NO2 (g)

Answer 1

Solution:

Balanced chemical equation:

N₂O₄(g) ⇆ 2NO₂(g)

For the given reaction,

ICE Table:

According to the ICE Table:

[NO₂] = 2x

[N₂O₄] = 0.152 − x

K_c = 0.50

Therefore,

0.50 = (2x)² / (0.152 − x) = (4x²) / (0.152 − x)

4x² + 0.50x − 0.076 = 0

Solving this quadratic equation, we get:

x = 0.089

[NO₂] = 2x = 2 × 0.089 = 0.178 M

[NO₂] = 0.178 M

Answer: The equilibrium concentration of NO₂ is 0.178 M