Related Problems with Solution
Problem 2 : For the reaction:
N₂O₄(g) ⇌ 2NO₂(g) Kc = 0.040
Given:
- Initial concentration of N₂O₄ = 0.10 M
Calculate the equilibrium concentrations of N₂O₄ and NO₂.
Solution :
Let x be the change in concentration at equilibrium: [N₂O₄] = 0.10 - x [NO₂] = 2x
Using the equilibrium constant expression: Kc = ([NO₂]²) / [N₂O₄] 0.040 = (2x)² / (0.10 - x)
Now, we can solve for x. Since Kc is relatively small, we can make an approximation that x will be small compared to the initial concentration of N₂O₄:
0.040 ≈ (4x²) / 0.10
4x² ≈ 0.040 * 0.10
4x² ≈ 0.004
x² ≈ 0.004 / 4
x² ≈ 0.001
x ≈ √0.001 ≈ 0.0316
So, the equilibrium concentrations are: [N₂O₄] ≈ 0.10 - 0.0316 ≈ 0.0684 M [NO₂] ≈ 2(0.0316) ≈ 0.0632 M
