### Related Problems with Solution

##### Problem 1 : For the reaction:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) Kc = 0.10

Given:

- Initial concentration of N₂ = 0.20 M
- Initial concentration of H₂ = 0.50 M

Calculate the equilibrium concentrations of N₂, H₂, and NH₃.

##### Solution :

Let x be the change in concentration at equilibrium: [N₂] = 0.20 - x [H₂] = 0.50 - 3x [NH₃] = 2x

Using the equilibrium constant expression: Kc = ([NH₃]²) / ([N₂][H₂]³) 0.10 = (2x)² / ((0.20 - x)(0.50 - 3x)³)

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 concentrations:

0.10 ≈ (4x²) / ((0.20)(0.50)³)

4x² ≈ 0.10 * 0.0125

4x² ≈ 0.00125

x² ≈ 0.00125 / 4

x² ≈ 0.0003125

x ≈ √0.0003125 ≈ 0.0177

So, the equilibrium concentrations are: [N₂] ≈ 0.20 - 0.0177 ≈ 0.1823 M [H₂] ≈ 0.50 - 3(0.0177) ≈ 0.447 M [NH₃] ≈ 2(0.0177) ≈ 0.0354 M