Question

The first step in HNO3 production is the catalyzed oxidation of NH3. Without a catalyst, a different reaction predominates:

4 NH3(g) + 3 O2(g) reverse reaction arrow 2 N2(g) + 6 H2O(g).
When 0.0150 mol of NH3(g) and 0.0150 mol of O2(g) are placed in a 1.00-L container at a certain temperature, the N2 concentration at equilibrium is 1.96 multiplied by 10-3 M. Calculate Kc.

Answer 1

Kc = 6x10⁻⁶

Step-by-step explanation:

For the reaction:

4NH₃(g) + 3O₂(g) ⇄ 2N₂(g) + 6H₂O(g)

Kc is defined as:

Kc =[N₂]² [H₂O]⁶ / [NH₃]⁴ [O₂]³

The equilibrium concentrations of the gases is -Because volume of the container is 1.00L-:

[N₂] = 2X = 1.96x10⁻³; X = 9.8x10⁻⁴

[H₂O] = 6X; 6ₓ9.8x10⁻⁴ = 5.88x10⁻³

[NH₃] = 0.0150M - 4X = 0.01108M

[O₂] = 0.0150M - 3X = 0.01206M

Replacing in Kc expression:

Kc =[1.96x10⁻³]² [5.88x10⁻³]⁶ / [0.01108M]⁴ [0.01206M]³

Kc = 6x10⁻⁶