Question

Reaction between \(N_2\) and \(O_2^–\) takes place as follows: 
\(2N_2 (g) + O_2 (g) ⇋ 2N_2O (g)\)
If a mixture of \(0.482 \ mol\) \(N_2\) and \(0.933\  mol\) of \(O_2\) is placed in a \(10\  L\) reaction vessel and allowed to form \(N_2O\) at a temperature for which \(K_c = 2.0 × 10^{–37}\), determine the composition of equilibrium mixture.

Answer 1

Given:

Reaction:
2N₂(g) + O₂(g) ⇌ 2N₂O(g)

Initial moles:
N₂ = 0.482 mol
O₂ = 0.933 mol

Volume of vessel = 10 L
Equilibrium constant, K_c = 2.0 × 10⁻³⁷

---

Step 1: Convert initial moles into concentrations

[N₂]₀ = 0.482 / 10 = 0.0482 M
[O₂]₀ = 0.933 / 10 = 0.0933 M
[N₂O]₀ = 0

---

Step 2: Assume change in concentration

Let the equilibrium concentration of N₂O formed be x M.

From stoichiometry:
Decrease in N₂ = x
Decrease in O₂ = x / 2

Equilibrium concentrations:
[N₂] = (0.0482 − x) M
[O₂] = (0.0933 − x/2) M
[N₂O] = x M

---

Step 3: Write equilibrium constant expression

K_c = [N₂O]² / ( [N₂]² [O₂] )

2.0 × 10⁻³⁷ = x² / [ (0.0482 − x)² (0.0933 − x/2) ]

---

Step 4: Approximation

Since K_c is extremely small, the value of x is negligible compared to initial concentrations.

So,

2.0 × 10⁻³⁷ ≈ x² / (0.0482² × 0.0933)

x² = 2.0 × 10⁻³⁷ × (0.0482² × 0.0933)

x ≈ 6.6 × 10⁻²¹ M

---

Step 5: Equilibrium composition

[N₂O] ≈ 6.6 × 10⁻²¹ M (negligible)
[N₂] ≈ 0.0482 M
[O₂] ≈ 0.0933 M

---

Final Answer:

Because K_c is extremely small, formation of N₂O is negligible.

Equilibrium mixture contains almost the same amounts of reactants as initially:
N₂ ≈ 0.0482 M
O₂ ≈ 0.0933 M
N₂O ≈ 6.6 × 10⁻²¹ M