At equilibrium, the concentrations of N2=3.0 × 10–3M, O2 = 4.2 × 10–3M and NO= 2.8 × 10–3M in a sealed vessel at ?

To determine the equilibrium constant (K) for the reaction involving nitrogen (N2), oxygen (O2), and nitric oxide (NO), we first need to understand the balanced chemical equation for the formation of nitric oxide. The reaction can be represented as follows:

N2(g) + O2(g) ⇌ 2NO(g)

In this reaction, one mole of nitrogen gas reacts with one mole of oxygen gas to produce two moles of nitric oxide. At equilibrium, we have the concentrations of each species given as:

• N2 = 3.0 × 10^–3 M
• O2 = 4.2 × 10^–3 M
• NO = 2.8 × 10^–3 M

Calculating the Equilibrium Constant (K)

The equilibrium constant expression for this reaction is derived from the concentrations of the products and reactants at equilibrium. It is formulated as follows:

K = [NO]² / ([N2][O2])

Now, we can substitute the equilibrium concentrations into this expression:

K = (2.8 × 10^–3)² / ((3.0 × 10^–3)(4.2 × 10^–3))

Step-by-Step Calculation

Let’s break this down step by step:

1. Calculate [NO]²:

(2.8 × 10^–3)² = 7.84 × 10^–6 M²

2. Calculate [N2][O2]:

(3.0 × 10^–3)(4.2 × 10^–3) = 1.26 × 10^–5 M²

3. Now, substitute these values into the K expression:

K = 7.84 × 10^–6 / 1.26 × 10^–5

4. Perform the division:

K ≈ 0.621

Interpreting the Result

The calculated equilibrium constant, K ≈ 0.621, indicates the ratio of products to reactants at equilibrium. A K value less than 1 suggests that at equilibrium, the concentrations of the reactants (N2 and O2) are higher than that of the product (NO), meaning the reaction favors the reactants under the given conditions.

This understanding of equilibrium constants is crucial in predicting the behavior of chemical reactions and can help in adjusting conditions to favor the formation of desired products in industrial applications or laboratory settings.