Reaction between N2 and O2– takes place as follows: 2N2 (g) + O2 (g) ? 2N2O (g) If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc= 2.0 × 10–37, determine the composition of equilibrium mixture. ?

To determine the composition of the equilibrium mixture for the reaction between nitrogen (N2) and oxygen (O2) to form nitrous oxide (N2O), we need to analyze the initial amounts of reactants, the equilibrium constant (Kc), and how the reaction proceeds. The reaction can be represented as follows:

Understanding the Reaction

The balanced chemical equation is:

2N2 (g) + O2 (g) ⇌ 2N2O (g)

This indicates that two moles of nitrogen gas react with one mole of oxygen gas to produce two moles of nitrous oxide gas. Given the equilibrium constant (Kc) of 2.0 × 10^-37, we can infer that the reaction heavily favors the reactants at equilibrium, indicating that very little product will form.

Initial Conditions

We start with:

• 0.482 moles of N2
• 0.933 moles of O2

Since the reaction vessel has a volume of 10 L, we can calculate the initial concentrations:

• [N2] = 0.482 mol / 10 L = 0.0482 M
• [O2] = 0.933 mol / 10 L = 0.0933 M

Setting Up the ICE Table

To analyze the changes in concentration as the reaction reaches equilibrium, we can set up an ICE (Initial, Change, Equilibrium) table:

Applying the Equilibrium Constant Expression

The equilibrium constant expression for this reaction is:

Kc = [N2O]² / ([N2]² * [O2])

Substituting the equilibrium concentrations from the ICE table into the Kc expression gives:

2.0 × 10^-37 = (2x)² / ((0.0482 - 2x)² * (0.0933 - x))

Assuming x is Small

Given the very small value of Kc, we can assume that x will be negligible compared to the initial concentrations. Thus, we can simplify:

• 0.0482 - 2x ≈ 0.0482
• 0.0933 - x ≈ 0.0933

Now, substituting these approximations into the Kc expression:

2.0 × 10^-37 = (2x)² / (0.0482² * 0.0933)

Solving for x

Calculating the denominator:

• 0.0482² = 0.002321
• 0.002321 * 0.0933 ≈ 0.000216

Now, substituting back into the equation:

2.0 × 10^-37 = (4x²) / 0.000216

Rearranging gives:

4x² = 2.0 × 10^-37 * 0.000216

4x² = 4.32 × 10^-41

Thus, solving for x:

x² = 1.08 × 10^-41

x ≈ 1.04 × 10^-21

Final Concentrations at Equilibrium

Now we can find the equilibrium concentrations:

• [N2] = 0.0482 - 2(1.04 × 10^-21) ≈ 0.0482 M
• [O2] = 0.0933 - (1.04 × 10^-21) ≈ 0.0933 M
• [N2O] = 2(1.04 × 10^-21) ≈ 2.08 × 10^-21 M

Summary of Equilibrium Composition

At equilibrium, the concentrations are approximately:

• [N2] ≈ 0.0482 M
• [O2] ≈ 0.0933 M
• [N2O] ≈ 2.08 × 10^-21 M

This analysis shows that due to the very small Kc value, the reaction does not proceed significantly towards the formation of products, resulting in a mixture that is predominantly reactants. This is a classic example of how equilibrium constants can indicate the extent of a reaction.