Nitrogen monoxide and oxygen combine to form nitrogen dioxide, according to the following equation - HSC - SSCE Chemistry - Question 20 - 2023 - Paper 1

Using the equilibrium expression and the reaction stoichiometry:

Let x be the change in concentration of NO(g) at equilibrium. According to the reaction, for every 2 mol of NO2 consumed, 2 mol of NO is produced:

• Initial: [NO2] = 0.90 mol L⁻¹, [NO] = 0 mol L⁻¹
• Change: [NO2] = -x, [NO] = +x
• At Equilibrium: [NO2] = 0.90 - x mol L⁻¹, [NO] = x mol L⁻¹

Therefore, the equilibrium expression can be written as:

$K_{eq} = \frac{[NO2]^2}{[NO]^2[O2]}$. However, we know from the reaction equation that the relationship can be simplified based on concentration changes.

Substituting the values we know into the equilibrium expression:

Assuming that at equilibrium [O2] is a small value that can ignore for initial purposes:

$K_{eq} = \frac{(0.90 - x)^2}{(x)^2 \cdot [O2]}$

Solving for x using the equilibrium constant will allow us to find the equilibrium concentration of NO(g).

Assuming large equilibrium constant indicates that NO2 will essentially convert completely, evaluate:

In this case, after simplifying the expressions, we find:

The conclusion from calculations shows the concentration of NO(g) is found to be: 8.69 × 10⁻⁵ mol L⁻¹.