Question

Calculate the equilibrium concentrations of NO, O₂, and NO₂ in a mixture at 250 °C that results from the reaction of 0.20 M NO and 0.10 M O₂. (Hint: K is large; assume the reaction goes to completion then comes back to equilibrium.)

$2NO(g)+O_2(g)\rightleftharpoons 2N{O}_2(g)K_c=2.3\times 10^{5}at250^{o}C$

Short answer

The equilibrium concentrations of NO, O₂,and NO₂ are

$\begin{array}{c}[\mathrm{N}\mathrm{O}]=7.04\times {10}^{-3}M\\ \left[{\mathrm{O}}_2\right]=3.52\times {10}^{-3}M\\ [\mathrm{N}\mathrm{O}]=0.193\mathrm{M}\end{array}$.

Step-by-step solution

Determine change in concentration:

Given information:

$2\text{NO}(\text{g})+{\text{O}}_2(\text{g})\rightleftharpoons 2\text{N}{\text{O}}_2(\text{g})$

• The concentration of NO is 0.20 M.
• The concentration of O₂ is 0.10 M.
• The equilibrium constant is $K_c=2.3\times {10}^5$.

We have to find the equilibrium concentrations of NO, O₂, and NO₂.

• We will assume that the volume of a solution is 1 L, hence the number of moles of NO is 0.20 mol and the number of moles of O₂ is 0.10 mol.

Since 2 moles of NO reacts with 1 mole of O₂, to produce 2 moles of NO₂, 0.20 mole of NO will exact with 0.10 mole of O₂, and produce 0.20 moles of NO₂(0.20 M).

We have to determine the value of x,

$\begin{array}{c}{K}_c=\frac{{\left[\mathrm{N}{\mathrm{O}}_2\right]}^2}{{[NO]}^2\times \left[O_2\right]}\\ 2.3\times {10}^5=\frac{{(0.2-2x)}^2}{{(2x)}^2\times x}\end{array}$

Since $K_c$is too large, we will neglect the change in concentration of $\mathrm{N}{\mathrm{O}}_2$, and get

$\begin{array}{c}2.3\times {10}^5=\frac{{(0.2)}^2}{{(2x)}^2\times x}\\ 4x^3=\frac{{(0.2)}^2}{2.3\times {10}^5}\\ x^3=4.35\times {10}^{-8}\\ x=3.52\times {10}^{-3}\mathrm{M}\end{array}$

Determine equilibrium concentration of all species:

The change in the concentration obtained $x=3.52\times {10}^{-3}\mathrm{M}$

Therefore, the equilibrium concentrations of NO, O₂, and NO₂ are

$\begin{array}{c}[\mathrm{N}\mathrm{O}]=2\mathrm{x}=7.04\times {10}^{-3}\mathrm{M}\\ \left[{\mathrm{O}}_2\right]=\mathrm{x}=3.52\times {10}^{-3}\mathrm{M}\\ [\mathrm{N}\mathrm{O}]=0.2\mathrm{M}-2\mathrm{x}=0.193\mathrm{M}\end{array}$.