Question

Consider the following equilibrium:

Answer 1

Answer:

1. The equilibrium lies to the left.  

$\large \boxed{2. \, \, 5.41 \times 10^{-2} \text{ mol/L}}$

Explanation:

N₂O₄ ⇌2NO₂; Keq = 5.85 × 10⁻³  

1. Position of equilibrium

Keq < 1, so the position of equilibrium lies to the left.

2. Equilibrium concentration

(a) Set up an ICE table.

$\begin{array}{rcc}\rm \text{N _{2} O}_{4} & \, \rightleftharpoons \, & \text{2NO}_{2} \\\text{E/M}:\qquad \qquad& & 1.78 \times 10^{-2} \\\end{array}$

(b) Calculate the equilibrium concentration

$K_{\text{c}} = \dfrac{\text{[NO _{2} ]}^{2}}{\text{[N _{2} O _{4} ]}} = \dfrac{(1.78 \times 10^{-2})^{2}}{\text{[N _{2} O _{4} ]}} = 5.85 \times 10^{-3}\\\\\begin{array}{rcl}\\(1.78 \times 10^{-2})^{2}&=&\text{[N _{2} O _{4} ]} \times 5.85 \times 10^{-3}\\\text{[N _{2} O _4 ]}&=& \dfrac{(1.78 \times 10^{-2})^{2}}{5.85 \times 10^{-3}}\\\\& = & \mathbf{5.41 \times 10^{-2}}\textbf{ mol/L}\\\end{array}$

$\text{The equilibrium concentration of N _{2} O _{4} is \large \boxed{\mathbf{5.41 \times 10^{-2}} \textbf{ mol/L}} }$

Check:

$\begin{array}{rcl}\dfrac{(1.78 \times 10^{-2})^{2}}{5.41 \times 10^{-2}} & = & 5.85 \times 10^{-3}\\\\5.85 \times 10^{-3} & = & 5.85 \times 10^{-3}\\\end{array}$

It checks.