Question

1. is a mixture of 0.0205 mol no₂(g) and 0.750 mol n₂o₄(g) in a 5.25 l flask at 25 °c at equilibrium? if not, in which direction will the reaction proceed?

n₂o₄(g) ⇌ 2 no₂(g) kₑ = 4.61×10⁻³ at 25 °c.

Explanation:

Step1: Calculate the initial concentrations

The concentration formula is $c=\frac{n}{V}$. For $NO_2$, $c_{NO_2}=\frac{0.0205\ mol}{5.25\ L}$, and for $N_2O_4$, $c_{N_2O_4}=\frac{0.750\ mol}{5.25\ L}$.
$c_{NO_2}=\frac{0.0205}{5.25}\ mol/L\approx0.00391\ mol/L$
$c_{N_2O_4}=\frac{0.750}{5.25}\ mol/L\approx0.143\ mol/L$

Step2: Calculate the reaction - quotient $Q_c$

The expression for the reaction - quotient for the reaction $N_2O_4(g)
ightleftharpoons2NO_2(g)$ is $Q_c=\frac{[NO_2]^2}{[N_2O_4]}$.
Substitute the calculated concentrations into the $Q_c$ formula:
$Q_c=\frac{(0.00391)^2}{0.143}=\frac{1.53\times10^{-5}}{0.143}\approx1.07\times10^{-4}$

Step3: Compare $Q_c$ and $K_c$

Given $K_c = 4.61\times10^{-3}$.
Since $Q_c(1.07\times10^{-4})

Answer:

The reaction is not at equilibrium and will proceed in the forward direction.