Question

Calculation the equilibrium concentration of NO2 in a solution prepared by dissolving 0.2 mol of N2O4 in 400ml of cloroform. For the reaction N2O4 NO, in chloroform, k=1.07×10^-5
(b) what will be the percentage decomposition of the original N2O4

Answer 1

N_2O_4 \rightleftharpoons 2NO_2 \\ [N_2O_4]_{in}= \frac{0.2 \;mol}{0.400\;L} = 0.5 \;M

$K= \frac{[NO_2]^2}{[N_2O_4]} \\ K = \frac{(2x)^2}{0.5-x} = 1.07 \times 10^{-5}$

Assume that the change in concentration of N₂O₄ is small enough to be neglected.

$0.5-x ≈0.5 \\ \frac{4x^2}{0.5}=1.07 \times 10^{-5} \\ x = \sqrt{1.3375 \times 10^{-6}} \\ = 1.15 \times 10^{-3} = 0.00115 \;M$

The equilibrium concentration of NO₂

$[NO_2]_{eq}= 2 \times 0.00115 = 0.0023 \;M$

b.

$[N_2O_4]_{final} = 0.5-0.00115 = 0.49885 \;M$

Proportion:

0.5 M – 100 %

0.49885 M – x

$x = \frac{0.49885 \times 100}{0.5}=99.77 \; \%$

The percentage decomposition of the original N₂O₄ will be 100-99.77=0.23 %