Answer to Question #228806 in Inorganic Chemistry for Kelechi

Question #228806
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
1
Expert's answer
2021-08-24T02:23:16-0400

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


"K= \\frac{[NO_2]^2}{[N_2O_4]} \\\\\n\nK = \\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 \u22480.5 \\\\\n\n\\frac{4x^2}{0.5}=1.07 \\times 10^{-5} \\\\\n\nx = \\sqrt{1.3375 \\times 10^{-6}} \\\\\n\n= 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 %


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Comments

Wilson Samuel
24.08.21, 09:27

Nice one, thanks alot I appreciate greatly..

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