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KnudsenEr KnudsenEr May 11, 2022

A sample of pure NO2 is heated to 337?C at which temperature it partially dissociates according to the equation2NO2(g)?2NO(g)+O2(g)At equilibrium the density of the gas mixture is 0.525g/L at 0.745atm . Calculate Kc for the reaction.

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Answer:Kc =5.915.10^{-3}

Explanation: Kc is an equilibrium constant of a chemical reaction and is dependent upon the reagents and products concentration. For this reaction, Kc is:  Kc = \frac{[NO]^{2} .[O2]}{[NO2]^{2} }

To determine each concentration, we have to find each relation:

From the reaction, we know that [NO] = 2[O2] (1)

From the Ideal Gas Law,



\frac{P}{RT} = \frac{n(NO2)}{V} + \frac{n(NO)}{V} + \frac{n(O2)}{V}

[NO2]+{NO}+[O2] = \frac{0.745}{0.082.610} = 0.015 mol/L

As [NO] = 2[O2]


[NO2] = 0.015 - 3[O2] (2)

To resolve this equation, we will turn towards density of the mixture:

ρ = \frac{m(NO2)+m(NO)+m(O2)}{V}

Substituting mass for molar mass and number of molar,

ρ = M(NO2)·\frac{n(NO2)}{V}+M(NO)·\frac{n(NO)}{V}+M(O2)·\frac{n(O2)}{V}

Knowing the molar mass of each molecule:

ρ = 46·[NO2]+30·[NO]+32·[O2] = 0.525 g/L (3)

Substituting (1), (2) and (3) we have:

46(0.015 - 3[O2])+30(2[O2])+32.[O2]=0.525


-48[O2]= -0.6375


Calculating for [NO]




And finding [NO2],

[NO2]=0.015 - 3[O2]

[NO2]=0.015 -^{-3}


So, to calculate Kc:

Kc = (26.6.10^{-3})²· 13.3.10^{-3} / (39.885.10^{-3}

Kc= 5.915.10^{-3}

The Kc is 5.915.10^{-3}.

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May 11, 2022
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