Question

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.

Answer 1

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,

P·V=ntotal·R·T

$\frac{P}{RT}=\frac{ntotal}{V}$

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

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

As [NO] = 2[O2]

[NO2]+2[O2]+[O2]=0.015

[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

32[O2]+60[O2]-138[O2]=0.0525-0.69

-48[O2]= -0.6375

[O2]=13.3.$10^{-3}$M

Calculating for [NO]

[NO]=2[O2]

[NO]=2.13.3.$10^{-3}$

[NO]=26.6.$10^{-3}$M

And finding [NO2],

[NO2]=0.015 - 3[O2]

[NO2]=0.015 - 3.13.3.$10^{-3}$

[NO2]=39.885.$10^{-3}$M

So, to calculate Kc:

Kc = (26.6.$10^{-3}$)²· 13.3.$10^{-3}$ / (39.885.$10^{-3}$)²

Kc= 5.915.$10^{-3}$

The Kc is 5.915.$10^{-3}$.