Respuesta :
Answer: Kc =5.915.[tex]10^{-3}[/tex]
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 = [tex]\frac{[NO]^{2} .[O2]}{[NO2]^{2} }[/tex]
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
[tex]\frac{P}{RT}=\frac{ntotal}{V}[/tex]
[tex]\frac{P}{RT} = \frac{n(NO2)}{V} + \frac{n(NO)}{V} + \frac{n(O2)}{V}[/tex]
[NO2]+{NO}+[O2] = [tex]\frac{0.745}{0.082.610}[/tex] = 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 = [tex]\frac{m(NO2)+m(NO)+m(O2)}{V}[/tex]
Substituting mass for molar mass and number of molar,
ρ = M(NO2)·[tex]\frac{n(NO2)}{V}[/tex]+M(NO)·[tex]\frac{n(NO)}{V}[/tex]+M(O2)·[tex]\frac{n(O2)}{V}[/tex]
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.[tex]10^{-3}[/tex]M
Calculating for [NO]
[NO]=2[O2]
[NO]=2.13.3.[tex]10^{-3}[/tex]
[NO]=26.6.[tex]10^{-3}[/tex]M
And finding [NO2],
[NO2]=0.015 - 3[O2]
[NO2]=0.015 - 3.13.3.[tex]10^{-3}[/tex]
[NO2]=39.885.[tex]10^{-3}[/tex]M
So, to calculate Kc:
Kc = (26.6.[tex]10^{-3}[/tex])²· 13.3.[tex]10^{-3}[/tex] / (39.885.[tex]10^{-3}[/tex])²
Kc= 5.915.[tex]10^{-3}[/tex]
The Kc is 5.915.[tex]10^{-3}[/tex].
