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The chemical equilibrium is given:

2NO (g) + O2(g) <—>2N O2, (g)

In a container of volume 10L is in equilibrium a mixture consisting of 10 mol NO (g), 10 mol O2, (g) and 20 mol NO2, (g).

The volume of the container changes under constant temperature and after the restoration of equilibrium the amount of NO has increased by 25%. Calculate the volume change in L.

How do I find the change in volume?

P.s sorry if my English isn’t perfect.

Respuesta :

ardni313

The volume of NO₂=6.25 L

Further explanation

Avogadro's stated :

In the same T,P and V, the gas contains the same number of molecules  

So the ratio of gas volume = the ratio of gas moles  

[tex]\tt \dfrac{V_1}{V_2}=\dfrac{n_1}{n_2}[/tex]

the volume of each gas (in a container of volume 10L) :

Total mol=10+10+20=40 mol

  • NO :

[tex]\tt \dfrac{10}{40}\times 10=2.5~L[/tex]

  • O₂ :

[tex]\tt \dfrac{10}{40}\times 10=2.5~L[/tex]

  • NO₂ :

[tex]\tt \dfrac{20}{40}\times 10=5~L[/tex]

NO₂ (has increased by 25%) :

[tex]\tt 0.25\times 5~L=1.25\\\\new~volume=5+1.25=6.25[/tex]

  • O₂ :

[tex]\tt \dfrac{10}{20}\times 6.25=3.125[/tex]

  • NO :

[tex]\tt \dfrac{10}{20}\times 6.25=3.125[/tex]

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