please help me within five munite

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 particular temperature, Kp = 0.260 reaction

N2O4(g)  2NO2(g)

With no change  amount material in flask, the volume container question increased to 5.000 times original. Assuming constant temperature,calculate (new) total pressure, at equilibrium.

Feb 1st, 2015

Thank you for the opportunity to help you with your question!

It makes no difference whether you increase 5times the volume after at the equlibrium established in (1) of you start start the reaction from initial pure nitrogen dioxide in a 5times bigger volume flask.

  increasing the volume 5 times  original means halving the initial pressure to
p₀/592 inHg

so normal atmosphere pressure= 29.92 in Hg

so initial pressure becomes = 29.92/5 =5.984

The partial fulfill equilibrium condition:
Kp = (p(NO2))² / p(N2O4)  

Kp = (p₀ - 5·Δp)² / Δp 

now we have Kp=0.26 and p0=5.984

now find out  Δp

0.26 = p0^2 +25 Δp^2 -1op0 Δp

25 Δp^2 -  10.26 p0 Δp + p0^2=0

25 Δp^2  -  61.4 Δp   +35.81 =0

find the value of Δp

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 13th, 2015

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