how to find equilibrium pressure when volume changes

**dnt** · Jul20-08, 05:41 PM

question: an equilibrium mixture contains N2O4 (P=0.27 atm) and NO2 (P=1.2 atm) at 350K. the volume is doubled at constant temperature. calculate the equilibrium pressure of both at new equilibrium.

what i know: when you double volume, pressure is halved and le chatliers principle says the equation will shift to the side with more moles of gas:

N2O4 <--> 2NO2

which in this case will shift to the right.

and i also know you cannot simply just half the pressures of each.

how do i start the rest of this problem? is it an ICE problem? any help to get started would be appreciated. thanks.

**janhaa** · Jul20-08, 08:10 PM

Originally Posted by dnt

question: an equilibrium mixture contains N2O4 (P=0.27 atm) and NO2 (P=1.2 atm) at 350K. the volume is doubled at constant temperature. calculate the equilibrium pressure of both at new equilibrium.
what i know: when you double volume, pressure is halved and le chatliers principle says the equation will shift to the side with more moles of gas:
N2O4 <--> 2NO2
which in this case will shift to the right.
and i also know you cannot simply just half the pressures of each.
how do i start the rest of this problem? is it an ICE problem? any help to get started would be appreciated. thanks.

That's right, when the volume doubles, the pressure decreases. and the nb of particles increases too. But after a short time, the system is back to equilibrium according to Le Chatliers principle. It means that the equilibrium constant (K) remains unchanged.
Remember that the equilibrium constant is only dependent of the temperature (T), according

$LaTeX Code: \\ln(\\frac{K_2}{K_1})=-\\frac{\\Delta H}{R}({1\\over T_2}\\,-\\,{1\\over T_1})$

$LaTeX Code: \\Delta H:\\,\\,\\,is \\,\\,entalpy$

**dnt** · Jul20-08, 09:41 PM

still a little confused. does this mean the partial pressure does not change as volume changes? i dont understand why the number pf particles increases too - where do these extra particles come from?

and therefore, if volume and pressure changes do not change equilibrium partial pressures (if i am reading that correctly) what does? is temperature the only thing?

**janhaa** · Jul21-08, 08:03 AM

Originally Posted by dnt

still a little confused. does this mean the partial pressure does not change as volume changes? i dont understand why the number pf particles increases too - where do these extra particles come from?
and therefore, if volume and pressure changes do not change equilibrium partial pressures (if i am reading that correctly) what does? is temperature the only thing?

OK, I will try to explain. Despite english is not my language. Of course if volume (V) increases, the nb of particles (N) increases according pV = nRT and you know that n = N/N_A, and pressure will diminish. But it's easy to calculate the new equilibrium pressure with this method, Given the reaction:

$LaTeX Code: N_2O_4 \\leftrightharpoons 2NO_2$

$LaTeX Code: P(start): P(N_2O_4)=0,27 \\,\\, and \\,\\, P(NO_2)=1,2$

$LaTeX Code: \\Delta P(N_2O_4)=-x\\,\\,and\\,\\,\\Delta P(NO_2)=2x$

$LaTeX Code: \\Delta P_{eq}(N_2O_4)=0,27-x\\,\\,and\\,\\,\\Delta P_{eq}(NO_2)=2x$

$LaTeX Code: K=5,33=\\frac{(P(NO_2))^2}{P(N_2O_4}=\\frac{(2x)^2}{ 0,27-x}$

solve this eqation, and we get x = 0,23
that is

$LaTeX Code: P(N_2O_4)=0,04\\,atm$
and
$LaTeX Code: P(NO_2)=0,46\\,atm$

**dnt** · Jul21-08, 12:33 PM

where did you get K = 5.33? and is that Kp or Kc? and why isnt Peq for NO2 = 1.2 + 2x? where did the 1.2 go?

thanks for all your help.