How to find equilibrium pressure when volume changes

In summary, an equilibrium mixture of N2O4 (P=0.27 atm) and NO2 (P=1.2 atm) at 350K has its volume doubled at constant temperature. Using Le Chatelier's principle, the equilibrium will shift to the right, increasing the number of particles and decreasing the pressure. The equilibrium constant (K) remains unchanged and can be calculated using the equation \ln(\frac{K_2}{K_1}) = -\frac{\Delta H}{R}({1\over T_2}\,-\,{1\over T_1}). The new equilibrium pressures of N2O4 and NO2 can be calculated using the reaction N2O4 <--> 2
  • #1
dnt
238
0
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.
 
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  • #2
dnt said:
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

[tex]\ln(\frac{K_2}{K_1})=-\frac{\Delta H}{R}({1\over T_2}\,-\,{1\over T_1})[/tex]

[tex]\Delta H:\,\,\,is \,\,entalpy [/tex]
 
  • #3
still a little confused. does this mean the partial pressure does not change as volume changes? i don't 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?
 
  • #4
dnt said:
still a little confused. does this mean the partial pressure does not change as volume changes? i don't 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/NA, and pressure will diminish. But it's easy to calculate the new equilibrium pressure with this method, Given the reaction:

[tex]N_2O_4 \leftrightharpoons 2NO_2[/tex]

[tex]P(start): P(N_2O_4)=0,27 \,\, and \,\, P(NO_2)=1,2[/tex]

[tex]\Delta P(N_2O_4)=-x\,\,and\,\,\Delta P(NO_2)=2x[/tex]

[tex]\Delta P_{eq}(N_2O_4)=0,27-x\,\,and\,\,\Delta P_{eq}(NO_2)=2x[/tex]

[tex]K=5,33=\frac{(P(NO_2))^2}{P(N_2O_4}=\frac{(2x)^2}{0,27-x}[/tex]

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

[tex]P(N_2O_4)=0,04\,atm[/tex]
and
[tex]P(NO_2)=0,46\,atm[/tex]
 
  • #5
where did you get K = 5.33? and is that Kp or Kc? and why isn't Peq for NO2 = 1.2 + 2x? where did the 1.2 go?

thanks for all your help.
 

1. How does volume affect equilibrium pressure?

Volume and pressure are directly proportional in a closed system at constant temperature. This means that as the volume decreases, the pressure increases and vice versa. Therefore, changes in volume can significantly impact the equilibrium pressure in a system.

2. What is the formula for finding equilibrium pressure when volume changes?

The formula for finding equilibrium pressure when volume changes is P1V1 = P2V2, where P1 and V1 represent the initial pressure and volume, and P2 and V2 represent the final pressure and volume. This equation is known as Boyle's Law.

3. Can you use the ideal gas law to find equilibrium pressure when volume changes?

Yes, the ideal gas law, PV = nRT, can also be used to find equilibrium pressure when volume changes. In this equation, P represents pressure, V represents volume, n represents the number of moles of gas, R is the ideal gas constant, and T represents temperature in Kelvin.

4. How do you solve for equilibrium pressure when only the volume and temperature are known?

To solve for equilibrium pressure when only the volume and temperature are known, you can use the combined gas law, P1V1/T1 = P2V2/T2. In this equation, P1 and V1 represent the initial pressure and volume, P2 and V2 represent the final pressure and volume, and T1 and T2 represent the initial and final temperatures in Kelvin.

5. What are the units for equilibrium pressure and volume in the equations used to find them?

Equilibrium pressure is typically measured in units of atmospheres (atm) or kilopascals (kPa), while volume is measured in liters (L) or cubic meters (m^3). It is important to make sure that the pressure and volume are both measured in compatible units when using the equations to find equilibrium pressure.

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