It is the second way. And one problem I am having (I looked at your post last night but didn't post a response) is I can't really tell which way the reaction is going since the arrows seem to be backwards to me. If I write an equation I usually write it forward. Of course I understand that equilibrium reactions can go either way; it's just that this is a little unusual for me. Can you clarify that for me; i.e., is the reaction
2NO2 ==> N2O4 with Kc = 5.83 x 10^-3 or is it N2O4 ==> 2NO2 with Kc = 5.83 x 10^-3?
Given: N2O4 (g) « 2NO2 (g) @ 25 degrees celcius, Kc is 5.84 x 10^-3.
(A) Calculate the equilibrium concentrations of both gases when 4.00 grams of N2O4 is placed in a 2.00 L flask at 25 degrees celcius.
(B) What will be the new equilibrium concentrations if the volume of the system is suddenly increased to 3.00 L at 25 degrees celcius.
Would you be able to find the moles of N2O4 by taking 4.0g/92.02 g/mol. Then in order to find mol NO2 can you just multiple mol N2O4 by 2 because the equation is 2:1. Then divide both by by liters to get their concentrations.
Or do you have to Calculate the number of moles of N2O4 and hence the concentration (C) in moles/L. Then say that at equilibrium let the concentration of NO2=x so Kc=[NO2]^2/[N2O4] Kc=(x)^2/(C-x)= 5.84x10^-3 we know C so solve for x
If it is the second way can you please guide me on how to solve for X
5 answers
It is N2O4 ==> 2NO2 with Kc= 5.83 x 10^-3 (the arrow is actually a double headed arrow to symbolize equilibrium) Sorry for the misunderstanding! I didn't realize the arrows were like that
Thank you for clarifying.
We can't control spacing on the board. You can write it as you usually do but I shall write mine in a column instead of a row.
initial:
N2O4 = 4.00/92.01 = moles and that divided by 2L = 0.02174M
NO2 = 0
change:
N2O4 = -xM
NO2 = +2xM
equilibrium:
N2O4 = 0.02174-x M
NO2 = 2x M
Set up (2x)^2/(0.02174-x) = 5.84*10^-3 and solve for x. Don't forget that NO2 is 2x. I usually check myself when I'm finished by substituting (NO2) and (N2O4) into Kc expression and see if I get Kc. Tells me if my math is ok. I think this one is. If you want to check yourself, I found x = 0.00495 to make (NO2) = 2x that and (N2O4) = 0.02174-0.00495. I hope this helps.
We can't control spacing on the board. You can write it as you usually do but I shall write mine in a column instead of a row.
initial:
N2O4 = 4.00/92.01 = moles and that divided by 2L = 0.02174M
NO2 = 0
change:
N2O4 = -xM
NO2 = +2xM
equilibrium:
N2O4 = 0.02174-x M
NO2 = 2x M
Set up (2x)^2/(0.02174-x) = 5.84*10^-3 and solve for x. Don't forget that NO2 is 2x. I usually check myself when I'm finished by substituting (NO2) and (N2O4) into Kc expression and see if I get Kc. Tells me if my math is ok. I think this one is. If you want to check yourself, I found x = 0.00495 to make (NO2) = 2x that and (N2O4) = 0.02174-0.00495. I hope this helps.
Thank you!
For part B I did the same thing only this time I used 3.00 L. However I can't seem to find the value of x... I am coming up with 0.0191, but when I substitue that in I do not come up with 5.84 x 10^ -3
For part B I did the same thing only this time I used 3.00 L. However I can't seem to find the value of x... I am coming up with 0.0191, but when I substitue that in I do not come up with 5.84 x 10^ -3
Given: N2O4 (g) « 2NO2 (g) @ 25 degrees celcius, Kc is 5.84 x 10^-3.
(A) Calculate the equilibrium concentrations of both gases when 4.00 grams of N2O4 is placed in a 2.00 L flask at 25 degrees celcius.
(B) What will be the new equilibrium concentrations if the volume of the system is suddenly increased to 3.00 L at 25 degrees celcius.
For part B I did the same thing as part (a) only this time I used 3.00 L. However I can't seem to find the value of x... I am coming up with 0.0191, but when I substitue that in I do not come up with 5.84 x 10^ -3
(A) Calculate the equilibrium concentrations of both gases when 4.00 grams of N2O4 is placed in a 2.00 L flask at 25 degrees celcius.
(B) What will be the new equilibrium concentrations if the volume of the system is suddenly increased to 3.00 L at 25 degrees celcius.
For part B I did the same thing as part (a) only this time I used 3.00 L. However I can't seem to find the value of x... I am coming up with 0.0191, but when I substitue that in I do not come up with 5.84 x 10^ -3
1 answer