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
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?
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.
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
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
To solve part A, you can indeed calculate the moles of N2O4 by dividing the mass (4.0g) by the molar mass of N2O4 (92.02 g/mol). This will give you the number of moles of N2O4.
Next, you can multiply the moles of N2O4 by 2 (as indicated by the balanced equation) to find the moles of NO2. Since the equation is 2:1, for every mole of N2O4 that reacts, 2 moles of NO2 are formed.
After obtaining the moles of N2O4 and NO2, you can divide the moles by the volume of the flask (2.00 L) to find their concentrations in moles per liter (M).
To solve part B, when the volume of the system is suddenly increased to 3.00 L, you cannot directly use the given concentrations in part A. You need to recalculate the new equilibrium concentrations.
To do this, you'll first need to determine the total number of moles of both N2O4 and NO2 in the system using the initial concentrations you calculated in part A. Then, divide the total moles of each species by the new volume (3.00 L) to obtain the new equilibrium concentrations.
To solve for X (the equilibrium concentration of NO2) in part B, you can use the formula you mentioned: Kc = [NO2]^2 / [N2O4]. Rearranging this equation to solve for X, you get:
X^2 / (C - X) = Kc.
Substitute the values you calculated in part A for C and solve the quadratic equation for X.