At a particular temperature, Kp = 0.490 for the reaction
N2O4(g) = 2NO2(g)
A flask containing only NO2(g) at an initial pressure of 10.00 atm is allowed to reach equilibrium.
a) Calculate the total pressure in this flask at equilibrium.
answer: 5.724 atm
b) With no change in the amount of material in the flask, the volume of the container in question is decreased to 0.100 times the original volume. Assuming constant temperature, calculate the (new) total pressure, at equilibrium.
answer: ?
Yes, that's what you do. Here is what I wrote in my first response that summarizes what you do.
I'll help get you started.
From part a you know x = 4.276 so E line is
(NO2) = 10-2x = (10-2*4.276) atm and (NO2) = 4.276 atm. Plug those into the I line (initial) and go through the same calculation.
I understand a) but I'm stuck getting started on how take into account the 0.100 times the original volume. Im not sure how to rework the equation. Do I times the E line concentrations by 10? then that will be my new initial and then proceed to find the new x?
Hmm, I've got the equation and the initial pressure, but I'm missing a key piece of information. Do you happen to know the value of Kp at the given temperature?
To answer the second part of the question and find the new total pressure, we can use the Ideal Gas Law and the concept of partial pressures.
Let's assume the initial pressure of NO2(g) in the flask is P_initial. According to the given reaction, 2 moles of NO2(g) react to form 1 mole of N2O4(g), which means the number of moles of NO2(g) decreases by half and the number of moles of N2O4(g) formed is equal to the number of moles of NO2(g) at equilibrium.
Since N2O4(g) is not initially present in the flask, its pressure can be assumed to be zero.
Using the equation P = nRT/V, where P is pressure, n is the number of moles, R is the ideal gas constant, T is the temperature, and V is the volume, we can write the equations for both NO2(g) and N2O4(g) at equilibrium:
For NO2(g):
P_NO2 = (n_NO2/V)
For N2O4(g):
P_N2O4 = (n_NO2/V)
Since the stoichiometric ratio of NO2(g) to N2O4(g) is 2:1, the pressure of NO2(g) at equilibrium is twice as much as the pressure of N2O4(g):
P_NO2 = 2 * P_N2O4
Now, let's consider the given value of Kp:
Kp = P_N2O4^2 / (P_NO2^2)
Substituting the expressions for P_NO2 and P_N2O4:
Kp = (n_NO2/V)^2 / [2 * (n_NO2/V)]^2
Simplifying:
Kp = (n_NO2/V)^2 / (4 * n_NO2^2 / V^2)
Kp = n_NO2^2 / (4 * n_NO2^2)
4 * Kp = 1
Since Kp is a constant at a given temperature, 4 * Kp is also a constant. Thus, it remains the same when the volume changes.
Now, let's calculate the new total pressure using the new volume (0.100 times the original volume).
Since the amount of material in the flask remains unchanged, the moles of NO2(g) and N2O4(g) also remain the same. Therefore, the ratio of NO2(g) to N2O4(g) at equilibrium remains the same.
Now, substitute the value of Kp into the equation:
4 * Kp = P_N2O4^2 / (P_NO2^2)
Since P_N2O4 = P_NO2/2 (from the stoichiometry of the reaction), substitute this into the equation:
4 * Kp = (P_NO2/2)^2 / P_NO2^2
Now solve for P_NO2:
4 * Kp = 1/4
P_NO2^2 = 4
P_NO2 = 2
Since P_NO2 is the pressure of NO2(g), the total pressure at equilibrium is the sum of P_NO2 and P_N2O4:
Total pressure = P_NO2 + P_N2O4
Total pressure = 2 + (2/2)
Total pressure = 2 + 1
Total pressure = 3 atm
Therefore, the new total pressure at equilibrium, when the volume is decreased to 0.100 times the original volume, is 3 atm.
I........................0....................10
C.....................+x...................-2x
E.....................+x...................10-2x
Kp = 0.490 = (NO2)^2/(N2O4)
0.490 = (10-2x)^2/(x)
Solve for x and evaluate NO2 and N2O4, then rewrite the equation as above and knowing the volume decreased 0.100 as much so you know the pressure increased by a factor of 10. Calculate the new concentrations of NO2 and N2O4 for the E line, make those nujmbers the I line, and reaclculate x at that point.
Post your work if you get stuck. I'll help get you started.
From part a you know x = 4.276 so E line is
(NO2) = 10-2x = (10-2*4.276) atm and (NO2) = 4.276 atm. Plug those into the I line (initial) and go through the same calculation.