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.
(C) What effect would increasing the pressure at 25 degrees celcius have on the equilibrium?
You have not shown the units for Kc, but let us assume that the units are moles/L
1. Calculate the number of moles of N2O4 and hence the concentration (C) in moles/L.
2. At equilibrium let the concentration of NO2=x
Thus Kc=[NO2]^2/[N2O4]
Kc=(x)^2/(C-x)= 5.84x10^-3
we know C so solve for x
for part B repeat the calculation using the new volume.
for part C by Le Chateliers Principle the equilibrium will move in the direction to accommodate the change so will move to smaller volume, which is the N2O4 side.
To determine the equilibrium concentrations of the gases in each scenario, we need to use the stoichiometry of the reaction and the given equilibrium constant.
(A) Calculate the equilibrium concentrations when 4.00 grams of N2O4 is placed in a 2.00 L flask at 25 degrees Celsius:
1. Convert the given mass of N2O4 to moles. The molar mass of N2O4 is 92.02 g/mol.
moles of N2O4 = mass / molar mass = 4.00 g / 92.02 g/mol = 0.04342 mol
2. Initially, there are no NO2 molecules, so the concentration of NO2 is 0 M.
Using the stoichiometry of the reaction, we have:
N2O4 (g) ⇌ 2NO2 (g)
At equilibrium: [N2O4] = (initial moles - moles reacted) / volume = (0.04342 mol - 2x) / 2.00 L
For the 2NO2, since 2 moles of NO2 are produced per mole of N2O4 reacted:
[NO2] = 2x / 2.00 L
We can assume that x is small compared to the initial moles of N2O4, so we can neglect its contribution to the initial moles. Therefore, we can simplify the expressions for [N2O4] and [NO2] at equilibrium.
Since Kc = [NO2]^2 / [N2O4], we can substitute the equilibrium concentrations into the expression for Kc:
Kc = (2x / 2.00 L)^2 / [(0.04342 mol - 2x) / 2.00 L]
Solve this equation to find the value of x, which represents the change in moles of N2O4 and NO2 at equilibrium.
(B) To determine the new equilibrium concentrations if the volume of the system is suddenly increased to 3.00 L, we use the same equilibrium expression as in part (A):
Kc = (2x / 3.00 L)^2 / [(0.04342 mol - 2x) / 3.00 L]
Using this equation, solve for the new value of x.
(C) To determine the effect of increasing the pressure at 25 degrees Celsius, we can use Le Chatelier's principle.
Since the forward reaction produces two molecules of gas (2NO2), increasing the pressure will shift the equilibrium towards the side with fewer gas molecules (N2O4). As a result, the equilibrium concentration of N2O4 will increase while the concentration of NO2 will decrease.
It's important to note that while we can predict the effect of changes in pressure on equilibrium concentrations using Le Chatelier's principle, we may need to recalculate the equilibrium concentrations using the equation and the new value of Kc if the pressure change is substantial.