A flask is charged with 1.596 atm of N2O4(g) and 1.008 atm of NO2(g) at 25°C, and the following equilibrium is achieved.
N2O4(g)-> 2 NO2(g)
After equilibrium is reached, the partial pressure of NO2 is 0.504 atm.
(a) What is the equilibrium partial pressure of N2O4?
for this quetsion i set up an ice box and got 1.85 as my answer but it was wrong.
(c) Calculate the value of Kc for the reaction
If you had typed in your work I could have found the error.
N2O4 -> 2NO2
I 1.596 1.008
C .252 -.504
E 1.845 .504
1.596+0.252 = 1.848.
that would still be 1.85 with 3 sig figs
To answer both parts (a) and (c) of your question, we need to use the concept of equilibrium and the expression for equilibrium constant, Kc.
(a) To find the equilibrium partial pressure of N2O4, we can use the stoichiometry of the reaction and the equilibrium partial pressure of NO2. Since the balanced equation shows that 1 mole of N2O4 produces 2 moles of NO2, we can set up the following relation:
2 * (partial pressure of NO2) = partial pressure of N2O4
Substituting the given value of the partial pressure of NO2 (0.504 atm), we get:
2 * 0.504 atm = 1.008 atm
Therefore, the equilibrium partial pressure of N2O4 is 1.008 atm, not 1.85 atm as you mentioned.
(c) To calculate the value of Kc, we need to use the equilibrium concentrations of the species involved. However, in this case, we only have partial pressures provided instead of concentrations. But we can still find the value of Kc by using the relation between partial pressures and concentrations known as the ideal gas law.
The ideal gas law is given as:
PV = nRT
Where P is the partial pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in kelvin.
At equilibrium, we have:
[P(NO2)]^2 / [P(N2O4)] = Kc
Substituting the values of partial pressures into this expression, we get:
[(0.504 atm)^2] / [1.008 atm] = Kc
Simplifying, we find:
Kc = 0.252
Therefore, the value of Kc for the reaction is 0.252.