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

4 answers

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
Similar Questions
  1. N2O4 --->2NO2a 1 liter flask is charged with .4 mols of N2O4. at equilibrium at 373 k, .0055 mols of N2O4 remain. what is the Kc
    1. answers icon 2 answers
    1. answers icon 6 answers
  2. i already posted this question, but im still confused on it.Dinitrogentetraoxide partially decomposes according to the following
    1. answers icon 2 answers
  3. I need help. I have no clue how to find this.A flask is charged with 1.680 atm of N2O4(g) and 1.220 atm of NO2(g) at 25 degrees
    1. answers icon 1 answer
more similar questions