Question

N2O4(g)<->2NO2 , kp= 0.15
A gaseous mixture contains 0.600 ATM pressure of n2o4 is mixed with 0.75. ATM pressure of no2. Calculate the equillibrium pressures of each gas ???

Answer 1

Do we need ICE table for this?

Answer 2

I was actually confused how to make ice chart

Answer 3

I don't think yoiu need an ICE table for this.

Answer 4

How do I solve it then ?

Answer 5

ok, ICE stands for I-initial C-change and E-equilibrium, for each reactant and products has these states,
N2O4 <-> 2NO2
I 0.600 0.75
C +x -2x
E 0.6+x 0.75-2x
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\[K_p=\frac{p(NO_2)^2}{p(N_2O_4)} \\ \\ 0.15=\frac{(0.75-2x)^2}{0.6+x} \\ \\ x=0.20162~~~and~~~0.585\]

Answer 6

I don't remember ICE tables at all. Lol i just remember they were long and tedious :p

Answer 7

Also , how do I determine where is - x and where is + x ?

Answer 8

Then use x=0.20162 on equilibriums,
N2O4=0.6+0.20162=0.80162
N2O2= 0.75-2(0.20162)=0.34676

Answer 9

N2O2 started at higher amount than N2O4, so it will be reduced

Answer 10

woohoo, go @.Sam.

Answer 11

Whoops its NO2, lol

Answer 12

thanks @abb0t tried my best :)

Answer 13

Got it! Thank you for your help a lot !!

Answer 14

yw :)