## S 2 years ago 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 ???

1. .Sam.

Do we need ICE table for this?

2. S

I was actually confused how to make ice chart

3. abb0t

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

4. S

How do I solve it then ?

5. .Sam.

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 ---------------------------------------------- $K_p=\frac{p(NO_2)^2}{p(N_2O_4)} \\ \\ 0.15=\frac{(0.75-2x)^2}{0.6+x} \\ \\ x=0.20162~~~and~~~0.585$

6. abb0t

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

7. S

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

8. .Sam.

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

9. .Sam.

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

10. abb0t

woohoo, go @.Sam.

11. .Sam.

Whoops its NO2, lol

12. .Sam.

thanks @abb0t tried my best :)

13. S

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

14. .Sam.

yw :)

Find more explanations on OpenStudy