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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 ???

Chemistry
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Do we need ICE table for this?
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I was actually confused how to make ice chart
I don't think yoiu need an ICE table for this.

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Other answers:

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How do I solve it then ?
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 ---------------------------------------------- \[K_p=\frac{p(NO_2)^2}{p(N_2O_4)} \\ \\ 0.15=\frac{(0.75-2x)^2}{0.6+x} \\ \\ x=0.20162~~~and~~~0.585\]
I don't remember ICE tables at all. Lol i just remember they were long and tedious :p
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Also , how do I determine where is - x and where is + x ?
Then use x=0.20162 on equilibriums, N2O4=0.6+0.20162=0.80162 N2O2= 0.75-2(0.20162)=0.34676
N2O2 started at higher amount than N2O4, so it will be reduced
woohoo, go @.Sam.
Whoops its NO2, lol
thanks @abb0t tried my best :)
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Got it! Thank you for your help a lot !!
yw :)

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