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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 ???

  • 11 months ago
  • 11 months ago

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  1. .Sam.
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    Do we need ICE table for this?

    • 11 months ago
  2. S
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    I was actually confused how to make ice chart

    • 11 months ago
  3. abb0t
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    I don't think yoiu need an ICE table for this.

    • 11 months ago
  4. S
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    How do I solve it then ?

    • 11 months ago
  5. .Sam.
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    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 ---------------------------------------------- \[K_p=\frac{p(NO_2)^2}{p(N_2O_4)} \\ \\ 0.15=\frac{(0.75-2x)^2}{0.6+x} \\ \\ x=0.20162~~~and~~~0.585\]

    • 11 months ago
  6. abb0t
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    I don't remember ICE tables at all. Lol i just remember they were long and tedious :p

    • 11 months ago
  7. S
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    Also , how do I determine where is - x and where is + x ?

    • 11 months ago
  8. .Sam.
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    Then use x=0.20162 on equilibriums, N2O4=0.6+0.20162=0.80162 N2O2= 0.75-2(0.20162)=0.34676

    • 11 months ago
  9. .Sam.
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    N2O2 started at higher amount than N2O4, so it will be reduced

    • 11 months ago
  10. abb0t
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    woohoo, go @.Sam.

    • 11 months ago
  11. .Sam.
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    Whoops its NO2, lol

    • 11 months ago
  12. .Sam.
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    thanks @abb0t tried my best :)

    • 11 months ago
  13. S
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    Got it! Thank you for your help a lot !!

    • 11 months ago
  14. .Sam.
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    yw :)

    • 11 months ago
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