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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: daylight on March 14, 2009, 08:29:27 AM

Title: Partial pressures and equilibrium
Post by: daylight on March 14, 2009, 08:29:27 AM
 If one starts with pure NO2(g) at a pressure of 0.500 atm, the total pressure inside the reaction vessel when 2NO2(g) →2NO(g) + O2(g) reaches equilibrium is 0.674 atm. Calculate the equilibrium partial pressure of NO2.
Choose one answer. 
 a. 0.152 atm   
  b. 0.174 atm   
  c. 0.200 atm   
  d. 0.326 atm   
  e. The total pressure cannot be calculated because Kp is not given
I'm having a hard time setting this  up.
     2NO2          2NO          O2
I    .500             0                 0
C     -x              +x               +x
E    .500-x         2X               x
P total= .674=P2NO2 + P2NO + PO2
????this doesn't come out right....
Title: Re: Partial pressures and equilibrium
Post by: ARGOS++ on March 14, 2009, 09:14:57 AM

Dear daylight;

Your line c) is wrong, because it doesn’t represents the stoichiometric factors of your RxN.
Finally express the total pressure (0.674 atm) exclusively with x.

Please try again.
Good Luck!
                    ARGOS++