there is a question from chemical equilibrium,plz answer it as soon as possibleQ) when NO & NO2 r mixed ther is an equilibria2NO2=N2O4 Kp=6.8/atmNO+NO2=N2O3 Kp=?in an experiment when NO & NO2 are mixedin the ratio of 1: 2 ,the total final presssure was 5.05atm & the partial pressure of N2O4 is 1.7 atm.calculate1) equilibrium partial pressure of NO2)Kp for reaction no. 2

CONSIDER PRESSURE OF NO BE 1 ATM THEREFORE PRESSURE OF NO2 WILL BE 2 ATM NOW CONSIDER ONLY ''X'' ATM PRESSURE OF NO2 IS USED IN 1 ST EQUATION 

                              2NO2---------------------> N2O4

 INITIAL                     X                                   0 

EQUILIBRIUM             X-Y                                 Y/2                             ASSUME Y ATM PRESSURE IS DEDUCTED

   KP= PRESSURE N2O4/(PRESSURE NO2)^2 

   6.8=(Y/2)/(X-Y)^2.................................(1)

  PARTIAL PRESSURE OF N2O4=1.7

  (Y/2)=1.7..........................................(2)

 FROM (1) AND (2) YOU WILL KNOW VALUE OF ''X'' AND ''Y''

                                          NO+ NO2----------------> N2O3

INITIAL                                1     (2-X)                        0

EQUILIBRIUM                        1-Z  (2-X)-Z                    Z

   NOW ACCORDING TO THE INFORMATION 

 (1-Z)+[(2-X)-Z]+Z=5.05.................................(3)

YOU KNOW THE VALUE OF ''X'' SO SUBSTITUTE IN (3) AND GET THE VALUE OF Z

  THEREFORE EQILIBRIUM PRESSURE OF NO IS (1-Z)ATM

  KP=(Z)/(1-Z)[(2-X)-Z] .....................................ANSWER SUBSTITUTE THE VALUE OF Z AND GET THE ANSWER

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