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there is a question from chemical equilibrium,plz answer it as soon as possible
Q) when NO & NO2 r mixed ther is an equilibria
2NO2=N2O4 Kp=6.8/atm
NO+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.calculate
1) equilibrium partial pressure of NO
2)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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