NO2 AND NO WAS MIXED IN THE RATIO OF 2 :1 AT 300K, IN ONE LITRE VESSEL AND MIXTURE WAS BROUGHT TO EQUILIBIRIUM FORMING TWO EQUILIBRIA
2NO2 <-----> N2O4 Kp=8.48 ATM-1
NO +NO2 <-----> N2O3
THE EQUILIBRIUM PRESSURE WAS 2.31ATM. HALF THE EQUILBRIUM MOLES OF NO2 WAS FOUND TO REACT WITH NaOH soln to give disproportion rxn, yielding one mole of NO-13 from every two moles of NO2 . if NaOH soln was prepared by 0.138gm of sodium. caculate kp for the second rxn
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1 Answers
13572NO2 <-----> N2O4 Kp=8.48 ATM-1
2p-2x-y x
NO +NO2 <-----> N2O3
p-y 2p-2x-y y
at equilibrium total pressure, 2p-2x-y+x+p-y+y=3p-x-y=2.31 (1)
x/(2p-2x-y)2=8.48 (2)
Also
find the moles of NaOH, 0.138/23
moles of NaOH=moles of NO2 reacted with NaOH
or [(2p-2x-y)/2]*V/RT=0.138/23 (3)
solve equations (1), (2), (3) to get p,x,y
