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Question

If Kp for the reaction N2O42NO2 is 0.66 then what is the equilibrium pressure of N2O3 (Total pressure at equilibrium is 0.5 atm):

A
0.168
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B
0.322
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C
0.1
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D
0.5
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Solution

The correct option is A 0.168
N2O42NO2
KP=0.66
Peq=0.5 atm
N2O42NO2
At t=0 1 0
At t=eq 1x 2x
Mole fraction of N2O4=1x1+x
Mole fraction of NO2=2x1+x
Partial pressure of N2O4=1x1+x×0.5
Partial pressure of NO2=2x1+x×0.5
Kp=[P(NO)2]2[P(N2O4)]0.66=2x1+x2x1+x×0.5×0.51x1+x×0.50.66=4x2×0.51x20.660.66 x2=2x22.66x2=0.66x=0.5
Now put the value of x in partial pressure of N2O4)
=10.51+0.5×0.5=0.5×0.51.5=0.166atm

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