Q.

A flask is initially filled with pure N2Og(g) having pressure 2 bar and following equilibria are established.

N2O3(g)NO2(g)+NO(g)       KP1=2.5bar ?

2NO2(g)N2O4(g)           KPy=?

If at equilibrium partial pressure of NO(g) was found to be 1.5 bar, then:

 

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a

Equilibrium partial pressure of N2O3(g) is 0.5 bar

b

Equilibrium partial pressure of N2O4 is 0.33 bar

c

Value ofKP2 is 0.48 bar

d

Equilibrium partial pressure of NO2(g) is 0.83 bar

answer is A, B, C, D.

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Detailed Solution

Here, we have,

N2O3NO2+NO

2x    xy x2NO2    N2O4xy    y/2

x=1.5  (given)

KP1=(3y)×1.50.5PN2O5=21.5=0.5barPNO2=0.83bary=1.50.83=0.67PN2O4=0.3barKP2=0.33(0.82)2=0.48

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