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Q.

In a 3L vessel, the following equilibrium partial pressures are measured; 

N2= 100  mm  of Hg,H2= 400  mm  of Hg and NH3= 1000  mm of Hg.Nitrogen is removed from the vessel at constant temperature until the pressure of hydrogen at equilibrium is equal to 700  mm of Hg . The new equilibrium partial pressure (in mm of Hg) of N2 is

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a

18.66

b

1194

c

11.94

d

200

answer is A.

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Detailed Solution

Given, 

PN2=  100  mm

PH2=  400  mm

PN3=  1000  mm

At equilibrium, PH2=  700  mm

New equilibrium PN2=  ?

N2100+3H24002NH31000KP=100024003×1002

After removal of N2, eqm shifts backward and N2, H2 are formed

PH2(new)=700PH2(formal)=300mm

PNH3 decreases by 200mm

PH2=700,PNH3=800, If PN2=P

Then, 1000240031002=80027003.PP=82×4373

 

 

 

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