Q.

Ammonia gas at a pressure of 20 atm and 270C  is heated in a constant volume in a container to a temperature of 3270C at which new pressure becomes 50 atm and the following equilibrium  2NH3(g) N2(g)+3H2(g) is established. The percentage dissociation of ammonia at this temperature is 

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a

60

b

90

c

10

d

25

answer is C.

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Detailed Solution

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 2NH3(g) N2(g)+3H2(g)
   1α                                        α2                           3α2     Total moles  =1+α

PV=nRT 20V50V=1R300nR600 n=1.25               α=0.25                  %=25

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