Q.19. Ammonia gas at 15 atm is introduced in a rigid vessel at 300 K. At equilibrium the total pressure of the vessel is found to be 40.11 atm at 300 C. The degree of dissociation of NH3 will be:
(A) 0.6
(B) 0.4
(C) unpredictable
(D) none of these
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Answer:
Dear Student,
Dissociation of NH3 is given in the reaction below :
2NH3 ⇌ N2 + 3H2
2 0 0
2 - 2α α 3α
Total number of moles initially = 2
Total number of moles at equilibrium = 2 + 2α
As we know
PV = nRT
V = nRT / P
Since the volume is constant on decomposition of NH3 we can write,
n1RT1P1=n2RT2P2
We have given in the question
Pressure, P1 = 15 atm (before decomposition)
Temperature, T1= 270C or 300 K (before decomposition)
Pressure P2 = 40.11 atm (after decomposition)
Temperature, T2 = 300oC or 573 oK (after decomposition)
n1=2 and n2= 2+2α
2×30015=(2+2α) ×57340.1160015=2(1+α) 57340.111+α = 300 ×40.1115×5731+α = 1.4α=0.4
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