Chemical EquilibriumHard
Question
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 300oC. The degree of dissociation of NH3 will be:
Options
A.0.6
B.0.4
C.Unpredictable
D.None of these
Solution
P1 = 15 atm ; T1 = 300 K.
Equilibrium temperature is 300oC that is 573 K.
So first of all we have to calculate pressure of NH3 at 573 K.
P2 = 28.65 atm at 300oC.
NH3 (g) ⇋
N2(g) + 
H2(g).
t = 0 28.65 atm 0 0
t = teq. [28.65-x]
atm 
x
But according to question.
Ptotal = 28.65 - x +
x
= 28.65 = x = 40.11.
x = 11.46.
Degree of dissociation of NH3 =
= 0.4.
Equilibrium temperature is 300oC that is 573 K.
So first of all we have to calculate pressure of NH3 at 573 K.
P2 = 28.65 atm at 300oC.
NH3 (g) ⇋
N2(g) + H2(g).t = 0 28.65 atm 0 0
t = teq. [28.65-x]
atm xBut according to question.
Ptotal = 28.65 - x +
x= 28.65 = x = 40.11.
x = 11.46.
Degree of dissociation of NH3 =
= 0.4.Create a free account to view solution
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