The equilibrium constant for the ammonia synthesis reaction at 500 K: N2 +3H2 <—

Question

The equilibrium constant for the ammonia synthesis reaction at 500 K:
N2 +3H2 <—> 2NH3 Kp = 0.10 atm^-2
If the ratio of N2 and H2 partial pressures is maintained at 1:3, what total pressure in the reactor is necessary to achieve a 1 atm partial pressure of ammonia at 500 K? The equilibrium constant for the ammonia synthesis reaction at 500 K:
N2 +3H2 <—> 2NH3 Kp = 0.10 atm^-2
If the ratio of N2 and H2 partial pressures is maintained at 1:3, what total pressure in the reactor is necessary to achieve a 1 atm partial pressure of ammonia at 500 K?
N2 +3H2 <—> 2NH3 Kp = 0.10 atm^-2
If the ratio of N2 and H2 partial pressures is maintained at 1:3, what total pressure in the reactor is necessary to achieve a 1 atm partial pressure of ammonia at 500 K?

Explanation / Answer

Solution :-

N2 + 3H2 ----> 2 NH3

Kp=0.10 atm^-2

Partial pressure of NH3 = 1.0 atm

Lets calculate the equilibrium partial pressures of the H2 and N2

Kp = [NH3]^2 /[N2][H2]^3

0.10 atm^-2= [1 atm]^2 / [x][3x]^3

0.10 = 1/[27x^4]

x^4= 1/[27*0.1]

x^4 = 0.370

x= (0.370)^(1/4)

x= 0.78 atm

Therefore since partial pressure of N2 = x = 0.78 atm

and H2 = 3x = 3*0.78 atm= 2.34 atm

Total pressure = N2+H2+NH3

                           = 0.78 atm + 2.34 atm + 1.0 atm

                           = 4.12 atm

Therefore total pressure neccessary in the reactor is 4.12 atm

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