the equilibrium NH4HS =NH3 +h2s is followed to set up at 127 degree Celsius in a closed vessel the total pressure at equilibrium was 20 atm the kc for the reaction is
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Given,
NH₄HS (s) ⇔ NH₃ (g) + H₂S(g)
∴ Kp = P(NH₃).P(H₂S)
Here you can see that , Ratio of formation of NH₃ and H₂S is 1 : 1
so, at equilibrium , ratio of pressure of NH₃ and H₂S is 1 : 1
∴ P(NH₃) = P(H₂S)
But P(NH₃) + P(H₂S) = 20 atm
∴ P(NH₃) = 10atm and P(H₂S) = 10 atm
Now, Kp = (10)² = 100 atm
∵ Kp = Kc(RT)^{∆n}
Here, Kp = 100 atm , R = 0.082 L.atm/mol/K and T = 127°C = 400K
∆n = 2 - 0 = 2 [ 1mole of NH₃ and 1 mole of H₂S gases ]
∴ 100 = Kc(0.082 × 400)^2
Kc = 100/(32.8)² ≈0.09
NH₄HS (s) ⇔ NH₃ (g) + H₂S(g)
∴ Kp = P(NH₃).P(H₂S)
Here you can see that , Ratio of formation of NH₃ and H₂S is 1 : 1
so, at equilibrium , ratio of pressure of NH₃ and H₂S is 1 : 1
∴ P(NH₃) = P(H₂S)
But P(NH₃) + P(H₂S) = 20 atm
∴ P(NH₃) = 10atm and P(H₂S) = 10 atm
Now, Kp = (10)² = 100 atm
∵ Kp = Kc(RT)^{∆n}
Here, Kp = 100 atm , R = 0.082 L.atm/mol/K and T = 127°C = 400K
∆n = 2 - 0 = 2 [ 1mole of NH₃ and 1 mole of H₂S gases ]
∴ 100 = Kc(0.082 × 400)^2
Kc = 100/(32.8)² ≈0.09
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