Question

Ammonium hydrogen sulphide dissociated according to the equation NH4HS (s) = NH3(g ) + H2S (g) (Here by = I mean reversible process) If observed pressure of the mixture is 1.12 atm at 106°C , what is the equilibrium constant Kp of the reaction? Answer is 0.3136 atm^2

Answer 1

We know, that For Solid the Value of the Concentration or Pressure is taken as 1, while calculation the Kc or Kp.

Now, the Observed Pressure is given, which will be equal to the sum of the Pressure of the Gases. [Acc. to Dalton law of Partial Pressure.]

∴ Pressure of Ammonia + Pressure of Hydrogen sulfide = 1.12

∴ Pressure of Ammonia = Pressure of Hydrogen sulfide = 1.12/2 = 0.56

Now, For Kp,

Kp = (NH₃)(H₂S)

⇒ Kp = 0.56 × 0.56

∴ Kp = 0.3136

For units,

Units of Kp = (NH₃)(H₂S)

⇒ Units of Kp = atm × atm = atm²

∴ Kp = 0.3136 atm²

Hope it helps.