Chemistry - Chemical Equilibrium Question with Solution | TestHub
For the reaction: NH4HS(s) ⇌ NH3(g) + H2S(g); Kp = 0.09 atm2 at 300 K. The reaction is performed in a rigid vessel of 8.2 L capacity maintained at 300 K, starting with the following amounts of NH4HS(s) (given in Column I). Match the correct information given in Column II for the amount given in Column I.
Column I | Column II |
|---|---|
P) 10.2 g | i) 50% dissociation of NH4HS |
Q) 5.1 g | ii) 100% dissociation of NH4HS |
R) 2.55 g | iii) Addition of small amount of NH3(g) may result shift of equilibrium in backward direction |
| iv) Addition of 5.1 g of NH4HS(s) may result increase in moles of NH3(g) |
Options:
Answer:
Solution:
At equilibrium pNH3 = pH2S and Kp = pNH3 × pH2S = 0.09 atm2, so each partial pressure is 0.3 atm n = pV/(RT) = (0.3 × 8.2)/(0.0821 × 300) = 0.10 mol of each gas, needing 0.10 mol = 5.1 g of NH4HS to dissociate
P: 10.2 g is twice 5.1 g, so only half dissociates, i.e. 50% dissociation (i); solid remains, so adding a little NH3 shifts the equilibrium backward (iii)
Q: exactly 5.1 g dissociates fully, 100% dissociation (ii); a small addition of NH3 raises the reaction quotient above Kp, so the reaction shifts backward (iii); the equilibrium pressure is already reached, so extra solid cannot raise the moles of gas
R: 2.55 g dissociates fully (ii); with the solid exhausted and the pressure below Kp, adding 5.1 g more NH4HS lets more dissociation occur, increasing moles of NH3 (iv)
The correct mapping is option (c).
