For the equilibrium: LiCl·3NH₃(s) ⇌ LiCl·NH₃(s) + 2NH₃(g); K_p = 9 atm² at 27°C. A 8.21 L vessel contains 0.1 mole of LiCl.NH₃(s). How many moles of NH₃(g) should be added to the flask at this tempera

Question

For the equilibrium: LiCl·3NH₃(s) ⇌ LiCl·NH₃(s) + 2NH₃(g); K_p = 9 atm² at 27°C. A 8.21 L vessel contains 0.1 mole of LiCl.NH₃(s). How many moles of NH₃(g) should be added to the flask at this temperature to derive the backward reaction for completion?

Prashant Kotian · Accepted Answer

Correct answer: C. K_p = (P_NH₃)² = 9 => P_NH₃ = 3 atm. The equilibrium requires 3 atm of NH₃ gas. Moles of NH₃ gas = PV/RT = (3 × 8.21) / (0.0821 × 300) = 1.0 mol. To drive the reaction completely backward, 0.1 mol of LiCl·NH₃ must react, consuming 0.2 mol of NH₃. Total NH₃ needed = 1.0 + 0.2 = 1.2 mol. Therefore, correct answer is C.

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