See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

   4 3 2 NH HS s NH g +H S g   Since 4 NH HS is a solid, hence 4 NH HS a 1  and its un-dissociated amount will not effect the total pressure (due to gaseous NH3 and H2S only). Let x be its moles decomposed at equilibrium and P be the equilibrium pressure. Moles NH4HS NH3 H2S Initial a 0 0 At equilibrium a x  x x Total moles at equilibrium = Moles of   3 2 NH +H S =2x (only gaseous moles) P=? K