NH4Cl(s) ⇌ NH3(g) + HCl(g) When solid ammonium chloride is heated, it decomposes as represented abov — Equilibrium Chemistry Question
Problem Context
NH4Cl(s) ⇌ NH3(g) + HCl(g)
When solid ammonium chloride is heated, it decomposes as represented above. The value of Kp for the reaction is 0.0792 at 575 K. A 10.0 g sample of solid ammonium chloride is placed in a rigid, evacuated 3.0 L container that is sealed and heated to 575 K. The system comes to equilibrium with some solid NH4Cl remaining in the container.
Model Answer
Kp = P(NH3)·P(HCl). One point is earned for correct equation. Brackets '[]' earn zero points. Parentheses are acceptable.
Model Answer
0.0792 = P(NH3)·P(HCl) = x²; x = 0.281 atm. One point is earned for correct numerical result.
Model Answer
Increase. Upon addition of NH3(g) the reaction will proceed to the left to return to equilibrium. This will result in an increase in NH4Cl(s). One point is earned for correct choice with explanation. || The same. The equilibrium constant is unaffected by change in concentration or pressure; only changes in temperature affect K. One point is earned for correct choice with explanation.
Model Answer
Endothermic. A decrease in T causes a reaction originally at equilibrium to proceed in the exothermic direction. Since the decrease in T in this case causes the reaction to proceed toward reactants, the forward reaction must be endothermic. One point is earned for correct choice with explanation.
Model Answer
Ka = Kw/Kb = (1.00×10⁻¹⁴)/(1.8×10⁻⁵) = 5.6×10⁻¹⁰. One point is earned for correct numerical result.
Model Answer
Basic: Ka = [H3O+][NH3]/[NH4+]; in this case [NH3]=[NH4+], so Ka=[H3O+]. Thus pH = pKa. pKa = -logKa = -(-9.26) = 9.26. Or: Kb for the weak base is larger than Ka for the weak acid, so the solution must be basic. One point is earned for indicating that pH = pKa. One point is earned for correct choice based on correctly calculating or noting the value of pH. || NH3(aq) + H+(aq) → NH4+(aq); I: 0.0160, 0.0200, 0.0160; C: -0.0160, -0.0160, +0.0160; E: 0, 0.0040, 0.0320. [H3O+] = 0.0040/0.040 = 0.10 M; pH = 1.00. One point is earned for correct calculation of moles or concentration of reactants prior to reaction. One point is earned for correct numerical value of pH.