Question

In a spontaneous gas-phase acid-base reaction represented as:

NH₃​(g)+HCl(g)⇌NH₄​Cl(s)

During an experiment, 10.00 g of solid NH₄Cl was introduced into a 500.0 mL flask at 25.0°C. Given that the equilibrium constant (Keq​) for this reaction at this temperature is 8.9×10⁻¹, determine the equilibrium partial pressures (in atm) of NH₃ and HCl.

Please assist with the calculations.

Answer 1

To determine the equilibrium partial pressures of NH₃ and HCl in the given acid-base reaction, we can use the given equilibrium constant (Keq) and the initial concentration of NH₄Cl. From the balanced equation, we can see that 1 mole of NH₄Cl is formed from 1 mole of NH₃ and 1 mole of HCl. Therefore, at equilibrium, the partial pressures of NH₃ and HCl are both equal to the moles of NH₄Cl.

Step-by-step explanation:

To determine the equilibrium partial pressures of NH₃ and HCl, we can use the given equilibrium constant (Keq) and the initial concentration of NH₄Cl.

Given that 10.00 g of solid NH₄Cl was introduced into a 500.0 mL flask at 25.0°C, we can convert this mass to moles using the molar mass of NH₄Cl. The molar mass of NH₄Cl is 53.49 g/mol. Therefore, we have:

moles of NH₄Cl = (10.00 g) / (53.49 g/mol) = 0.1870 mol

Since the reaction is in a gas-phase, the equilibrium partial pressures of NH₃ and HCl can be determined using the stoichiometry of the reaction. The balanced equation for the reaction is:

NH₃(g) + HCl(g) ⇌ NH₄Cl(s)

From the balanced equation, we can see that 1 mole of NH₄Cl is formed from 1 mole of NH₃ and 1 mole of HCl. Therefore, at equilibrium, the partial pressures of NH₃ and HCl are both equal to the moles of NH₄Cl.

So, the equilibrium partial pressures of NH₃ and HCl are both 0.1870 atm.