What is the partial pressure of HCl?

Question

Calculate the partial pressure (in atm) of HCl at equilibrium when 22.9 g of NH₄Cl decomposes at 650 K according to the following chemical equation:

NH₄Cl(s) ⇌ NH₃(g) + HCl(g)

K_p = 7.95

Answer 1

First, calculate the number of moles of NH₄Cl that decompose:

Molar mass of NH₄Cl = 14.01 g/mol + 4 * 1.01 g/mol + 35.45 g/mol = 53.49 g/mol

Number of moles of NH₄Cl = 22.9 g / 53.49 g/mol ≈ 0.4286 mol

Since NH₄Cl decomposes to form NH₃ and HCl in a 1:1 ratio, the number of moles of NH₃ and HCl formed will also be approximately 0.4286 mol.

Now, consider the equilibrium expression for the given reaction:

Kp = (P_NH₃ * P_HCl) / P_NH₄Cl

Since P_NH₃ and P_HCl are both equal to 0.4286 mol and P_NH₄Cl is initially 0.4286 mol, we can substitute these values into the equation to find P_HCl:

7.95 = (0.4286 * 0.4286) / 0.4286

7.95 = 0.4286

Therefore, the partial pressure of HCl at equilibrium is 0.4286 atm.