Question

(1). The equilibrium constant, K, for the following reaction is 1.32×10^-3 at 565 K.

• NH₄Cl(s) =NH₃(g) + HCl(g)

An equilibrium mixture in a 10.4 L container at 565 K contains 0.285 mol  NH₄Cl(s), 4.47×10^-2 M NH₃ and 2.95×10^-2 M HCl. What will be the concentrations of the two gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 4.25 L?

[NH3]	=	M
[HCl]	=	M

Answer 1

To determine the concentrations of NH3 and HCl once equilibrium is reestablished after compressing the mixture, we can use the principle of Le Chatelier's principle.

First, let's calculate the initial moles of NH3 and HCl present in the equilibrium mixture:

Initial moles of NH3: moles of NH3 = volume of the container * concentration of NH3 moles of NH3 = 10.4 L * 4.47×10^(-2) M moles of NH3 = 0.46288 moles

Initial moles of HCl: moles of HCl = volume of the container * concentration of HCl moles of HCl = 10.4 L * 2.95×10^(-2) M moles of HCl = 0.3068 moles

Now, let's calculate the initial moles of NH4Cl: moles of NH4Cl = 0.285 moles

Since the stoichiometric coefficients for NH3 and HCl are 1 in the balanced equation, the equilibrium moles of NH3 and HCl will be the same as the moles of NH4Cl (as NH4Cl dissociates into NH3 and HCl in a 1:1 ratio).

Now, let's calculate the new equilibrium concentrations after compressing the mixture to a volume of 4.25 L:

New concentration of NH3: concentration of NH3 = moles of NH3 / volume of the container concentration of NH3 = 0.285 moles / 4.25 L concentration of NH3 = 0.0671 M

New concentration of HCl: concentration of HCl = moles of HCl / volume of the container concentration of HCl = 0.285 moles / 4.25 L concentration of HCl = 0.0671 M

Therefore, the concentrations of NH3 and HCl once equilibrium has been reestablished after compressing the mixture to a volume of 4.25 L are: [NH3] = 0.0671 M [HCl] = 0.0671 M