Suppose #"2.83 g"# of ammonium chloride solid is placed into a closed, evacuated rigid container, and allowed to decompose at a certain temperature. If 40% of it decomposed into ammonia and hydrogen chloride gas, what is #K_c# for this reaction?

1 Answer

Answer:

The equilibrium constant, #K_c=9.05 times 10^-5# is calculated from the equilibrium concentrations of gases. If more #NH_4Cl(s)# is added, nothing happens because the gases are already at their equilibrium concentrations.

Explanation:

The first step in any equilibrium problem like this is to write the complete balanced equation:

#NH_4Cl(s) harr NH_3(g) + HCl(g)#

The form of #K_c# for this reaction (the subscript c means concentration, and the standard concentration is in units of #"mol/L"#) is:

#K_c = [NH_3][HCl]#

The #NH_4Cl# does not appear in the equilibrium expression because it is a solid and has a thermodynamic activity of 1.

The starting amount of reactant is #(2.83g)/(59.49 g/(mol))=0.0476 mol#

Because 40% of the reactant decomposed, the equilibrium concentrations are:
#[NH_3]_e = [HCl]_e = (0.0476mol times 0.40)/(2L)=0.00951 M#

The value of #K_c# is therefore #K_c=(0.00951)^2=9.05 times 10^-5#