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

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