Question

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?

Answer 1

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`