Question

use the references to access important values if needed for this question.
the equilibrium constant, ( k_{c} ), for the following reaction is ( 5.10 \times 10^{-6} ) at ( 548 mathrm{~k} ).
mathrm{nh}_{4} mathrm{cl}(s)
ightleftharpoons mathrm{nh}_{3}(g)+mathrm{hcl}(g)
if an equilibrium mixture of the three compounds in a ( 6.03 mathrm{~l} ) container at ( 548 mathrm{~k} ) contains ( 2.82 mathrm{~mol} ) of ( mathrm{nh}_{4} mathrm{cl}(s) ) and ( 0.448 mathrm{~mol} ) of ( mathrm{nh}_{3} ), the number of moles of ( mathrm{hcl} ) present is ( square ) mol.

Explanation:

Step1: Write the expression for \(K_c\)

For the reaction \(NH_4Cl(s)
ightleftharpoons NH_3(g)+HCl(g)\), the equilibrium constant expression \(K_c = [NH_3][HCl]\) (since the concentration of a solid (\(NH_4Cl\)) is omitted from the \(K_c\) expression).

Step2: Calculate the concentrations of \(NH_3\)

The concentration of \(NH_3\), \([NH_3]=\frac{n_{NH_3}}{V}\), where \(n_{NH_3} = 0.448\space mol\) and \(V=6.03\space L\). So \([NH_3]=\frac{0.448}{6.03}\space mol/L\approx0.0743\space mol/L\)

Step3: Solve for \([HCl]\) using the \(K_c\) expression

We know \(K_c = 5.10\times 10^{-6}\) and \(K_c=[NH_3][HCl]\). Substituting the value of \([NH_3]\) into the \(K_c\) expression: \([HCl]=\frac{K_c}{[NH_3]}\)

$$ [HCl]=\frac{5.10\times 10^{-6}}{0.0743}\space mol/L\approx6.86\times 10^{-5}\space mol/L $$

Step4: Calculate the moles of \(HCl\)

Since \([HCl]=\frac{n_{HCl}}{V}\), then \(n_{HCl}=[HCl]\times V\). Substituting \([HCl]=6.86\times 10^{-5}\space mol/L\) and \(V = 6.03\space L\)

$$ n_{HCl}=6.86\times 10^{-5}\times6.03\space mol\approx4.14\times 10^{-4}\space mol $$

Answer:

\(4.14\times 10^{-4}\)