Question

the equilibrium constant, ( k_{c} ), for the following reaction is ( 5.10\times10^{-6} ) at ( 548k ).
( nh_{4}cl(s)
ightleftharpoons nh_{3}(g)+hcl(g) )
this reaction is
enter product or reactant.

the concentrations of ( nh_{3} ) and ( hcl ) will be
enter high or low.

Explanation:

Step1: Analyze the equilibrium constant value

The equilibrium constant \(K_{c}=\frac{[NH_{3}][HCl]}{[NH_{4}Cl]}\). Since \(NH_{4}Cl\) is a solid, its concentration is considered as 1. So \(K_{c}=[NH_{3}][HCl]\). Given \(K_{c} = 5.10\times10^{-6}\) which is a very small value.

Step2: Determine the favored side

When \(K_{c}\ll1\), the concentration of products is much less than the concentration of reactants (in terms of non - solid species here). So the reactant side is favored.

Step3: Determine the product concentrations

Because \(K_{c}=[NH_{3}][HCl]=5.10\times 10^{-6}\), and \(K_{c}\) is small, the concentrations of \(NH_{3}\) and \(HCl\) (products) will be low.

Answer:

REACTANT
LOW