To answer this question, let's analyze the chemical equilibrium reaction:
[tex]\[
\left( NH_4 \right)_2 CO_3(s) \rightleftharpoons 2 NH_3(g) + CO_2(g) + H_2O(g)
\][/tex]
When considering the effect of adding water [tex]\(\left(H_2O\right)\)[/tex] to this reaction, we have to refer to Le Chatelier's Principle. This principle states that if a system at equilibrium is disturbed by an external change, the equilibrium will shift in a direction that counteracts the disturbance in order to re-establish equilibrium.
In this reaction, adding water [tex]\(\left(H_2O\right)\)[/tex] would increase the concentration of one of the products (water vapor). To counteract this change, the system will shift its equilibrium position to favor the reverse reaction, which will consume the added water [tex]\(\left(H_2O\right)\)[/tex] along with ammonia [tex]\(\left(NH_3\right)\)[/tex] and carbon dioxide [tex]\(\left(CO_2\right)\)[/tex] to form more ammonium carbonate [tex]\(\left((NH_4)_2CO_3\right)\)[/tex].
Therefore, the concentration of the products [tex]\(NH_3(g)\)[/tex], [tex]\(CO_2(g)\)[/tex], and [tex]\(H_2O(g)\)[/tex] will decrease, and the concentration of the reactants [tex]\(\left((NH_4)_2CO_3\right)\)[/tex] will increase.
Hence, the correct answer is:
C. The concentration of products would decrease, and the concentration of reactants would increase.
