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What effect would adding water (H₂O) have on this reaction over time?

(NH₄)₂CO₃(s) ⇌ 2 NH₃(g) + CO₂(g) + H₂O(g)

A. The concentration of products would increase, and the concentration of reactants would decrease.
B. The concentrations of both products and reactants would increase.
C. The concentration of products would decrease, and the concentration of reactants would increase.
D. The concentrations of both products and reactants would decrease.

Sagot :

GinnyAnswer
To determine the effect of adding water ([tex]\(H_2O\)[/tex]) to the given equilibrium reaction:

[tex]\[ \left( NH_4\right)_2 CO_3(s) \rightleftharpoons 2 NH_3(g) + CO_2(g) + H_2O(g), \][/tex]

we need to consider Le Chatelier's principle. Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

In this reaction, adding water ([tex]\(H_2O\)[/tex]) increases the concentration of one of the products (water vapor, [tex]\(H_2O (g)\)[/tex]).

By Le Chatelier's principle, the system will respond by favoring the reactants to reduce the disturbance. Therefore, the reaction will shift to the left to counter the added water. This shift to the left will result in a decrease in the concentrations of the gaseous products ([tex]\(NH_3(g)\)[/tex], [tex]\(CO_2(g)\)[/tex], and [tex]\(H_2O(g)\)[/tex]) while increasing the amount of solid reactant, [tex]\(\left( NH_4\right)_2 CO_3(s)\)[/tex].

Consequently, the concentration of products (which are [tex]\(NH_3(g)\)[/tex], [tex]\(CO_2(g)\)[/tex], and [tex]\(H_2O(g)\)[/tex]) would decrease, and the concentration of reactants would increase.

Thus, the correct answer is:

C. The concentration of products would decrease, and the concentration of reactants would increase.
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