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What effect would adding water [tex][tex]$\left( H_2O \right)$[/tex][/tex] have on this reaction over time?

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

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 correct answer for the effect of adding water ([tex]\(H_2O\)[/tex]) on the 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 understand the principle of chemical equilibrium, specifically Le Chatelier's Principle. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

Here, water ([tex]\(H_2O\)[/tex]) is one of the products in the reaction. If we add more [tex]\(H_2O\)[/tex], the system will adjust to counter the increase in [tex]\(H_2O\)[/tex] by shifting the equilibrium position to the left. This means the reaction will produce more reactants and fewer products.

Therefore:

- The concentration of products ([tex]\(NH_3(g)\)[/tex], [tex]\(CO_2(g)\)[/tex], and [tex]\(H_2O(g)\)[/tex]) would decrease.
- The concentration of reactants ([tex]\((NH_4)_2CO_3(s)\)[/tex]) would increase.

Given this information, the correct answer is:

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