At \(1130^{\circ} \mathrm{C}\) the equilibrium constant \(\left(K_{\mathrm{c}}\right)\) for the reaction $$ 2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) $$ is \(2.25 \times 10^{-4}\). If \(\left[\mathrm{H}_{2} \mathrm{~S}\right]=4.84 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=\) \(1.50 \times 10^{-3} M,\) calculate \(\left[\mathrm{S}_{2}\right].\)

The Law of Mass Action is a principle that governs the relationship between the concentrations of reactants and products at a state of chemical equilibrium. It explains how the rate of a chemical reaction is proportional to the product of the concentrations of the reactants, each raised to the power of their respective coefficients as expressed in the balanced chemical equation.

For example, consider a simple reaction where reactants A and B combine to form product AB:
\[\text{A} + \text{B} \rightleftharpoons \text{AB}\]
The corresponding Law of Mass Action would state that the rate at which AB is formed is proportional to the concentrations of A and B:
\[\text{Rate} \propto [\text{A}][\text{B}]\]
Applying this concept to the provided chemical equation,\[2 \text{H}_2 \text{S}(g) \rightleftharpoons 2 \text{H}_2(g) + \text{S}_2(g)\],
the Law of Mass Action helps us derive the equilibrium constant expression as: \[K_c = \frac{[\text{H}_2]^2[\text{S}_2]}{[\text{H}_2 \text{S}]^2}\].

This is foundational for understanding equilibrium behavior in chemical systems and crucial for calculating equilibrium concentrations, as seen in the exercise.

Chemical equilibrium occurs when a chemical reaction and its reverse reaction occur at the same rate, leading to no net change in the concentration of reactants and products over time. It is a dynamic state, meaning that the reactions are still proceeding, but the concentrations remain constant.

In the context of the textbook problem, we're examining a system at a given temperature where the decomposition of hydrogen sulfide into hydrogen gas and sulfur occurs at the same rate as their recombination.

At equilibrium, the ratio of these substances’ concentrations is constant and is represented by the equilibrium constant, \( K_c \). The value of \( K_c \) is specific to a particular reaction at a given temperature. If we know the equilibrium constant and the concentration of some of the reactants or products, we can calculate the unknown concentrations, which is precisely what the exercise demonstrates.

Stoichiometry is the quantitative relationship between the amounts of reactants and products in a chemical reaction, based on the balanced chemical equation. It allows us to predict the amount of products that will form from a given amount of reactants, or conversely, the amount of reactants needed to form a specific amount of product.

In the example given, stoichiometry tells us that two moles of \(\text{H}_2 \text{S}(g)\) decompose to produce two moles of \(\text{H}_2(g)\) and one mole of \(\text{S}_2(g)\), as indicated by the coefficients in the balanced equation. This stoichiometric ratio is crucial for applying the Law of Mass Action correctly, as it determines the exponents in the expression for the equilibrium constant \(K_c\).

Understanding stoichiometry is essential for solving problems related to chemical equilibrium, as it ties together the mole concept, the balanced chemical equation, and the Law of Mass Action.