A \(0.100-\mathrm{mol}\) sample of \(\mathrm{H}_{2} \mathrm{~S}\) is placed in a \(10.0\) - \(\mathrm{L}\) reaction vessel and heated to \(1132^{\circ} \mathrm{C}\). At equilibrium, \(0.0285 \mathrm{~mol} \mathrm{H}_{2}\) is present. Calculate the value of \(K_{c}\) for the reaction \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) \rightleftharpoons\) \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{S}_{2}(\mathrm{~g})\) at \(1132^{\circ} \mathrm{C}\).

Understanding how to calculate the equilibrium constant, denoted as \( K_c \), is crucial for analyzing chemical reaction equilibria. The equilibrium constant is a numerical value that reflects the relative concentrations of products to reactants at equilibrium for a reversible reaction under a given set of conditions, like temperature.

For the equilibrium reaction \(2 \text{H}_2S(g) \rightleftharpoons 2 \text{H}_2(g) + \text{S}_2(g)\), the constant is expressed as \( K_c = \frac{[\text{H}_2]^2[\text{S}_2]}{[\text{H}_2S]^2} \), with square brackets denoting the molarity of each substance. When calculating \( K_c \), it's important to start with the balanced chemical equation and use the stoichiometric coefficients from the equation directly in the equilibrium constant expression. Also, the molar concentrations of gases are based on their amounts (in moles) divided by the volume of the container (in liters).

Errors in calculations, such as incorrect stoichiometry, not only lead to wrong results but can contradict physical realities, such as negative concentrations. In the given problem, ensuring the correct equilibrium concentrations and acknowledging the stoichiometry were vital steps. After the mathematical operations, which must be checked for accuracy, the \( K_c \) was determined to be approximately 0.2259, which reflects the state of the system at chemical equilibrium at \(1132^{\text{o}}C\).

The stoichiometry of a chemical reaction refers to the quantitative relationship between the reactants and products. It's based on the balanced chemical equation and dictates the proportions at which substances react and form. In the context of equilibrium, stoichiometry plays a crucial role in determining the changes in concentrations of reactants and products as the reaction proceeds to reach equilibrium.

In the example exercise, stoichiometry tells us that two molecules of hydrogen sulfide (\( \text{H}_2S \)) produce two molecules of hydrogen gas (\( \text{H}_2 \)) and one molecule of sulfur (\( \text{S}_2 \)). This means that for every mole of \( \text{H}_2 \), half a mole of \( \text{S}_2 \) is formed, which is essence to correctly calculate the equilibrium constant. A common mistake is to overlook the stoichiometric ratios during calculations, leading to incorrect conclusions. By carefully applying stoichiometric principles, the correct equilibrium concentrations are found and any possible contradiction, like the initially calculated negative concentration of \( \text{H}_2S \), is avoided.

Chemical reaction equilibrium occurs when a reversible chemical reaction proceeds in such a way that the rate of the forward reaction (formation of products) is equal to the rate of the reverse reaction (reformation of reactants). As a result, the concentrations of reactants and products remain constant over time, though not necessarily equal.

The concept of equilibrium does not mean that the reactants and products are present in the same amounts but that their ratios do not change. In the provided exercise, the presence of \(0.0285 \text{ moles of H}_2\) at equilibrium indicates that the system has reached a state where the decomposition of \( \text{H}_2S \) and the formation of \( \text{H}_2 \) and \( \text{S}_2 \) has balanced out. The calculation of the equilibrium constant \( K_c \) provides a quantitative measure of this balance and can predict the direction in which the reaction mixture will shift if conditions change, such as the addition of more reactants or the removal of products.