Hydrogen sulfide decomposes according to the following reaction, for which \(K_{c}=9.30 \times 10^{-8}\) at \(700^{\circ} \mathrm{C}:\) $$ 2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) $$ If \(0.45 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{~S}\) is placed in a 3.0 - \(\mathrm{L}\) container, what is the equilibrium concentration of \(\mathrm{H}_{2}(g)\) at \(700^{\circ} \mathrm{C} ?\)

In chemical equilibrium, the equilibrium constant (\text{K}_c) is a crucial value that helps in understanding the ratio of product concentrations to reactant concentrations at equilibrium. It is specific to a given reaction at a specific temperature. For the decomposition of hydrogen sulfide (\text{H}_2\text{S}):

$$2 \text{H}_2 \text{S} (g) \rightleftharpoons 2 \text{H}_2(g) + \text{S}_2(g),$$

The expression for the equilibrium constant (\text{K}_c) is given by:

$$K_c = \frac{[\text{H}_2]^2[\text{S}_2]}{[\text{H}_2\text{S}]^2}. $$

Here, \text{K}_c = 9.30 \times 10^{-8} at 700°C. This small value of \text{K}_c signifies that at equilibrium, the concentration of the reactants (\text{H}_2\text{S}) is much higher than the concentration of the products (\text{H}_2 and \text{S}_2).

Calculating concentrations involves using the initial conditions and changes that occur as the system reaches equilibrium. To find the equilibrium concentration of hydrogen gas (\text{H}_2) at 700°C, you can start by determining the initial concentration of \text{H}_2\text{S}. Given:

• 0.45 mol of \text{H}_2\text{S}
• Volume = 3.0 L

The initial concentration is calculated as: $$[\text{H}_2\text{S}]_0 = \frac{0.45 \text{ mol}}{3.0 \text{ L}} = 0.15 \text{ M}.$$

At equilibrium, let

Le Chatelier's Principle is fundamental in predicting how an equilibrium will shift when subjected to a change in conditions. If the system at equilibrium is disturbed by a change in concentration, pressure, volume, or temperature, the equilibrium will shift in a direction that counteracts the disturbance, re-establishing equilibrium. For our reaction:

$$2 \text{H}_2 \text{S} (g) \rightleftharpoons 2 \text{H}_2(g) + \text{S}_2(g),$$

Increasing the concentration of \text{H}_2\text{S} will shift the equilibrium to the right, producing more \text{H}_2 and \text{S}_2. Conversely, increasing the concentration of \text{H}_2 or \text{S}_2 will shift the equilibrium to the left, forming more \text{H}_2\text{S}.

The Reaction Quotient (Q) provides a snapshot of the concentrations of reactants and products at any point in time, not just at equilibrium. For the reaction:

$$2 \text{H}_2 \text{S}(g) \rightleftharpoons 2 \text{H}_2(g) + \text{S}_2(g),$$

• The expression for Q is similar to \text{K}_c:
• Q = \frac{[\text{H}_2]^2[\text{S}_2]}{[\text{H}_2\text{S}]^2}.

Comparing Q to \text{K}_c helps predict the direction in which the reaction will proceed:

• If Q < \text{K}_c, the reaction moves forward, producing more products.
• If Q > \text{K}_c, the reaction goes in reverse, producing more reactants.

For your calculations, you can use the initial concentrations to find Q and understand whether the system is in equilibrium or needs adjustment.