Problem 49 - Chemistry and Chemical Reactivity

Question

\(K_{c}\) for the decomposition of ammonium hydrogen sulfide is \(1.8 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\) $$ \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g}) $$ (a) When the pure salt decomposes in a flask, what are the equilibrium concentrations of \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S} ?\) (b) If \(\mathrm{NH}_{4} \mathrm{HS}\) is placed in a flask already containing \(0.020 \mathrm{mol} / \mathrm{L}\) of \(\mathrm{NH}_{3}\) and then the system is allowed to come to equilibrium, what are the equilibrium concentrations of \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S} ?\)

Step-by-Step Solution

Answer

(a) [NH₃] = [H₂S] = 0.0134 mol/L. (b) [NH₃] = 0.02778 mol/L, [H₂S] = 0.00778 mol/L.

Step 1: Write the Equilibrium Expression

The reaction given is \(\mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\). The equilibrium constant \(K_c\) can be written in terms of the concentrations of the gaseous products: \[ K_{c} = [\mathrm{NH}_3][\mathrm{H}_2 \mathrm{S}] \] since \(\mathrm{NH}_{4} \mathrm{HS}\) is a solid and is not included in the \(K_c\) expression.

Step 2: Solve for Equilibrium Concentrations when Initial Concentration is Zero

Let \(x\) be the equilibrium concentration of both \(\mathrm{NH}_{3}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) since they are produced in a 1:1 ratio. Substituting into the \(K_c\) expression yields: \[ K_c = x \cdot x = x^2 \]Given \(K_c = 1.8 \times 10^{-4}\), solve for \(x\): \[ x^2 = 1.8 \times 10^{-4} \]\[ x = \sqrt{1.8 \times 10^{-4}} = 0.0134 \]Thus, the equilibrium concentrations of \(\mathrm{NH}_3\) and \(\mathrm{H}_2 \mathrm{S}\) are both \(0.0134 \ \text{mol/L}\).

Step 3: Solve for Equilibrium with Initial NH3 Concentration

Now consider the scenario with \(0.020 \ \text{mol/L}\) initial \(\mathrm{NH}_3\):- Initial concentrations: \([\mathrm{NH}_3] = 0.020\ \text{mol/L}\), \([\mathrm{H}_2 \mathrm{S}] = 0\ \text{mol/L}\).- Change: Let \(\Delta x\) be the change; \([\mathrm{NH}_3] = 0.020 + \Delta x\), \([\mathrm{H}_2 \mathrm{S}] = \Delta x\).- Substituting into the \(K_c\) expression gives:\[ 1.8 \times 10^{-4} = (0.020 + \Delta x)(\Delta x) \]Let's simplify the equation:\[ 1.8 \times 10^{-4} = 0.020\Delta x + (\Delta x)^2 \]This is a quadratic equation in the form \((\Delta x)^2 + 0.020\Delta x - 1.8 \times 10^{-4} = 0\). Solving gives \(\Delta x \approx 0.00778 \ \text{mol/L}\).Thus, the equilibrium concentrations are \([\mathrm{NH}_3] = 0.020 + 0.00778 = 0.02778 \ \text{mol/L}\), and \([\mathrm{H}_2 \mathrm{S}] = 0.00778 \ \text{mol/L}\).

Key Concepts

Equilibrium Concentration

In the study of chemical equilibrium, equilibrium concentration refers to the molarity of reactants and products present in a chemical reaction when the rate of the forward reaction equals the rate of the reverse reaction. This state is characterized by no net change in the concentrations of the chemicals involved. To find the equilibrium concentration, we start by setting up an expression based on the balanced chemical equation.

For the decomposition of ammonium hydrogen sulfide (NH₄HS), the equilibrium concentrations are determined by defining the initial conditions and using the equilibrium constant expression. When NH₄HS decomposes, it forms ammonia (NH₃) and hydrogen sulfide (H₂S) in a 1:1 ratio. This means that for every mole of NH₄HS that decomposes, one mole each of NH₃ and H₂S is produced. By applying the initial conditions and stoichiometrically balancing the equation, you can find these equilibrium concentrations. For instance, given an initial condition of zero for both gaseous products, you denote their equilibrium concentration with 'x'. This allows you to substitute into the equilibrium expression to solve for 'x', which represents the molarity of NH₃ and H₂S at equilibrium.

Equilibrium Constant (Kc)

The equilibrium constant, denoted as \(K_c\), is a vital component in understanding chemical equilibrium. It quantitatively expresses the ratio of the concentration of the products to the reactants at equilibrium. For a reaction at a given temperature, \(K_c\) remains constant, underscoring its significance in predicting the behavior of the system.

Consider the decomposition reaction: \[ \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g}) \] For this equation, \(K_c\) is formulated using the concentrations of the gaseous products only, as NH₄HS is a solid and does not appear in the equation.
- At equilibrium, \(K_c\) can be expressed as \([\mathrm{NH}_3][\mathrm{H}_2 \mathrm{S}]\).
Understanding \(K_c\) helps in determining the extent of the reaction; a small \(K_c\) like 1.8 × 10⁻⁴ indicates that the products are not favored, and only a small amount of ammonia and hydrogen sulfide are present at equilibrium.

Ammonium Hydrogen Sulfide Decomposition

The decomposition of ammonium hydrogen sulfide \((\mathrm{NH}_4\mathrm{HS})\) is a reversible process where the solid decomposes into ammonia \((\mathrm{NH}_3)\) and hydrogen sulfide \((\mathrm{H}_2\mathrm{S})\). This reaction is a classical example of a heterogeneous equilibrium since it involves both solid and gaseous phases.

At equilibrium, both the forward decomposition reaction and the reverse reformation of NH₄HS occur simultaneously at equal rates, ensuring no net change in concentrations. This balance forms the basis of studying chemical equilibrium, specifically how changes in conditions (like initial concentrations) affect equilibrium states.
- When NH₄HS is placed in a closed system with pre-existing NH₃, the equilibrium must readjust to account for the initial concentration of NH₃. This adjustment can be calculated using modifications to the equilibrium constant expression, incorporating this pre-existing concentration.
Through understanding the factors affecting this decomposition, one can predict how changes in system conditions, such as pressure and temperature, might influence the equilibrium position, showcasing the practical application of chemical equilibrium principles in real-world scenarios.