Question

\(K_{c}\) for the decomposition of ammonium hydrogen sulfide is \(1.8 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\) $$ \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftarrows \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g}) $$ (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 \(\mathrm{NH}_{1} \mathrm{HS}\) 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 \(\mathrm{H}_{2} \mathrm{S} ?\)

Short answer

(a) [NH₃] = [H₂S] = 0.0134 M; (b) [NH₃] = 0.0278 M, [H₂S] = 0.0078 M.

Step-by-step solution

Establish the Equilibrium Expression

For the decomposition reaction \(\mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g}) + \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\), the equilibrium constant \(K_c\) is given by \[ K_c = [\mathrm{NH}_{3}][\mathrm{H}_{2} \mathrm{S}] \] since the reactant is a solid and does not appear in the expression.

Calculate Equilibrium Concentrations for Part (a)

At equilibrium, let \(x\) be the concentration of \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S}\), as they are produced in a 1:1 ratio. Thus, \([\mathrm{NH}_{3}] = x\) and \([\mathrm{H}_{2} \mathrm{S}] = x\).Substituting these into the expression for \(K_c\), we have:\[ x^2 = 1.8 \times 10^{-4} \]Solving for \(x\):\[ x = \sqrt{1.8 \times 10^{-4}} \approx 0.0134 \]Therefore, the equilibrium concentrations of \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S}\) are \(0.0134 \text{ M}\) each.

Determine Initial Conditions for Part (b)

If \(\mathrm{NH}_{4} \mathrm{HS}\) is added to a flask containing \(0.020 \text{ mol/L}\) of \(\mathrm{NH}_{3}\), then the initial concentrations are:- \([\mathrm{NH}_{3}] = 0.020 \text{ M}\)- \([\mathrm{H}_{2} \mathrm{S}] = 0\)At equilibrium, let the change in concentration be \(+x\) for both \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S}\) produced, making \([\mathrm{NH}_{3}] = 0.020 + x\) and \([\mathrm{H}_{2} \mathrm{S}] = x\).

Solve for Equilibrium Concentrations for Part (b)

Substitute the expressions for equilibrium concentrations into the \(K_c\) equation:\[ K_c = (0.020 + x)x = 1.8 \times 10^{-4} \]Expanding this:\[ 0.020x + x^2 = 1.8 \times 10^{-4} \]Solve this quadratic equation using methods such as factoring or the quadratic formula:\[ x^2 + 0.020x - 1.8 \times 10^{-4} = 0 \]Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 0.020\), \(c = -1.8 \times 10^{-4}\):\[ x = \frac{-0.020 \pm \sqrt{(0.020)^2 - 4 \times 1 \times (-1.8 \times 10^{-4})}}{2 \times 1} \approx 0.0078 \]Thus, \([\mathrm{NH}_{3}] = 0.020 + 0.0078 = 0.0278 \text{ M}\) and \([\mathrm{H}_{2} \mathrm{S}] = 0.0078 \text{ M}\).

Key concepts

Equilibrium Constant

In a chemical reaction, the equilibrium constant, often denoted as \(K_c\), is a key figure that tells us about the ratio of the concentrations of the products to the reactants when the system is at equilibrium. For gaseous reactions, we use the concentrations of gases in the equilibrium expression.
When dealing with solid reactants, such as ammonium hydrogen sulfide (\(\text{NH}_4\text{HS}(\text{s})\)), the concentration of a pure solid is considered constant and is not included in the equilibrium expression.
The expression for the equilibrium constant \(K_c\) in the decomposition of \(\text{NH}_4\text{HS}\) is as follows:
\[ K_c = [\text{NH}_3][\text{H}_2\text{S}] \]
Since the concentration of solid reactant does not appear in the expression, this process focuses entirely on the products which, in this case, are the gaseous substances, ammonia \((\text{NH}_3)\) and hydrogen sulfide \((\text{H}_2\text{S})\).
Understanding \(K_c\) is crucial as it provides insight into the dynamics of the reaction equilibrium, aiding in predicting which direction the reaction will favor under certain conditions.

Concentration Calculations

Calculating concentrations at equilibrium involves setting up an equation using the known equilibrium constant \(K_c\). When given initial concentrations, or even zero concentrations for some reactants or products, the equilibrium condition shows the change over time.
For this reaction, we consider the decomposition process of \(\text{NH}_4\text{HS}\) into \(\text{NH}_3\) and \(\text{H}_2\text{S}\).
Initially, if the flask contains only the pure solid \(\text{NH}_4\text{HS}\), both \(\text{NH}_3\) and \(\text{H}_2\text{S}\) will start from zero concentration. As they form equally over time and come to equilibrium, let \(x\) be the concentration of either gas. The equilibrium equation is then:

• \( [\text{NH}_3] = x \)
• \( [\text{H}_2\text{S}] = x \)

Placing these into the expression for \(K_c\):
\[ x^2 = 1.8 \times 10^{-4} \]
Solving gives \(x \approx 0.0134\) mol/L. Therefore, each gas has a concentration of 0.0134 mol/L at equilibrium, demonstrating how these calculations reveal the balance a system reaches.

Ammonium Hydrogen Sulfide Decomposition

The decomposition of ammonium hydrogen sulfide, represented by \(\text{NH}_4\text{HS}(\text{s}) \rightleftharpoons \text{NH}_3(\text{g}) + \text{H}_2\text{S}(\text{g})\), involves breaking down the solid \(\text{NH}_4\text{HS}\) into ammonia and hydrogen sulfide gases. This process is in equilibrium, meaning the products can react to form the original solid as well.
The extent of this decomposition at equilibrium is influenced by the equilibrium constant \(K_c\), temperature, and initial concentrations of the products involved. Since \(\text{NH}_4\text{HS}\) is the only solid and natives the equilibrium expression, tracking the production of \(\text{NH}_3\) and \(\text{H}_2\text{S}\) is sufficient for our calculations.
In scenarios where initial gaseous concentrations are present, such as when \(0.020\) mol/L of \(\text{NH}_3\) exists initially, the system adjusts to reach equilibrium. This involves a dynamic balance of forward and reverse reactions, highlighting the reversible nature of chemical reactions and the interactions within a closed system.

Quadratic Equation in Chemistry

Solving quadratic equations is a fundamental part of chemical equilibrium problems, especially when initial concentrations are involved, and you need to account for changes as the system reaches equilibrium.
When ammonia \((\text{NH}_3)\) is initially present at 0.020 mol/L and \(\text{NH}_4\text{HS}\) is also introduced, the equilibrium concentrations need to reflect the changes via the equation:
\[ (0.020 + x)x = 1.8 \times 10^{-4} \]
This expands into:
\[ x^2 + 0.020x - 1.8 \times 10^{-4} = 0 \]
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 0.020\), \(c = -1.8 \times 10^{-4}\), we solve for \(x\) giving approximately \(x \approx 0.0078\). This shows the increase in concentration for \(\text{H}_2\text{S}\) and the adjusted concentration for \(\text{NH}_3\), demonstrating how changes in initial conditions impact the equilibrium state of the reaction.