Question

Ammonium hydrogen sulfide decomposes according to the following reaction, for which \(K_{\mathrm{p}}=0.11\) at \(250^{\circ} \mathrm{C}\) $$ \mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{NH}_{3}(g) $$ If \(55.0 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{HS}(s)\) is placed in a sealed 5.0 -L container, what is the partial pressure of \(\mathrm{NH}_{3}(g)\) at equilibrium?

Short answer

The partial pressure of \( \text{NH}_3 \) is \( 0.33 \text{ atm} \).

Step-by-step solution

Write the equilibrium expression

For the given reaction \( \text{NH}_{4} \text{HS}(s) \rightleftharpoons \text{H}_{2} \text{S}(g) + \text{NH}_{3}(g) \) write the equilibrium expression based on the given equilibrium constant: \( K_p = P_{\text{H}_2\text{S}} \times P_{\text{NH}_3} \).

Assume initial conditions

Assume initially there is no \( \text{H}_2\text{S} \) and \( \text{NH}_3 \) present in the container. Therefore, let the partial pressures of \( \text{H}_2\text{S} \) and \( \text{NH}_3 } \) at equilibrium be \( P_{\text{H}_2\text{S}} = x \) and \( P_{\text{NH}_3} = x \).

Calculate equilibrium pressures

Since both \( \text{H}_2\text{S} \) and \( \text{NH}_3 \) are produced in a 1:1 ratio, we can use the equilibrium constant expression as: \( K_p = x \times x = 0.11 \), thus \( x^2 = 0.11 \).

Solve for x

Solving for \( x \) gives \( x = \sqrt{0.11} \approx 0.33 \text{ atm} \). Therefore, the partial pressures of \( \text{H}_2\text{S} \) and \( \text{NH}_3 \) at equilibrium are both 0.33 \text{ atm.

Key concepts

Equilibrium Constant

Chemical equilibrium is when the rates of the forward and reverse reactions are equal, so the concentrations of the reactants and products do not change with time. The equilibrium constant, denoted as either \( K_c \) for concentrations or \( K_p \) for partial pressures, measures the ratio of product concentrations to reactant concentrations at equilibrium.
For gas reactions, the equilibrium constant \( K_p \) is used and depends on the partial pressures of the involved gases.
In our example, the reaction: \[ \text{NH}_{4}\text{HS}(s) \rightleftharpoons \text{H}_{2}\text{S}(g) + \text{NH}_{3}(g) \] has an equilibrium constant \( K_p = 0.11 \) at 250°C.

The equilibrium expression for this reaction is:
\[ K_p = P_{\text{H}_2\text{S}} \times P_{\text{NH}_3} \]
Here, \( P_{\text{H}_2\text{S}} \) and \( P_{\text{NH}_3} \) are the partial pressures of \( \text{H}_2\text{S} \) and \( \text{NH}_3 \) at equilibrium.
When writing the equilibrium constant expression, only gases and aqueous solutions are included. Solids and pure liquids are omitted because their concentrations do not change.
Understanding \( K_p \) helps predict the direction in which a reaction will shift to reach equilibrium.

Partial Pressure

Partial pressure is the pressure exerted by a single gas in a mixture of gases.
It is a key concept in understanding gas reactions at equilibrium.
The total pressure of a gas mixture is the sum of the partial pressures of all individual gases.

In our decomposition reaction, the partial pressures of \( \text{H}_2\text{S} \) and \( \text{NH}_3 \) are initially zero, since they are not present at the start.
As the reaction reaches equilibrium, both gases are produced in equal amounts because their formation occurs in a 1:1 ratio.
If the partial pressure of \( \text{H}_2\text{S} \) is \ (x) \, then the partial pressure of \( \text{NH}_3 \) will also be \ (x) \.
Using the equilibrium constant expression, we have \[ K_p = x \times x = 0.11 \]
To find the equilibrium partial pressures, solve the equation to get \[ x = \sqrt{0.11} \] which approximately equals 0.33 atm.
Thus, the partial pressures of both \( \text{H}_2\text{S} \) and \( \text{NH}_3 \) are 0.33 atm at equilibrium.

Decomposition Reaction

A decomposition reaction involves a single compound breaking down into two or more simpler substances.

In the given exercise, ammonium hydrogen sulfide (\text{NH}_{4}\text{HS}) decomposes into hydrogen sulfide (\text{H}_2\text{S}) and ammonia (\text{NH}_3). The balanced chemical equation is:
\[ \text{NH}_{4}\text{HS}(s) \rightleftharpoons \text{H}_2\text{S}(g) + \text{NH}_3(g) \]

Decomposition reactions are common in chemistry, especially when analyzing the thermal stability of compounds or during the production of gases from solid reactants.
The key to solving such problems is determining the proportions in which the products are formed and understanding that decomposition often reaches an equilibrium state.
The amount of each product formed is governed by the equilibrium constant and the initial conditions of the reactants.
In this example, the solid \text{NH}_{4}\text{HS} decomposes in a sealed container, and the equilibrium is established when the rate of decomposition equals the rate of recombination of \text{H}_2\text{S} and \text{NH}_3}. This is why the equilibrium partial pressures are equal for both gases, showing a dynamic but stable system at 250°C.