Problem 81 - Chemistry The Central Science

Question

Solid \(\mathrm{NH}_{4} \mathrm{SH}\) is introduced into an evacuated flask at \(24^{\circ} \mathrm{C} .\) The following reaction takes place: $$\mathrm{NH}_{4} \mathrm{SH}(s) \Longrightarrow \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g)$$ At equilibrium, the total pressure (for \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S}\) taken together) is 0.614 atm. What is \(K_{p}\) for this equilibrium at \(24^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The equilibrium constant \(K_p\) for the given reaction at \(24^{\circ} \mathrm{C}\) is approximately 0.0942.

Step 1: Write the balanced equation for the decomposition

The given balanced equation for the decomposition of solid \(\mathrm{NH}_4\mathrm{SH}\) is: \[\mathrm{NH}_{4} \mathrm{SH}(s) \Longrightarrow \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g)\]

Step 2: Calculate the partial pressures of \(\mathrm{NH}_3\) and \(\mathrm{H}_2\mathrm{S}\) at equilibrium

Since the system was evacuated initially, the pressure of the two gases (\(\mathrm{NH}_3\) and \(\mathrm{H}_2\mathrm{S}\)) is considered to be the same at equilibrium. The given total pressure for both gases taken together is 0.614 atm. Therefore, the partial pressure of each gas at equilibrium is half the total pressure. \[P_{\mathrm{NH}_3} = P_{\mathrm{H}_2\mathrm{S}} = \frac{0.614 \ \text{atm}}{2} = 0.307 \ \text{atm}\]

Step 3: Calculate the equilibrium constant, \(K_p\)

To determine the equilibrium constant \(K_p\), we use the formula: \[K_p = \frac{P_{\mathrm{NH}_3} \cdot P_{\mathrm{H}_2\mathrm{S}}}{P_{\mathrm{NH}_4\mathrm{SH}}}\] Since the reactant is a solid, its pressure is not considered, and we can rewrite the formula as: \[K_p = P_{\mathrm{NH}_3} \cdot P_{\mathrm{H}_2\mathrm{S}}\] Now, substitute the calculated values of the partial pressures into the equation: \[ K_p = (0.307 \ \text{atm}) \cdot (0.307 \ \text{atm}) = 0.0942\] Therefore, the equilibrium constant \(K_p\) for this reaction at \(24^{\circ} \mathrm{C}\) is approximately 0.0942.

Key Concepts

Equilibrium Constant Calculation

Understanding the equilibrium constant, often denoted as \(K_p\) or \(K_c\), is vital when studying chemical reactions that reach a state of dynamic equilibrium. This constant provides insight into the proportion of products to reactants at equilibrium, and its value is determined by the specific conditions, such as temperature, under which the reaction occurs. It is essential to grasp that \(K_p\) refers to equilibrium involving gaseous substances where pressure is the determining factor, while \(K_c\) involves concentration.

To calculate \(K_p\), one must employ the partial pressures of the gaseous reactants and products. For a simple decomposition reaction like \(\text{NH}_4\text{SH}(s) \rightleftharpoons \text{NH}_3(g) + \text{H}_2\text{S}(g)\), the solid reactant's pressure is not considered in the calculation. By squaring the partial pressure of one product, as they are identical and represented by the same variable in the equilibrium expression, one can find the value of \(K_p\). Why square? Because the expression for equilibrium must reflect the stoichiometry of the balanced chemical equation, and in our example, each product has a coefficient of one. Thus, the simplicity of this particular reaction allows a straightforward calculation of the equilibrium constant.

Partial Pressure

The concept of partial pressure is a cornerstone of gas law calculations and it plays a critical role in determining the equilibrium constant for reactions involving gases. Partial pressure refers to the pressure one component of a gaseous mixture would exert if it occupied the entire volume of the mixture alone at the same temperature. It's a way of describing how much each gas contributes to the total pressure of the mixture.

In our example, the total pressure at equilibrium was given, so the partial pressures of ammonia (\(\text{NH}_3\)) and hydrogen sulfide (\(\text{H}_2\text{S}\)) could be assumed equal, as they arise in a 1:1 ratio from the decomposition. Dividing the total pressure by the number of gas species gives the partial pressure of each. It is this equitable distribution of pressure that simplifies calculations, knowing that in a closed system, the sum of the partial pressures equals the total pressure, according to Dalton's Law of Partial Pressures. Understanding and accurately calculating partial pressures is crucial for correctly determining the \(K_p\) of a gaseous equilibrium system.

Ammonium Hydrosulfide Decomposition

The decomposition of ammonium hydrosulfide (\(\text{NH}_4\text{SH}\)) is a chemical reaction in which solid ammonium hydrosulfide breaks down into ammonia (\(\text{NH}_3\)) and hydrogen sulfide (\(\text{H}_2\text{S}\)), both in gaseous forms. This is an important chemical process often studied in the context of physical chemistry and environmental science due to the substances involved and the principles it illustrates.

This heterogenous equilibrium involves a solid reactant and gaseous products, showcasing the principle that only the gaseous components exert measurable pressure and thus, affect the equilibrium calculations. Since the solid's pressure does not change during the reaction, it is excluded from the equilibrium expression. The exercise highlights how temperature and pressure are essential factors that influence the position of equilibrium. Ultimately, comprehension of concepts like phase changes, stoichiometry, and the behavior of gases is crucial to fully understanding such decompositions and their significance in various fields, from industrial chemistry to environmental science.