Problem 42 - Chemistry and Chemical Reactivity

Question

Ammonium hydrogen sulfide decomposes on heating. $$ \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftarrows \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g}) $$ If \(K_{p}\) for this reaction is 0.11 at \(25^{\circ} \mathrm{C}\) (when the partial pressures are measured in atmospheres), what is the total pressure in the flask at equilibrium?

Step-by-Step Solution

Answer

The total pressure at equilibrium is approximately 0.663 atm.

Step 1: Understanding the Reaction

The chemical equation given is \( \mathrm{NH}_{4} \mathrm{HS} \rightleftarrows \mathrm{NH}_{3} + \mathrm{H}_{2} \mathrm{S} \). This represents a decomposition reaction where solid ammonium hydrogen sulfide decomposes into gaseous ammonia and hydrogen sulfide.

Step 2: Setting Up the Equilibrium Expression

The equilibrium constant \( K_{p} \) for the reaction is given as 0.11 at 25°C. For this reaction, the expression for \( K_{p} \) is \( K_{p} = P_{\mathrm{NH}_3} \cdot P_{\mathrm{H}_2\mathrm{S}} \), where \( P_{\mathrm{NH}_3} \) and \( P_{\mathrm{H}_2\mathrm{S}} \) are the partial pressures of \( \mathrm{NH}_3 \) and \( \mathrm{H}_2\mathrm{S} \) respectively.

Step 3: Using the Reaction Stoichiometry

Since the stoichiometry of the decomposition produces one mole of \( \mathrm{NH}_3 \) and one mole of \( \mathrm{H}_2\mathrm{S} \) from solid \( \mathrm{NH}_4 \mathrm{HS} \), their equilibrium partial pressures will be equal, denoted as \( x \). Thus, \( P_{\mathrm{NH}_3} = x \) and \( P_{\mathrm{H}_2\mathrm{S}} = x \).

Step 4: Substitute and Solve for Partial Pressure

Substitute into the equilibrium expression: \[ K_{p} = P_{\mathrm{NH}_3} \cdot P_{\mathrm{H}_2\mathrm{S}} = x \cdot x = x^2 = 0.11 \]Solve for \( x \):\[ x = \sqrt{0.11} \approx 0.3317 \text{ atm} \]

Step 5: Calculate Total Pressure in the Flask

The total pressure at equilibrium is the sum of the partial pressures of \( \mathrm{NH}_3 \) and \( \mathrm{H}_2\mathrm{S} \):\[ P_{\text{total}} = P_{\mathrm{NH}_3} + P_{\mathrm{H}_2\mathrm{S}} = x + x = 2x \]Substitute \( x = 0.3317 \text{ atm} \):\[ P_{\text{total}} = 2 \times 0.3317 = 0.6634 \text{ atm} \]

Key Concepts

Partial Pressure

In a gas mixture, partial pressure refers to the pressure that each individual gas would exert if it alone occupied the entire volume. This concept is crucial when dealing with chemical equilibria involving gases. For the decomposition of ammonium hydrogen sulfide, we consider the partial pressures of ammonia (\(\mathrm{NH}_3\)) and hydrogen sulfide (\(\mathrm{H}_2\mathrm{S}\)). These pressures are essential in calculating the equilibrium constant, \(K_p\), for the reaction.

Partial pressure calculation can be straightforward if the stoichiometry of the reaction is simple, as in our case. Each mole of \(\mathrm{NH}_{4} \mathrm{HS}\) decomposing gives one mole of \(\mathrm{NH}_3\) and one mole of \(\mathrm{H}_2\mathrm{S}\), meaning they will have equal partial pressures, denoted as \(x\). By knowing that the equilibrium constant \(K_p = 0.11\), we can setup:\[K_p = P_{\mathrm{NH}_3} \cdot P_{\mathrm{H}_2\mathrm{S}} = x^2\]. Then solving for \(x\) gives the partial pressure of each gas at equilibrium.

Equilibrium Constant

The equilibrium constant (\(K_p\)) is a critical value used to express the ratio of products to reactants at equilibrium at a specific temperature. It plays a pivotal role in helping predict the direction of the reaction and the extent to which a reaction will proceed. For the provided reaction, the equilibrium constant is given as 0.11 at 25°C.

For gases, this constant is commonly expressed in terms of partial pressures, different from the concentration-based equilibrium constant used for reactions in solution. The \(K_p\) expression involves the partial pressures of the gaseous products:

• \(K_p = P_{\mathrm{NH}_3} \cdot P_{\mathrm{H}_2\mathrm{S}}\)

Given \(K_p = 0.11\), solving for partial pressures helps determine how the pressures of ammonia and hydrogen sulfide balance out at equilibrium, thereby enabling the calculation of total pressure in the flask.

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where one compound breaks down into two or more simpler substances. In this scenario, solid ammonium hydrogen sulfide (\(\mathrm{NH}_{4} \mathrm{HS} (s)\)) decomposes into two gases: ammonia (\(\mathrm{NH}_3 (g)\)) and hydrogen sulfide (\(\mathrm{H}_2\mathrm{S} (g)\)).

• Initial compound: \(\mathrm{NH}_{4} \mathrm{HS}\) (solid)
• Products: \(\mathrm{NH}_3\) and \(\mathrm{H}_2\mathrm{S}\) (both gaseous)

Decomposition reactions often require heating or some form of energy input to proceed, which is why the decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\) occurs upon heating. Understanding the stoichiometry, each mole of the solid leads to one mole of ammonia and one mole of hydrogen sulfide. This stoichiometric relationship is straightforward, aiding in the calculation of partial pressures and total pressure of the gases involved in the equilibrium.
Overall, understanding decomposition reactions helps in anticipating the products formed and the conditions needed to achieve equilibrium.