Question

Ammonium iodide dissociates reversibly to ammonia and hydrogen iodide if the salt is heated to a sufficiently high temperature. $$ \mathrm{NH}_{4}\left[(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{HI}(\mathrm{g})\right. $$ Some ammonium iodide is placed in a flask, which is then heated to \(400^{\circ} \mathrm{C}\). If the total pressure in the flask when equilibrium has been achieved is \(705 \mathrm{mm} \mathrm{Hg},\) what is the value of \(K_{\mathrm{p}}\) (when partial pressures are in atmospheres)?

Short answer

\( K_\mathrm{p} \approx 0.214 \, \mathrm{atm}^2 \)

Step-by-step solution

Understanding the dissociation reaction

The given reaction is the dissociation of ammonium iodide (\( \mathrm{NH}_4\mathrm{I} \)) into ammonia (\( \mathrm{NH}_3 \)) and hydrogen iodide (\( \mathrm{HI} \)). The equilibrium equation is \( \mathrm{NH}_4\mathrm{I}_{(s)} \rightleftharpoons \mathrm{NH}_3_{(g)} + \mathrm{HI}_{(g)} \). At equilibrium, the partial pressures of the gaseous products are required to calculate \( K_\mathrm{p} \).

Relating total pressure to partial pressures

The total pressure at equilibrium is given as \( 705 \, \mathrm{mmHg} \), which converts to atmospheres as \( \frac{705}{760} \, \mathrm{atm} \). For each mole of \( \mathrm{NH}_3 \) produced, one mole of \( \mathrm{HI} \) is also produced, resulting in equal partial pressures of both gases.

Calculating equilibrium partial pressures

Let the equilibrium partial pressure of \( \mathrm{NH}_3 \) be \( P \). Then the partial pressure of \( \mathrm{HI} \) will also be \( P \). The total pressure is the sum of these partial pressures, thus: \( 2P = \frac{705}{760} \). Calculate \( P \) using this equation.

Solve for \( P \)

Solving \( 2P = \frac{705}{760} \), you get \( P = \frac{705}{1520} \). Compute this value.

Calculate \( K_\mathrm{p} \)

The equilibrium constant \( K_\mathrm{p} \) for the reaction is given by the expression \( K_\mathrm{p} = (P_{\mathrm{NH}_3})(P_{\mathrm{HI}}) = P^2 \). Substitute the value of \( P \) from Step 4 into this equation to find \( K_\mathrm{p} \).

Key concepts

Dissociation Reaction

A dissociation reaction involves the breaking apart of a compound into two or more components. In our example, ammonium iodide (\( \mathrm{NH}_4\mathrm{I} \)) undergoes dissociation when heated to form ammonia (\( \mathrm{NH}_3 \)) and hydrogen iodide (\( \mathrm{HI} \)). This is a reversible reaction, meaning that it can proceed in both forward and backward directions, reaching a point of equilibrium. At this point, the rate of dissociation equals the rate of re-formation of ammonium iodide. Understanding the characteristics of dissociation reactions is key:

• They often require energy, such as heat, to proceed.
• They are described by a chemical equation showing the conversion from a compound to simpler products.
• In equilibrium, the concentration of reactants and products remains constant over time.

The specific reaction in this exercise is represented as:\[ \mathrm{NH}_4\mathrm{I}_{(s)} \rightleftharpoons \mathrm{NH}_3_{(g)} + \mathrm{HI}_{(g)} \]Here, solid ammonium iodide dissociates into gaseous products. The balance of this reaction at equilibrium is important for determining various properties, such as the equilibrium constant.

Equilibrium Constant (Kp)

The equilibrium constant (\( K_p \)) provides a quantitative measure of the position of equilibrium in a chemical reaction involving gases, and it is expressed in terms of partial pressures. For our dissociation of ammonium iodide, \( K_p \) represents the ratio of the product of the partial pressures of the gases at equilibrium. It is expressed as:\[ K_p = (P_{\mathrm{NH}_3})(P_{\mathrm{HI}}) = P^2 \]Here, \( P \) represents the partial pressure of both ammonia and hydrogen iodide since they are produced in equal amounts during the reaction.To calculate \( K_p \):

• Understand that the total pressure at equilibrium is the sum of the partial pressures of \( \mathrm{NH}_3 \) and \( \mathrm{HI} \).
• Convert the given total pressure from mmHg to atmospheres for compatibility with \( K_p \) calculations. In this case, the total pressure is \( 705 \ \mathrm{mmHg} \) converted to \( \frac{705}{760} \ \mathrm{atm} \).
• Determine the partial pressure of each gas (\( P \)) using \( 2P = \frac{705}{760} \), resulting in \( P = \frac{705}{1520} \).
• Substitute back the value of \( P \) to find \( K_p \).

Having \( K_p \) is crucial as it tells us how far a reaction has proceeded and the relative amounts of reactants and products at equilibrium.

Partial Pressure

Partial pressure describes the pressure exerted by a single gas in a mixture of gases. It is a crucial concept in understanding gas mixtures, especially in equilibrium reactions like the one involving \( \mathrm{NH}_4\mathrm{I} \) dissociation. The total pressure in a container is the sum of the partial pressures of each gas present. To determine partial pressures at equilibrium:

• Recognize that each component in the gas mixture contributes to the total pressure based on its mole fraction.
• In the given reaction, ammonia and hydrogen iodide contribute equally to the pressure. Therefore, we let the partial pressure of \( \mathrm{NH}_3 \) be \( P \) and \( \mathrm{HI} \) will also have \( P \) as its partial pressure.
• Relate the total pressure to individual partial pressures using the equation for total pressure: \( P_{\text{total}} = P_{\mathrm{NH}_3} + P_{\mathrm{HI}} = 2P \).
• Use this relationship to solve for \( P \), which is \( \frac{705}{1520} \) in this exercise after converting total pressure to atmospheres.

For chemical reactions and calculations involving gases, knowing the partial pressures helps in determining other properties such as reaction rates and equilibrium constants.