Problem 94 - Chemistry The Molecular Science

Question

Solid ammonium iodide decomposes to ammonia and hydrogen iodide gases at sufficiently high temperatures. $$ \mathrm{NH}_{4} \mathrm{I}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{HI}(\mathrm{g}) $$ The equilibrium constant, \(K_{\mathrm{P}}\), for the decomposition at \(673 \mathrm{~K}\) is 0.215 . A \(15.0-\mathrm{g}\) sample of ammonium iodide is sealed in a \(5.00-\mathrm{L}\) flask and heated to \(673 \mathrm{~K}\). (a) Calculate the total pressure in the flask at equilibrium. (b) Calculate the amount (in moles) of ammonium iodide that decomposes.

Step-by-Step Solution

Answer

(a) Total pressure = 0.720 atm; (b) 0.0652 mol of NH₄I decomposes.

Step 1: Calculate Moles of NH₄I

First, find the molar mass of ammonium iodide (NH₄I), which is calculated as follows: Molar mass of N (14.01 g/mol) + 4 × H (1.01 g/mol) + I (126.90 g/mol) = 144.94 g/mol. Then, calculate moles of NH₄I: \[ \text{Moles of NH}_4\text{I} = \frac{15.0 \text{ g}}{144.94 \text{ g/mol}} \approx 0.1035 \text{ mol} \]

Step 2: Set Up Reaction Table

For the decomposition reaction \(\text{NH}_4\text{I (s)} \rightleftharpoons \text{NH}_3\text{(g)} + \text{HI (g)}\), the initial moles of NH₄I = 0.1035 mol. Assume \(x\) moles decompose, producing \(x\) moles of NH₃ and \(x\) moles of HI at equilibrium.

Step 3: Apply Equilibrium Expression

The equilibrium expression for \(K_P\) is given by: \[ K_P = P_{\text{NH}_3} \times P_{\text{HI}} \]At equilibrium, both partial pressures are \(\frac{x}{V}R T\), where \(V = 5.00\) L, \(R = 0.0821\) L·atm/(K·mol), and \(T = 673\) K. Calculate the equilibrium pressures:

Step 4: Calculate Pressures and Solve for x

Substitute into the expression for \(K_P = 0.215\): \[ 0.215 = \left(\frac{x}{5.00} \times 0.0821 \times 673\right)^2 \]This simplifies to: \[ x = \sqrt{\frac{0.215 \times 5.00^2}{0.0821^2 \times 673^2}} \approx 0.0652 \text{ mol} \]

Step 5: Find Total Pressure

The total pressure \(P_{\text{total}}\) at equilibrium is the sum of the partial pressures of NH₃ and HI:\[ P_{\text{total}} = 2 \times \frac{0.0652}{5.00} \times 0.0821 \times 673 \]\[ P_{\text{total}} \approx 0.720 \text{ atm} \]

Step 6: Calculate Remaining NH₄I

The total moles of NH₄I decomposed is \(0.0652\) mol. From the initial \(0.1035\) mol, the remaining NH₄I is:\[ 0.1035 - 0.0652 = 0.0383 \text{ mol} \]

Step 7: Verify Results

Check calculations to ensure pressure and decomposition are consistent with equilibrium constant and initial conditions.

Key Concepts

Ammonium Iodide Decomposition

Ammonium iodide (NH₄I) decomposes when heated, undergoing a chemical reaction that turns it into two gases: ammonia (NH₃) and hydrogen iodide (HI). This type of reaction is reversible, meaning it can occur in both directions under certain conditions. Decomposition reactions like this are significant in chemistry because they allow us to study the behavior of substances as they change state.
In terms of a balanced chemical equation, the decomposition of ammonium iodide can be represented as follows: \[ \text{NH}_4\text{I (s)} \rightleftharpoons \text{NH}_3\text{(g)} + \text{HI (g)} \]The term \( \rightleftharpoons \) represents the dynamic equilibrium that occurs in this reversible reaction. Here, NH₄I is a solid, while NH₃ and HI are gases at the elevated temperature where equilibrium is observed. Understanding this equilibrium, and how it behaves under changes in conditions, is crucial for predicting and controlling chemical processes.

Equilibrium Constant

The concept of an equilibrium constant \(K_{P}\) plays a central role in understanding chemical equilibria. It quantifies the ratio of the concentrations of products to reactants for a reaction at equilibrium. For gaseous reactions like the decomposition of ammonium iodide, \(K_{P}\) is expressed in terms of partial pressures rather than concentrations:
\[ K_{P} = \frac{P_{\text{NH}_3} \times P_{\text{HI}}}{P_{\text{NH}_4\text{I}}} \]However, since NH₄I is a solid, its pressure does not contribute to the equilibrium expression, simplifying it to:\[ K_{P} = P_{\text{NH}_3} \times P_{\text{HI}} \]With a given \(K_{P}\) value of 0.215 at 673 K, this tells us how far the reaction proceeds towards products under these specific conditions. To find the equilibrium pressures needed to solve practical problems, like determining decomposed amounts or total system pressure, we apply this expression to experimental setups.

Partial Pressure

Partial pressure is a key concept in studying reactions involving gases. It refers to the pressure exerted by a single type of gas in a mixture of gases. In the context of ammonium iodide decomposition, the gases of interest are NH₃ and HI, each contributing to the total pressure in a container.
Let's assume we have an equilibrium state where both gases are present at equal amounts due to the reaction stoichiometry:
- The partial pressures depend on the volume of the container, the temperature of the reaction, and the amount of each gas formed.
- According to the ideal gas law, we use the formula \( P = \frac{nRT}{V} \), where \( n \) is moles of gas, \( R \) is the gas constant, \( T \) is temperature, and \( V \) is volume.
This relationship allows us to calculate the pressure each gas exerts based on the moles of the gas present. These calculations are critical when determining total pressure or verifying experimental data.

Mole Calculations

Performing mole calculations is fundamental in predicting and verifying the outcome of chemical reactions. When ammonium iodide decomposes, we calculate how many moles transform into ammonia and hydrogen iodide by using stoichiometry.
Here's how it's typically done:

• First, calculate the molar mass of the compound (NH₄I in this case), which helps in determining how many moles are in the given mass. Molar mass is calculated by summing the atomic masses of each element within the compound.
• Then, divide the mass of the compound by its molar mass to find the number of moles, like so: \( \text{Moles} = \frac{\text{mass}}{\text{molar mass}} \).

The challenge is to do this accurately to predict the yield of the decomposition reaction. Further, knowing the initial moles helps in setting up an equilibrium table to find how much of the compound decomposes, which, in this case, is determined through derivation and solving equations related to the equilibrium constant.