Question

Solid ammonium iodide decomposes to ammonia and hydrogen gases at sufficiently high temperatures. $$ \mathrm{NH}_{4} \mathrm{I}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{HI}(g) $$ The equilibrium constant for the decomposition at \(400^{\circ} \mathrm{C}\) is 0.215. Twenty grams of ammonium iodide are sealed in a \(7.50-\mathrm{L}\) flask and heated to \(400^{\circ} \mathrm{C}\). (a) What is the total pressure in the flask at equilibrium? (b) How much solid \(\mathrm{NH}_{4} \mathrm{I}\) is left after the decomposition?

Short answer

Answer: The total pressure in the flask at equilibrium is 3.12 atm, and 9.28 grams of solid NH4I are left after the decomposition.

Step-by-step solution

(Step 1: Convert grams of NH4I to moles)

Given the initial mass of NH4I is 20 grams, we can convert it into moles using the molar mass of NH4I: NH4I = 1(N) + 4(H) + 1(I) = 14.01 + 4(1.008) + 126.90 = 144.92 g/mol moles of NH4I = (20 g) / (144.92 g/mol) = 0.138 moles

(Step 2: Set up the ICE table)

The ICE table helps us analyze the initial number of moles of the substances, the change in moles during the reaction, and the equilibrium numbers of moles of the substances. Initial: 0.138 moles (NH4I) | 0 moles (NH3) | 0 moles (HI) Change: -x moles (NH4I) | +x moles (NH3) | +x moles (HI) Equilibrium: (0.138-x) moles (NH4I) | x moles (NH3) | x moles (HI)

(Step 3: Write the equilibrium expression)

We will use the equilibrium constant (K) to relate the concentrations of the products and reactants at equilibrium. K is given as 0.215. The partial pressure of the gases can be calculated using the moles/volume in liters, and the volume of the flask is 7.5 Liters. K = [NH3][HI] / ([NH4I] * 7.5 L)

(Step 4: Solve for x)

Substitute the equilibrium moles into the equilibrium equation and solve for x: 0.215 = (x * x) / ((0.138 - x) * 7.5 L) x = 0.074 moles

(Step 5: Calculate the total pressure at equilibrium)

At equilibrium, we have x moles of both NH3 and HI. We can find the partial pressures of both gases, and then calculate the total pressure. Partial pressure of NH3 = x moles * R * T / V = (0.074 moles * 0.0821 L*atm/mol*K) * (400°C + 273.15) K / 7.5 L = 1.56 atm Partial pressure of HI = x moles * R * T / V = (0.074 moles * 0.0821 L*atm/mol*K) * (400°C + 273.15) K / 7.5 L = 1.56 atm Total pressure at equilibrium = partial pressure of NH3 + partial pressure of HI = 1.56 atm + 1.56 atm = 3.12 atm So, the total pressure in the flask at equilibrium is 3.12 atm.

(Step 6: Calculate the amount of solid NH4I left after decomposition)

At equilibrium, there are (0.138 - x) moles of NH4I left. Convert it back to grams: grams of NH4I = (0.138 - 0.074) moles * 144.92 g/mol = 9.28 g Thus, 9.28 grams of solid NH4I are left after the decomposition.