2. Solid ammonium iodide decomposes to yield ammonia gas(NH3) and hydrogen iodide gas at 400oF. The equilibrium constant for this reaction is 0.215. Ten grams of ammonium iodide are sealed in a 10.0 L flask at equilibrium. Complete the following:

a. Write a balanced reaction

b. What is the total pressure in the flask at equilibrium?

c. How much ammonium iodide decomposes?

NH4I ==> NH3 +HI

Is that 0.215 = Kc or Kp?

a. The balanced reaction for the decomposition of solid ammonium iodide is:

NH4I(s) → NH3(g) + HI(g)

b. To determine the total pressure in the flask at equilibrium, we need to use the ideal gas law which states:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature

Since the volume (V) and temperature (T) are given (10.0 L and 400oF), and the number of moles (n) can be calculated using the given mass of ammonium iodide, we can substitute these values into the equation to solve for the pressure (P).

c. To determine how much ammonium iodide decomposes, we need to use the equilibrium constant (K). The equilibrium constant is given as 0.215, which is the ratio of the product concentrations (NH3 and HI) to the reactant concentration (NH4I).

K = [NH3][HI] / [NH4I]

Let's assume x moles of NH4I decompose, then the moles of NH3 and HI will be x as well.

Therefore, the concentrations at equilibrium will be:
[NH3] = x moles / 10.0 L
[HI] = x moles / 10.0 L
[NH4I] = (10 - x) moles / 10.0 L

Substituting these into the equilibrium constant expression:
0.215 = (x / 10.0) * (x / 10.0) / ((10 - x) / 10.0)

Simplifying the equation and solving for x will give us the amount of ammonium iodide that decomposes.

a. The balanced reaction for the decomposition of solid ammonium iodide is:

NH4I (s) → NH3 (g) + HI (g)

b. To determine the total pressure in the flask at equilibrium, we need to use the ideal gas law, which states that the pressure (P) of a gas is equal to the product of its molar amount (n), the gas constant (R), and the temperature (T), divided by the volume (V):

P = nRT/V

The total pressure in the flask is the sum of the partial pressures of ammonia gas (NH3) and hydrogen iodide gas (HI). Since the reaction reaches equilibrium, the molar amounts of NH3 and HI will be equal.

Let's assume that x moles of NH3 and HI are produced at equilibrium. Then, the molar amount of NH4I decomposed will also be x moles.

According to the balanced reaction equation, the molar ratio between NH4I and NH3 (or HI) is 1:1. Therefore, the concentrations of NH3 and HI are also x/V.

The total concentration, or pressure, at equilibrium can be calculated by summing the partial pressures:

Ptotal = PNH3 + PHI = (x/V) + (x/V) = 2(x/V)

Given that the equilibrium constant (K) for the reaction is 0.215, we can set up an expression for K:

K = [NH3][HI] / [NH4I]

Since the concentrations of NH3 and HI are equal to x/V, and the concentration of NH4I is initially 10 g in a 10.0 L flask, we can substitute these values into the expression:

0.215 = (x/V)(x/V) / (10/V)

0.215 = (x^2 / V^2) / (10/V)

0.215 = x^2 / (10V)

x^2 = 2.15V

x = √(2.15V)

Now, we can substitute this value of x back into the equation for the total pressure at equilibrium:

Ptotal = 2(x/V) = 2(√(2.15V)/V) = 2√2.15 ≈ 2.94

Therefore, the total pressure in the flask at equilibrium is approximately 2.94 units.

c. To determine how much ammonium iodide decomposes at equilibrium, we need to calculate the value of x. Using the expression x = √(2.15V) derived in part b, we can substitute the volume provided in the question:

x = √(2.15 * 10) = √21.5 ≈ 4.64

Therefore, approximately 4.64 moles of ammonium iodide decompose at equilibrium.