Question

Solid ammonium iodide decomposes to ammonia and hydrogen iodide 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 \(673 \mathrm{~K}\) is \(0.215\). Fifteen grams of ammonium iodide are sealed in a \(5.0\) -L flask and heated to \(673 \mathrm{~K}\). (a) What is the total pressure in the flask at equilibrium? (b) How much ammonium iodide decomposes?

Short answer

Answer: The total pressure in the flask at equilibrium is 12.4 atm, and 0.463 moles of ammonium iodide decompose at equilibrium.

Step-by-step solution

Write the balanced chemical equation and the expression for equilibrium constant

The balanced chemical equation for the decomposition of solid ammonium iodide is given as: $$\mathrm{NH}_{4} \mathrm{I}(s) \rightleftharpoons \mathrm{NH}_{3}(g) + \mathrm{HI}(g)$$ The equilibrium constant expression for this reaction can be written as: $$K_c = \frac{\left[\mathrm{NH}_3\right]\left[\mathrm{HI}\right]}{\left[\mathrm{NH}_4 \mathrm{I}\right]}$$ Since the ammonium iodide is a solid, its concentration does not affect the equilibrium constant. We simplify the equilibrium constant expression: $$K_c = \left[\mathrm{NH}_3\right]\left[\mathrm{HI}\right]$$

Calculate the initial moles of ammonium iodide

We have 15g of ammonium iodide and need to find the initial moles. The molar mass of ammonium iodide is: $$\mathrm{(1\times14.01)+(1\times1.01)+(1\times126.9)\ \text{g/mol}}=142.92\ \text{g/mol}$$ Find the moles: $$\text{moles of}\ \mathrm{NH_{4}I} = \frac{15\ \text{g}}{142.92\ \text{g/mol}}=0.105\ \text{mol}$$

Set up the ICE (Initial, Change, Equilibrium) table

Below is our ICE table representing the initial moles of components, the change of moles at equilibrium and the equilibrium moles of components. NH4I(s) <=> NH3(g) + HI(g) Initial 0.105 0 0 Change -x +x +x Final 0.105-x x x

Calculate the molar equilibrium concentrations

We can use the simplified equilibrium constant expression and the equilibrium concentrations calculated from our ICE table to find x: $$K_c = \left[\mathrm{NH}_3\right] \left[\mathrm{HI}\right] = x^2$$ Plug in the value of Kc: $$0.215=x^2$$ Solve for x: $$x=\sqrt{0.215}=0.463$$

Calculate the total pressure at equilibrium

At equilibrium, there are x moles of \(\mathrm{NH}_3\) and \(\mathrm{HI}\). We can use the ideal gas law (PV=nRT) to calculate the total pressure at equilibrium: $$P_{total} = \frac{(n_{NH_{3}}+n_{HI})RT}{V}$$ Plug in the values for moles, temperature, and volume: $$P_{total} = \frac{(0.463+0.463)(0.0821)(673)}{5.0}$$ Calculate the total pressure: $$P_{total} = 12.4\ \text{atm}$$ (a): The total pressure in the flask at equilibrium is 12.4 atm.

Calculate the amount of ammonium iodide that decomposes

The change in moles of ammonium iodide, x, represents the amount of ammonium iodide that decomposes. Therefore: $$\text{Amount of NH}_{4}\text{I decomposed (mol)} = x = 0.463\ \text{mol}$$ (b): 0.463 moles of ammonium iodide decomposes at equilibrium.

Key concepts

Ammonium Iodide Decomposition

In the decomposition of ammonium iodide (\( \text{NH}_4\text{I} \)), the solid breaks down into two gases: ammonia (\( \text{NH}_3 \)) and hydrogen iodide (\( \text{HI} \)).This process occurs at high temperatures, leading to a dynamic equilibrium between the reactants and products. This means the reaction can move in both directions, forward and reverse, and the concentration of reactants and products remains constant over time.
The balanced chemical equation for this reaction is:

• \( \text{NH}_4 \text{I}(s) \rightleftharpoons \text{NH}_3(g) + \text{HI}(g) \)

The decomposition involves only the gaseous products when considering the equilibrium constant, as solid concentrations don't affect it. Understanding this reaction helps in calculating changes in amount and pressure during the process.

Equilibrium Constant

The equilibrium constant (\( K_c \)) represents the relationship between the concentrations of products and reactants at equilibrium. For ammonium iodide decomposition, it is defined as follows:

• \( K_c = [\text{NH}_3][\text{HI}] \)

Since ammonium iodide is a solid, its concentration doesn’t impact \( K_c \).
In our example, \( K_c = 0.215 \), indicating the level of product formation relative to the reactants when equilibrium is reached. A higher \( K_c \) would suggest more product formation at equilibrium.
The value of \( K_c \) is crucial for calculating concentrations at equilibrium using the ICE table and determining the equilibrium state.

ICE Table

The ICE table is a tool used to track changes in concentration for the components involved in a chemical reaction.
"ICE" stands for Initial, Change, and Equilibrium, representing the stages in the reaction:

• Initial: The beginning concentration of each substance. For this reaction, initial moles of \( \text{NH}_3 \) and \( \text{HI} \) are zero.
• Change: The change in moles as the reaction approaches equilibrium. Represented by \( x \), which is gained or lost as the reaction proceeds.
• Equilibrium: The concentration of components when the reaction is balanced. Calculated using initial values adjusted by changes.

The ICE table helps us solve for \( x \) in this case, revealing the concentration of gases at equilibrium and informing us about how much ammonium iodide decomposes.

Ideal Gas Law

The Ideal Gas Law connects pressure, volume, temperature, and moles of gas in a system. Expressed as:\[PV = nRT\]Where:

• \( P \) is pressure in atmosphere (atm)
• \( V \) is volume in liters (L)
• \( n \) is number of moles
• \( R \) is the gas constant \( 0.0821 \frac{L \, atm}{mol \, K} \)
• \( T \) is temperature in Kelvin (K)

This formula is used to find the total pressure at equilibrium once the moles of \( \text{NH}_3 \) and \( \text{HI} \) are known. In the problem, after solving for \( x \), we plug these values into the Ideal Gas Law to calculate the total pressure, illustrating the practical application of both equilibrium concepts and gas laws.