Question

Ammonium carbamate \(\left(\mathrm{NH}_{2} \mathrm{COONH}_{4}\right)\) is a salt of carbamic acid that is found in the blood and urine of mammals. At \(250 .{ }^{\circ} \mathrm{C}, K_{\mathrm{c}}=1.58 \times 10^{-8}\) for the following equilibrium: $$ \mathrm{NH}_{2} \mathrm{COONH}_{4}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) $$ If \(7.80 \mathrm{~g}\) of \(\mathrm{NH}_{2} \mathrm{COONH}_{4}\) is put into a \(0.500-\mathrm{L}\) evacuated container, what is the total pressure at equilibrium?

Short answer

Total pressure at equilibrium is approximately 0.12 atm.

Step-by-step solution

- Write the balanced chemical equation

The equilibrium process is given by:\[\mathrm{NH}_{2}\mathrm{COONH}_{4}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)\]

- Define the initial conditions

Initially, we have 7.80 g of \(\mathrm{NH}_{2}\mathrm{COONH}_{4}\) in a 0.500-L container. The molecular weight of \(\mathrm{NH}_{2}\mathrm{COONH}_{4}\) is 79.06 g/mol, so the number of moles of \(\mathrm{NH}_{2}\mathrm{COONH}_{4}\) is: \[\frac{7.80 \text{ g}}{79.06 \text{ g/mol}} = 0.0987 \text{ mol}\]

- Establish the expression for Kc

The equilibrium constant \(K_c\) is given by:\[K_c = \frac{[\mathrm{NH}_{3}]^2 [\mathrm{CO}_{2}]}{[\mathrm{NH}_{2}\mathrm{COONH}_{4}]}\]since \(\mathrm{NH}_{2}\mathrm{COONH}_{4}\) is a solid, its concentration is constant and its activity is taken as 1. Hence,\[K_c = [\mathrm{NH}_{3}]^2 [\mathrm{CO}_{2}]\]

- Set up the ICE table

Create an ICE (Initial, Change, Equilibrium) table:\[\begin{array}{c|c|c|c}& \mathrm{NH}_{2}\mathrm{COONH}_{4} (s) & \mathrm{NH}_{3} (g) & \mathrm{CO}_{2} (g) \hline\text{Initial} & 0.0987 \text{ mol} & 0 & 0 \text{Change} & -x & +2x & +x \text{Equilibrium} & 0.0987 - x & 2x & x \end{array}\]

- Apply the equilibrium expression

Using the ICE table values in the equilibrium expression:\[K_c = (2x)^2 (x) \]Given, \(K_c = 1.58 \times 10^{-8}\)\[1.58 \times 10^{-8} = 4x^3\]

- Solve for x

Solve for \(x\) to find the equilibrium concentrations of gases:\[4x^3 = 1.58 \times 10^{-8}\]\[x^3 = \frac{1.58 \times 10^{-8}}{4}\]\[x^3 = 3.95 \times 10^{-9}\]\[x = \sqrt[3]{3.95 \times 10^{-9}}\]\[x \approx 1.58 \times 10^{-3} \text{ M}\]

- Calculate the total pressure

Given the partial pressures at equilibrium as \(P_{NH_3} = 2x\) and \(P_{CO_2} = x\), use the ideal gas law to convert the equilibrium concentrations to pressures:\[P_{\text{total}} = P_{NH_3} + P_{CO_2}\]\[= 2(1.58 \times 10^{-3} \text{ M}) + 1.58 \times 10^{-3} \text{ M}\]Convert to pressure using PV=nRT, where R=0.0821 L·atm/(mol·K):\[P_{\text{total}} = 3(1.58 \times 10^{-3} \text{ mol}) \frac{0.0821 \text{ L·atm/(mol·K)}}{0.5 \text{ L}} (250 + 273)\]\[P_{\text{total}} \approx 0.12 \text{ atm}\]

Key concepts

Ammonium Carbamate Equilibrium

Ammonium carbamate, represented as \(\text{NH}_2\text{COONH}_4\), undergoes a decomposition reaction when heated. It breaks down into ammonia (NH\textsubscript{3}) and carbon dioxide (CO\textsubscript{2}). The balanced chemical equation for this reaction is: \(\text{NH}_2\text{COONH}_4(s) \rightleftharpoons 2 \text{NH}_3(g) + \text{CO}_2(g)\). This means that for each mole of ammonium carbamate decomposing, 2 moles of ammonia and 1 mole of carbon dioxide gas are produced.

ICE Table Method

The ICE table is a crucial tool in solving chemical equilibrium problems. The acronym ICE stands for Initial, Change, and Equilibrium, representing the different stages of the reactants and products concentration changes. Here's a breakdown:

• Initial: This is the starting concentration of the reactants and products before any reaction occurs.
• Change: This refers to the change in concentration as the system moves towards equilibrium. It’s usually represented with variable changes (for instance, -x, +2x).
• Equilibrium: These values represent the concentrations when the system is at equilibrium.

In the given exercise, the initial concentrations of ammonia and carbon dioxide are zero, and the ammonium carbamate is given in grams which is converted into moles. We then denote the change in concentration using the variable x.

Equilibrium Constant (\text{K}_c)

The equilibrium constant, \(K_c\), quantifies the ratio of the concentrations of products to reactants at equilibrium and is a fundamental concept in chemical equilibrium. For the decomposition of ammonium carbamate: \(\text{K}_c = \frac{[\text{NH}_3]^2 [\text{CO}_2]}{[\text{NH}_2\text{COONH}_4]}\). Here, because \(\text{NH}_2\text{COONH}_4\) is a solid, its concentration remains constant and is taken as 1 in the \(K_c\) expression: \(\text{K}_c = [\text{NH}_3]^2[\text{CO}_2]\). Given \(K_c = 1.58 \times 10^{-8}\), this relation helps us calculate the concentrations of NH\textsubscript{3} and CO\textsubscript{2} at equilibrium.

Ideal Gas Law

The Ideal Gas Law links pressure, volume, temperature, and the number of moles of gas through the expression \(PV = nRT\). Here:

• P: Total pressure of the gas
• V: Volume of the container
• n: Number of moles of the gas
• R: Ideal gas constant (0.0821 L·atm/(mol·K))
• T: Temperature in Kelvin

The total pressure at equilibrium can be calculated using the concentrations obtained from the ICE table and converting them using the Ideal Gas Law. Simply plug in the values: \(P_{\text{total}} = 3(1.58 \times 10^{-3} \text{ mol}) \frac{0.0821 \text{ L·atm/(mol·K)}}{0.5 \text{ L}} (250 + 273)\), yielding the total pressure.