Solid ammonium carbamate, \(\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}\), decomposes at \(25^{\circ} \mathrm{C}\) to ammonia and carbon dioxide. $$\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{CO}_{2}(g)$$ The equilibrium constant for the decomposition at \(25^{\circ} \mathrm{C}\) is \(2.3 \times 10^{-4}\), At \(25^{\circ} \mathrm{C}, 7.50 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}\) is sealed in a \(10.0\) - \(\mathrm{L}\) flask and allowed to decompose. (a) What is the total pressure in the flask when equilibrium is established? (b) What percentage of \(\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}\) decomposed? (c) Can you state from the data calculated that the decomposition took place slowly?

Step by step solution

01

Write down the given equilibrium constant

The equilibrium constant, Kc, for the decomposition at 25°C is given as \(2.3 \times 10^{-4}\).

02

Determine the initial moles of NH4CO2NH2 and the initial pressure

Calculate the initial moles of solid ammonium carbamate, NH4CO2NH2 in the flask. First determine the molar mass of NH4CO2NH2: Molar mass = (2 * 14.01) + 4 + (2 * 16) + 12.01 = 78.07 g/mol Next, find the number of moles of NH4CO2NH2: moles = 7.50 g / 78.07 g/mol ≈ 0.096 moles Since the flask starts with solid ammonium carbamate only, initially, there is no ammonia and carbon dioxide. Thus, the initial pressure in the flask is 0 atm.

03

Set up an equilibrium table using moles

To find the moles of ammonia and carbon dioxide formed at equilibrium, we can set up a table with the initial moles, the changes in moles during the reaction, and the moles at equilibrium. [NH4CO2NH2] [NH3] [CO2] Initial 0.096 0 0 Change - x +2x +x Equilibrium 0.096-x 2x x

04

Use the equilibrium constant expression to solve for the pressure at equilibrium

For the reaction: $\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{CO}_{2}(g)$, the given equilibrium constant Kc can be expressed as: Kc = [NH3]^2 / [CO2]. Substitute the moles in terms of x from our equilibrium table for the concentrations and solve for x: Kc = ((2x)/10)^2 / (x/10) 2.3 * 10^(-4) = (4x^2) / (x) x = 0.0114 moles Since the equilibrium constant is related to concentration ratios, we need to convert the moles of NH3 and CO2 back to pressures (in atm) by using the Ideal Gas Law: p = nRT/V P_NH3 = (2 * 0.0114 * 0.0821 * 298) / 10 ≈ 0.00564 atm P_CO2 = (0.0114 * 0.0821 * 298) / 10 ≈ 0.00282 atm Now, find the total pressure at equilibrium: P_total = P_NH3 + P_CO2 ≈ 0.00846 atm

05

Determine the percentage of ammonium carbamate decomposed

Calculate the percentage of ammonium carbamate that decomposed: % decomposition = (moles of NH4CO2NH2 decomposed / initial moles of NH4CO2NH2) * 100 % decomposition = (0.0114 / 0.096) * 100 ≈ 11.87%

06

Discuss whether data indicates the decomposition took place slowly

From the equilibrium constant, we can see that the decomposition of ammonium carbamate into ammonia and carbon dioxide is not favored under these conditions (Kc < 1). Additionally, the total pressure at equilibrium and the percentage of decomposition are both relatively low. However, this data does not provide enough information to definitively state whether the decomposition took place slowly, as the reaction rate is not the focus in this exercise.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!