Explanation Given Enthalpy of vaporization of NH 3 ( l ) is 23.2 kJ/mol . Enthalpy of formation of NH 3 ( g ) is − 46.2 kJ/mol . Density of NH 3 ( l ) is 0.81 g/cm 3 . Density of CH 3 OH ( l ) is 0.792 g/cm 3 . Enthalpy of formation of CH 3 OH ( l ) is − 239 kJ/mol . Refer Appendix C Enthalpy of formation of H 2 O ( g ) is − 241.82 kJ/mol . Enthalpy of formation of H 2 O ( l ) is − 285.83 kJ/mol . Enthalpy of formation of N 2 ( g ) and O 2 ( g ) is 0 kJ/mol . Enthalpy of formation of CO 2 ( g ) is − 393.5 kJ/mol . Vaporization of NH 3 ( l ) to give NH 3 ( g ) is represented as, NH 3 ( l ) → NH 3 ( g ) The enthalpy of vaporization of liquid ammonia is same as enthalpy change for the above reaction that is equal to difference between enthalpy of formation of products and reactants. So, enthalpy of formation of NH 3 ( l ) is calculated as, Δ H v a p = Δ H f ° ( NH 3 ( g ) ) − Δ H f ° ( NH 3 ( l ) ) Where, Δ H v a p is the enthalpy of vaporization of liquid ammonia. Δ H f ° ( NH 3 ( l ) ) is enthalpy of formation of NH 3 ( l ) . Δ H f ° ( NH 3 ( g ) ) is enthalpy of formation of NH 3 ( g ) . Substitute the values of Δ H f ° ( NH 3 ( g ) ) and Δ H v a p in the above formula. 23.2 kJ/mol = ( − 46.2 kJ/mol ) − Δ H f ° ( NH 3 ( l ) ) Δ H f ° ( NH 3 ( l ) ) = − 46.2 kJ/mol − 23.2 kJ/mol = − 69.4 kJ/mol Combustion of NH 3 ( l ) takes place in air to give N 2 ( g ) and H 2 O ( g ) according to the reaction, 2NH 3 ( l ) + 3 2 O 2 ( g ) → N 2 ( g ) + 3 H 2 O ( g ) Enthalpy change for the above reaction is calculated as, Δ H 1 = [ Δ H f ° ( N 2 ( g ) ) + 3 Δ H f ° ( H 2 O ( g ) ) ] − [ 2 Δ H f ° ( NH 3 ( l ) ) + 3 2 Δ H f ° ( O 2 ( g ) ) ] Where, Δ H 1 is the enthalpy change for the given reaction. Δ H f ° ( N 2 ( g ) ) is the enthalpy of formation of N 2 ( g ) . Δ H f ° ( H 2 O ( g ) ) is the enthalpy of formation of H 2 O ( g ) . Δ H f ° ( O 2 ( g ) ) is the enthalpy of formation of O 2 ( g ) . Substitute the values of Δ H f ° ( N 2 ( g ) ) , Δ H f ° ( H 2 O ( g ) ) , Δ H f ° ( O 2 ( g ) ) and Δ H f ° ( NH 3 ( l ) ) in the above formula. Δ H 1 = [ 0 kJ/mol + 3 × ( − 241.82 kJ/mol ) ] − [ 2 × ( − 69.4 kJ/mol ) + 3 2 × 0 kJ/mol ] = [ − 725.46 kJ/mol + 138.8 kJ/mol ] = − 586.66 kJ/mol Conversion of 1 L into cm 3 is done as, 1 L = 1 000 cm 3 Mass of NH 3 ( l ) is calculated as, m 1 = d 1 × V 1 Where, m 1 is the mass of NH 3 ( l ) . d 1 is the density of NH 3 ( l ) . V 1 is the volume of NH 3 ( l ) . Substitute the values of d 1 and V 1 in the above formula. m 1 = 0

Chemistry: The Central Science, Books a la Carte Edition & Modified Mastering Chemistry with Pearson eText -- ValuePack Access Card Package

1st Edition

ISBN: 9780133910919

Author: Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, Catherine Murphy, Patrick Woodward, Matthew E. Stoltzfus

Publisher: PEARSON

See similar textbooks

Concept explainers

Thermochemistry

Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that …

Exergonic Reaction

The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energ…

Videos

Question

Chapter 5, Problem 107AE

Interpretation Introduction

To determine:

The enthalpy change when 1 L of liquid NH3 is burned in air and its comparison with ΔH for complete combustion of 1 L of liquid methanol.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 5, Problem 106AE

Chapter 5, Problem 108AE

Students have asked these similar questions

Predict the organic products that form in the reaction below: OH H+ H+ + ☑ Y Note: You may assume you have an excess of either reactant if the reaction requires more than one of those molecules to form the products. In the drawing area below, draw the skeletal ("line") structures of the missing organic products X and Y. You may draw the structures in any arrangement that you like, so long as they aren't touching. Click and drag to start drawing a structure. ✓ m

Determine the structures of the missing organic molecules in the following reaction: + H₂O +H H+ Y Z ☑ ☑ Note: Molecules that share the same letter have the exact same structure. In the drawing area below, draw the skeletal ("line") structures of the missing organic molecules X, Y, and Z. You may draw the structures in any arrangement that you like, so long as they aren't touching. Molecule X shows up in multiple steps, but you only have to draw its structure once. Click and drag to start drawing a structure. AP +

Please help, this is all the calculations i got!!! I will rate!!!Approx mass of KMnO in vial: 3.464 4 Moss of beaker 3×~0. z Nax200: = 29.9219 Massof weacerv after remosimgain N2C2O4. Need to fill in all the missing blanks. ง ง Approx mass of KMnO4 in vials 3.464 Mass of beaker + 3x ~0-304: 29.9219 2~0.20 Miss of beaker + 2x- 29.7239 Mass of beaker + 1x~0.2g Naz (204 29-5249 Mass of beaver after removing as qa Na₂ C₂O T1 T2 T3 Final Buiet reading Initial butet reading (int)) Hass of NaOr used for Titration -reading (mL) calculation Results: 8.5ml 17mL 27.4mL Oml Om Oml T1 T2 T3 Moles of No CO Moles of KMO used LOF KM. O used Molenty of KMNO Averagem Of KMOWL

Chapter 5 Solutions

Chemistry: The Central Science, Books a la Carte Edition & Modified Mastering Chemistry with Pearson eText -- ValuePack Access Card Package

Ch. 5.2 - Prob. 5.1.1PE Ch. 5.2 - Prob. 5.1.2PE Ch. 5.3 - Prob. 5.2.1PE Ch. 5.3 - Prob. 5.2.2PE Ch. 5.3 - Prob. 5.3.1PE Ch. 5.3 - Prob. 5.3.2PE Ch. 5.4 - Prob. 5.4.1PE Ch. 5.4 - Prob. 5.4.2PE Ch. 5.5 - Prob. 5.5.1PE Ch. 5.5 - Prob. 5.5.2PE

Ch. 5.5 - Practice Exercise 1 Suppose you have equal masses...Ch. 5.5 - Prob. 5.6.2PE Ch. 5.5 - Prob. 5.7.1PE Ch. 5.5 - Prob. 5.7.2PE Ch. 5.6 - Prob. 5.8.1PE Ch. 5.6 - Prob. 5.8.2PE Ch. 5.6 - Prob. 5.9.1PE Ch. 5.6 - Prob. 5.9.2PE Ch. 5.7 - Prob. 5.10.1PE Ch. 5.7 - Prob. 5.10.2PE Ch. 5.7 - Prob. 5.11.1PE Ch. 5.7 - Prob. 5.11.2PE Ch. 5.7 - Prob. 5.12.1PE Ch. 5.7 - Practice Exercise 2 Use Table 5.3 to calculate the...Ch. 5.8 - Prob. 5.13.1PE Ch. 5.8 - Practice Exercise 2 Given the following standard...Ch. 5.8 - Prob. 5.14.1PE Ch. 5.8 - Prob. 5.14.2PE Ch. 5 - Prob. 1DE Ch. 5 - Prob. 1E Ch. 5 - Prob. 2E Ch. 5 - Prob. 3E Ch. 5 - Practice Exercise 2 Using Table 20.1, rank...Ch. 5 - Prob. 5E Ch. 5 - Prob. 6E Ch. 5 - Prob. 7E Ch. 5 - Prob. 8E Ch. 5 - Prob. 9E Ch. 5 - Prob. 10E Ch. 5 - Prob. 11E Ch. 5 - Prob. 12E Ch. 5 - Prob. 13E Ch. 5 - Prob. 14E Ch. 5 - Prob. 15E Ch. 5 - Prob. 16E Ch. 5 - Prob. 17E Ch. 5 - Prob. 18E Ch. 5 - Prob. 19E Ch. 5 - Prob. 20E Ch. 5 - Prob. 21E Ch. 5 - Prob. 22E Ch. 5 - Prob. 23E Ch. 5 - Prob. 24E Ch. 5 - Prob. 25E Ch. 5 - Prob. 26E Ch. 5 - Prob. 27E Ch. 5 - In chemical kinetics, the entropy of activation is...Ch. 5 - Prob. 29E Ch. 5 - Prob. 30E Ch. 5 - Prob. 31E Ch. 5 - The following data compare the standard enthalpies...Ch. 5 - Prob. 33E Ch. 5 - Prob. 34E Ch. 5 - Prob. 35E Ch. 5 - What is the reducing agent in the following...Ch. 5 - Prob. 37E Ch. 5 - Prob. 38E Ch. 5 - Prob. 39E Ch. 5 - Prob. 40E Ch. 5 - Prob. 41E Ch. 5 - Prob. 42E Ch. 5 - Prob. 43E Ch. 5 - The standard cell potential is 1.46 V for a...Ch. 5 - Prob. 45E Ch. 5 - Prob. 46E Ch. 5 - Prob. 47E Ch. 5 - Prob. 48E Ch. 5 - Prob. 49E Ch. 5 - Practice Exercise 1 Which of the following...Ch. 5 - Prob. 51E Ch. 5 - Prob. 52E Ch. 5 - Prob. 53E Ch. 5 - Prob. 54E Ch. 5 - Prob. 55E Ch. 5 - Prob. 56E Ch. 5 - Prob. 57E Ch. 5 - Prob. 58E Ch. 5 - Prob. 59E Ch. 5 - Prob. 60E Ch. 5 - What is the connection between Hess’s law and the...Ch. 5 - Prob. 62E Ch. 5 - 20.2 You may have heard that “antioxidants” are...Ch. 5 - Prob. 64E Ch. 5 - Prob. 65E Ch. 5 - Prob. 66E Ch. 5 - Prob. 67E Ch. 5 - Prob. 68E Ch. 5 - Prob. 69E Ch. 5 - Prob. 70E Ch. 5 - Prob. 71E Ch. 5 - Prob. 72E Ch. 5 - 20.13 What is meant by the term oxidation? On...Ch. 5 - Prob. 74E Ch. 5 - Prob. 75E Ch. 5 - Prob. 76E Ch. 5 - Prob. 77E Ch. 5 - Prob. 78E Ch. 5 - Prob. 79E Ch. 5 - Prob. 80E Ch. 5 - Prob. 81E Ch. 5 - Prob. 82E Ch. 5 - Prob. 83E Ch. 5 - Prob. 84E Ch. 5 - Prob. 85E Ch. 5 - The heat of combustion of ethanol, C2H5OH(l) is...Ch. 5 - Prob. 87E Ch. 5 - Prob. 88E Ch. 5 - Prob. 89E Ch. 5 - The automobile fuel called E85 consists of 85%...Ch. 5 - Prob. 91AE Ch. 5 - Prob. 92AE Ch. 5 - Prob. 93AE Ch. 5 - Prob. 94AE Ch. 5 - 5.95 Consider a system consisting of the following...Ch. 5 - A sample of gas is contained in a...Ch. 5 - Prob. 97AE Ch. 5 - Prob. 98AE Ch. 5 - A house is designed to have passive solar energy...Ch. 5 - Prob. 100AE Ch. 5 - Prob. 101AE Ch. 5 - Prob. 102AE Ch. 5 - Burning methane in oxygen can produce three...Ch. 5 - Prob. 104AE Ch. 5 - Prob. 105AE Ch. 5 - The hydrocarbons acetylene (C2H2) and benzene...Ch. 5 - Prob. 107AE Ch. 5 - Three common hydrocarbons that contain four...Ch. 5 - Prob. 109AE Ch. 5 - The Sun supplies about 1.0 kilowatt of energy for...Ch. 5 - It is estimated that the net amount of carbon...Ch. 5 - Prob. 112IE Ch. 5 - Prob. 113IE Ch. 5 - Prob. 114IE Ch. 5 - Prob. 115IE Ch. 5 - Prob. 116IE Ch. 5 - Prob. 117IE Ch. 5 - The methane molecule, CH4, has the geometry shown...Ch. 5 - Prob. 119IE

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

The enthalpy change when 1 L of liquid NH 3 is burned in air and its comparison with Δ H for complete combustion of 1 L of liquid methanol.

Concept explainers

Videos

To determine: The enthalpy change when 1 L of liquid NH3 is burned in air and its comparison with ΔH for complete combustion of 1 L of liquid methanol.

Want to see the full answer?

Chapter 5 Solutions

To determine:

The enthalpy change when 1 L of liquid NH3 is burned in air and its comparison with ΔH for complete combustion of 1 L of liquid methanol.