CHE-122 Written Assignment 1 PDF

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Name: College ID: Thomas Edison State University General Chemistry II with Labs (CHE-122) Section no.: OL009 Semester and year: JUN2018

Written Assignment 1: Intermolecular Forces and Liquids and Solids Answer all assigned questions and problems, and show all work. 1. Explain and give an example for each type of intermolecular force. (12 points) a. dipole-dipole interaction: Involves two polar molecules are aside one-another. A very common IMF due to the abundance of polar molecules available. An example is hydrogen bonding. b. dipole-induced dipole interaction: Weaker than ion-dipole forces but involve polar and nonpolar molecules. c. ion-dipole interaction: Occurs when an ion interacts with a polar molecule. When a sodium ion (Na+) and water (H2O) interact is an example. d. dispersion forces (London forces): The weakest of the intermolecular forces and is the force between two nonpolar molecules. An example is two methyl CH3 molecules. e. van der Waals forces: An interaction between uncharged atoms and molecules. An example would be many H2O molecules together. f. hydrogen bond: The same as a dipole-dipole interaction, but must include a Hydrogen. An example would be HF. (Reference: Chang 11.1)

2. The binary hydrogen compounds of the Group 4A elements and their boiling points are: CH4, –162ºC; SiH4, –112ºC; GeH4, –88ºC; and SnH4, –52ºC. Explain the increase in boiling points from CH4 to SnH4. (4 points) The strength of the dispersion force increases with the number of electrons in the molecule. (10 electrons, 18 electrons, 36 electrons, and 54 electrons respectively). The rise in boiling points increase with increasing dispersion forces as we see from the numbers provided. 1 Copyright © 2017 by Thomas Edison State University. All rights reserved.

(Reference: Chang 11.9)

3. Arrange the following in order of increasing boiling point: RbF, CO2, CH3OH, CH3Br. Explain your reasoning. (4 points)

CO2: CH3Br: CH3OH: RbF:

-109.2Fo 38.4Fo 148.5Fo 2,566Fo

Boiling points can be influenced by multiple factors. One factor obviously being the molecule or atoms involved. Another factor is the atmospheric pressure. The intermolecular forces help us to calculate what the boiling point will be of a particular compound, the stronger the attracting forces equates to higher boiling points. (Reference: Chang 11.13)

4. Which member of each of the following pairs of substances would you expect to have a higher boiling point? (6 points) a. O2 (-297F) and Cl2 (-29F): Since these two have London Forces and Cl2 has far more electrons than O2, it should be harder to boil i.e. higher boiling point. b. SO2 (14F) and CO2 (109F): Having a carbon in the carbon-dioxide molecule will make it more difficult to boil. c. HF (67F) and HI (-32F) The HF hydrogen bonding is so strong due to a smaller number of electrons to shield the bonds. This means while HF has fewer electrons in the same period, it is harder to boil than HI with significantly more electrons. (Reference: Chang 11.15)

5. Explain in terms of intermolecular forces why (a) NH3 has a higher boiling point than CH4 and (b) KCl has a higher melting point than I2. (4 points) a.) NH3 (-28F) utilizes hydrogen bonding which is much stronger than the London 2 Copyright © 2017 by Thomas Edison State University. All rights reserved.

forces that CH4 (-258F) uses. Since these bonds are significantly stronger, it is harder to break those bonds oh NH3 when heating up, therefore it’s boiling point is much higher. b.) KCl (2588F) uses ion-ion forces which are the strongest IMF. Meanwhile, I2 (364F) uses London forces which are the weakest IMF. This means that KCl will have a much higher boiling point than I2. (Reference: Chang 11.17)

6. What kind of attractive forces must be overcome in order to (a) melt ice, (b) boil molecular bromine, (c) melt solid iodine, and (d) dissociate F2 into F atoms? (4 points) a.) Overcoming dispersive forces and dipole-dipole interactions. b.) Overcoming dispersive forces due to being nonpolar. c.) Overcoming dispersive forces due to being nonpolar. d.) Overcoming covalent bonds holding together the F2. (Reference: Chang 11.18)

7. A glass can be filled slightly above the rim with water. Explain why the water does not overflow. (3 points) This can be explained due to surface tension. The water molecules are in essence sticking together, sort of like a magnet, and not allowing very small beads of water to overflow down the side of the glass. If you continued filling the glass with water though, it would overflow once that magnetic like force is overpowered by the weight of the water pushing over the rim. (Reference: Chang 11.25)

8. Outdoor water pipes have to be drained or insulated in winter in a cold climate. Why? (3 points) When water freezes, the molecules begin to string together and form a cyrstaline structure. This leaves a large amount of open space between them due to the arrangement of the hydrogens and oxygen. This is why water actually expands when 3 Copyright © 2017 by Thomas Edison State University. All rights reserved.

it freezes, and most other substances will shrink when frozen. When an amount of water is trapped in a pipe and subjected to cold enough temps to freeze, that expanding force is exerted on the pipe. Something has to give eventually, which may end up in rupturing a pipe or cracking a valve. It is wise to isolate external fittings in the fall before temperatures threaten below freezing temperatures. (Reference: Chang 11.30)

9. Explain and give an example for each type of crystal: (8 points) a. ionic crystal: i.e. NaCl. Form with alternating positively and negatively charged ions. b. covalent crystal: i.e. C. Very strong covalent bonds generate a lattice structure but only consist of single atoms and not molecules. c. molecular crystal: i.e. NH3. Held together with generally weak IMF and generate a lattice structure thus typically have low boiling points. d. metallic crystal: i.e. Au. Consist of metal cations and a free floating pool of electrons. This is why metals, such as gold, are such good conductors of electricity.

10. Classify the solid state of the following substances as ionic crystals, covalent crystals, molecular crystals, or metallic crystals. (a) CO2, (b) B12, (c) S8, (d) KBr, (e) Mg, (f) SiO2, (g) LiCl, (h) Cr. (8 points) a.) molecular crystals b.) covalent crystals c.) molecular crystals d.) ionic crystals e.) metallic crystals f.) covalent crystals g.) ionic crystals h.) metallic crystals (Reference: Chang 11.55)

11. What is an amorphous solid? How does it differ from a crystalline solid? (3 points)

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An amorphous solid consists of seemingly random arrangement of molecules. A crystalline solid has a repeatable and predictable manner of arrangement. (Reference: Chang 11.57)

12. What is a phase change? Construct a heating-cooling curve for a substance that freezes at –23.0 °C and has a boiling point of 67.0 °C. Name all possible changes that can occur among the vapor, liquid, and solid phases of this substance, and name the four phase change enthalpies involved. Indicate whether these changes are exothermic or endothermic. (6 points) Transitioning from one phase to another usually requires addition or subtraction of energy, likely in the form of heat, to restructure the physical molecular layout. Examples of exothermic phase changes are liquid to solid, gas to liquid or gas to solid. Examples of endothermic phase changes are solid to liquid, liquid to gas or solid to gas.

13. Why is solid carbon dioxide called dry ice? (3 points) Instead of melting like regular ice and leaving water behind, it sublimates straight form a solid to a gas. This leaves the contents of the cooler dry. (Reference: Chang 11.71)

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14. Which of the following phase transitions gives off more heat? (a) 1 mole of steam to 1 mole of water at 100ºC, or 1 mole of water to 1 mole of ice at 0.0ºC. (3 points) 1 mole of steam to 1 mole of water. (Reference: Chang 11.73)

15. Calculate the amount of heat (in kJ) required to convert 74.6 g of water to steam at 100ºC. (Specific heat of water = 4.184 J/g • ºC; ∆Hfus (H2O) = 6.01 kJ/mol; ∆Hvap (H2O) = 40.79 kJ/mol). (5 points) 74.6g / 18g/mol=4.14mol 4.14mol * 40.79kJ/mol= 168.7kJ (Reference: Chang 11.75)

16. How is the rate of evaporation of a liquid affected by (a) temperature, (b) the surface area of a liquid exposed to air, (c) intermolecular forces? (3 points) a.) More temperature means more energy to encourage the evaporation to occur faster. b.) The greater the surface area means the more area for the evaporation to occur in, resulting in faster evaporation. c.) The lower the IMF, the easier it is to evaporate. (Reference: Chang 11.77)

17. The following compounds, listed with their boiling points, are liquid at –10ºC: butane, –0.5ºC; ethanol, 78.3ºC; toluene, 110.6ºC. At –10ºC, which of these liquids would you expect to have the highest vapor pressure? Which the lowest? Explain. (4 points) Boiling points lower as vapor pressure rises. Butane would have the highest vapor pressure, then ethanol and finally toluene would have the lowest vapor pressure. (Reference: Chang 11.79) 6 Copyright © 2017 by Thomas Edison State University. All rights reserved.

18. A student hangs wet clothes outdoors on a winter day when the temperature is –15ºC. After a few hours, the clothes are found to be fairly dry. Describe the phase changes in this drying process. (2 points) The clothes may feel dry, but this is likely due to the water on the clothes being in a solid form. The freezing point of water is 0C. There could be some evaporation of water occurring at that temperature but it would likely be a very slow process. (Reference: Chang 11.81)

19. Determine the final state and its temperature when 150.0 kJ of heat are added to 50.0 g of water at 20 ºC. (Specific heat of water = 4.184 J/g ºC; Specific heat of steam = 1.99 J/g • ºC; ∆Hfus (H2O) = 6.01 kJ/mol; ∆Hvap (H2O) = 40.79 kJ/mol). (5 points) 50g * 80 * 4.184J/g / 1000= 16.7kJ to boil 50g of water from 20 degrees C. 50g / 18g/mol= 2.8mol 150kJ – 16.7kJ=133kJ remaining energy to calculate. 2.8mol * 40.8kJ/mol= 113.4kJ 133kJ - 113.4kJ= 19.6kJ or 19600J remaining energy to calculate. 19600J= (50T – 5000) * 1.99J/g 19600 / 1.99= 9849 9849= 50T – 5000 9849+5000=14849 14849=50T 297 * 50= approx. 14849 Final temperature is approximately 297 degrees celcius.

(Reference: Chang 11.118)

20. Referring to the phase diagram of CO2 below, determine the stable phase of CO2 at (a) 4 atm and –60ºC; and (b) 0.5 atm and –20ºC (2 points) a.) appears to be a Solid but is very close to the line for Vapor. It is challenging to tell from this graph. b.) Vapor 7 Copyright © 2017 by Thomas Edison State University. All rights reserved.

21. If water were a linear molecule, (a) would it still be polar, and (b) would the water molecules still be able to form hydrogen bonds with one another? (4 points) a.) No, it would not be polar. b.) No, the hydrogen bonds would not be able to bond with one another. (Reference: Chang 11.121)

22. A pressure cooker is a sealed container that allows steam to escape when it exceeds a predetermined pressure. How does this device reduce the time needed for cooking? (4 points) The water can be super heated because the pressure is stopping the water from converting to steam. This means the food can be cooked much faster and heat is more efficiently transferred to the food, also resulting in faster cooking. (Reference: Chang 11.131)

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