Intermolecular forces review

1. Intermolecular forces review

2. A Quick Review Critical Temp & Pressure = THE POINT OF NO RETURN EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 105 torr Beyond 374°C = will forever be a gas Critical temp + pressure are the points at which we can still liquefy gas

3. Question 1 • Of Br2, Ne, HCl, HBr, and N2, which is likely to have (a) the largest london dispersion forces; (b) the largest dipole-dipole attractive forces?

4. Answer 1 • A) Br2 = greatest molar mass • B) HBr = polar molecule with greatest mass

5. Question 2 • In which of the following substances is hydrogen bonding likely to play an important role in determining physical properties: • Methane (CH4) • Hydrazine (H2NNH2) • Methyl fluoride (CH3F) • Hydrogen sulfide (H2S)

6. Answer 2 • H2NNH2 = hydrogen bonding exists between molecules of this type – not the others!

7. Question 3 • In which of the following substances is significant hydrogen bonding possible: • Methylene chloride (CH2Cl2) • Phosphide (PH3) • Hydrogen peroxide (HOOH) • Acetone (CH3COCH3)

8. Answer 3 • HOOH

9. Question 4 • List the substances BaCl2, H2, CO, HF, and Ne in order of increasing boiling points.

10. Answer 4 • H2, Ne, CO, HF, BaCl2

11. Question 5 • (A)Identify the intermolecular forces present in the following substances, and • (B) select the substance with the highest boiling point • CH3CH3 • CH3OH • CH3CH2OH

12. Answer 5 • (A) CH3CH3 = dispersion forces; other substances have both dispersion forces and hydrogen bonds • (B) CH3CH2OH = hydrogen bond & molar mass

13. Question 6 • Calculate the enthalpy change (in kJ) upon converting 1.00 mol of ice at -25 ° C to water vapor at 125 ° C under constant pressure of 1 atm. The specific heats are • Ice = 2.09 J/g-K • Water = 4.18 J/g-K • Steam = 1.84 J/g-K • Heat of fusion = 6.01 kJ/mol • Heat of vaporization = 40.67 kJ/mol

14. Answer 6 • 56.0 kJ

15. Question 7 • What is the enthalpy change (in kJ) during the process in which 100.0 g of water at 50 ° C is cooled to -30 ° C? • The specific heats are • Ice = 2.09 J/g-K • Water = 4.18 J/g-K • Steam = 1.84 J/g-K • Heat of fusion = 6.01 kJ/mol • Heat of vaporization = 40.67 kJ/mol

16. Answer 7 • -60.6 kJ

17. What is this substance’s normal melting point? QUESTION 8

18. ANSWER 8 • 60 ° C

19. At what temperature and pressure do all three phases coexist? Question 9

20. ANSWER 9 • 45˚ C

21. Question 10 In Denver, we live approximately 5,280 feet above sea level, which means the normal atmospheric pressure is less than 1 atm. In Denver, will water boil at a higher or lower temperature, than at 1atmosphere?

22. Answer 10 • Lower temperature. • At 1atm, water boils at 100˚C.

23. Question 11 Water is an unusual substance because the slope of the boundary between solid and liquid is negative. What happens to solid water at 0˚C if you increase the pressure?

24. Answer 11 • It becomes a liquid! • All other substances become a solid, but water behaves differently!

25. Question 12 • Water vapor condenses on the outside of a soda can. • a. Is energy being released or absorbed by the water? • b. What phase change is the water going through? • c. If you wanted to calculate the heat transferred, what formula would you use and why?

26. Answer 12 • Water vapor condenses on the outside of a soda can. • a. Is energy being released or absorbed by the water? • Energy is released from the water. • b. What phase change is the water going through? • Gas to Liquid. • c. If you wanted to calculate the heat transferred, what formula would you use and why? • Q = mΔHvap

27. Question 13 • How much energy in joules does 28.5g of liquid sulfur lose when it lowers from 120°C to 115°C,then change into a solid? The specific heat of liquid sulfur is 0.71 J/g°C. • Melting point 115° • Boiling point 445° • Heat of fusion 54 J/g • Heat of vaporization 1406 J/g

28. Answer 13 • 1640.2 J Released energy.