Intermolecular forces review

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

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?

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

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)

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

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

Answer 3 HOOH

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

Answer 4 H2, Ne, CO, HF, BaCl2

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

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

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

Answer 6 56.0 kJ

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

Answer 7 -60.6 kJ

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

ANSWER 8 60 ° C

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

ANSWER 9 45˚ C

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?

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

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?

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

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?

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

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

Answer 13 1640.2 J Released energy.