Transcript Ch. 11 Powerpoint review with answers

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
 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?
 A)
Br2 = greatest molar mass
 B) HBr = polar molecule with greatest mass
 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)
 H2NNH2
= hydrogen bonding exists between
molecules of this type – not the others!
 In
which of the following substances is
significant hydrogen bonding possible:




Methylene chloride (CH2Cl2)
Phosphide (PH3)
Hydrogen peroxide (HOOH)
Acetone (CH3COCH3)
 HOOH
 List
the substances BaCl2, H2, CO, HF, and Ne
in order of increasing boiling points.
 H2 ,
Ne, CO, HF, BaCl2
 (A)Identify
the intermolecular forces present
in the following substances, and
 (B) select the substance with the highest
boiling point



CH3CH3
CH3OH
CH3CH2OH
 (A)
CH3CH3 = dispersion forces; other
substances have both dispersion forces and
hydrogen bonds
 (B) CH3CH2OH = hydrogen bond & molar mass
 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
 56.0
kJ
 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
 -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?


b. What phase change is the water going
through?


Energy is released from the water.
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
Boiling point
Heat of fusion
Heat of vaporization
115°
445°
54 J/g
1406 J/g
Answer 13
 1640.2
J Released energy.