Chemistry 122 [Tyvoll]
Quiz 2 Key
September 8, 2008
Part I. Multiple Choice (2 points each) See answer section reverse side.
1. When the pressure over most liquids is decreased, the boiling point of the liquid will
Answer 1. always decrease. The temperature required to provide energy for escape from the
liquid surface is less because, at the b.p., P = P(external) and the external pressure is less.
2. Which statement is not correct? Vapor pressure is equilibrium vapor pressure.
Answer 1. A volatile liquid has a high boiling point. Volatile means that the liquid has a
relatively high vapor pressure (and a relatively low boiling point) at any given temperature,
just the opposite of the statement.
3. Which of the following would have the highest melting point?
Answer 1. MgO The ions, Mg2+ and O2−, have the greatest charges and since interionic
forces of attraction are proportional to the square of the charges, the interionic forces (and the
melting point) for MgO are the highest of any of the salts listed.
4. The reason that some insects can walk on water is due to
Answer 2. surface tension. (see pp. 490-491, especially the figure, Moore)
5. Consider a closed container containing a liquid and its vapor. Which statement is incorrect?
Answer 3. Evaporation and condensation cease after a constant pressure has been attained.
Not true …it seems like it from a macroscopic observer’s perspective but, at the nanoscale,
the processes continue, albeit at the same rate (the definition of equilibrium).
Part II. Problems (10 points) Read carefully! Show all work for credit!
1. (7 points) What is the total quantity of heat energy transfer (in kJ) required to change 5.00
mol of ice at −10 OC to 5.00 mol (90.12 g )of liquid water at 20 OC? H2O data below:
H2O (s) m.p. = 0.00 0C
c(s) = 2.06 J/g 0C
ΔH(fusion) = +6.020 kJ/mol
H2O (l) b.p. = 100.00 0C c(l)) = 4.184 J/g 0C
ΔH(vaporization) = +40.7 kJ/mol
(a) sketch the heating curve and label its axes; indicate what process is occurring for each
“leg” of the curve.
20
Heat water (q3 = mctfinal − tinitial)
Temp., 0C
Melt ice (q2 = molΔH
0
Heat ice (q1 = mctfinal − tinitial)
- 10
Time
1. (b) Perform the appropriate calculations, showing all intermediate steps. Given: q = mcΔT
q1 = (90.12 g)(2.06 J/g 0C)(0 − (−10)) 0C = +1.86 x 103 J = +1.86 kJ
q2 = (5.00 mol)(+6.020 kJ/mol) = +30.1 kJ
q3 = (90.12 g)(2.06 J/g 0C)(20 − 0) 0C = +7.54 x 103 J = +7.54 kJ
q(total) = +1.86 kJ + (+30.1 kJ) + (+7.54 kJ) = +39.5 kJ
2. (3 points) Clearly and specifically, explain why you would expect the boiling point of 1propanol, CH3CH2CH2-OH, to be higher than that of ethylmethyl ether, CH3CH2-O-CH3.
1-propanol (32 electrons) has dispersion, dipole-dipole and H-bonding intermolecular forces;
ethylmethyl ether (34 electrons) has only dispersion and dipole-dipole forces; since b.p. is
proportional to intermolecular attractive forces, 1-propanol should have a higher b.p.