CHM 151 Dr. Hascall Spring 2012 Recitation Week 14 KEY
4/25/12
Unit conversions: 1 atm = 760 mmHg = 14.7 psi = 101325 Pa
PV = nRT; R = 0.0821
L • atm
mol• K
K = °C + 273.15
1. (a) How many grams of phosphorus would react with 35.5 L of O2 at STP?
!
4 P (s) + 5 O2 (g) → P4O10 (s)
(atomic masses: P 30.97, O 16.00; Volume of 1 mol ideal gas at STP = 22.4 L)
Moles O2 = 35.5 L / 22.4 mol / L = 1.585 mol (since it is at STP)
Moles P = 0.0446 mol × 4 mol P /5 mol O2 = 1.268 mol
Mass P = 1.268 mol × 30.97 g/mol
[ans 39.3 g]
(b) If 45 g of phosphorus and 83.0 L O2 at STP react, how many grams of P4O10
would form?
If all P reacted: 45 g P / 30.97 g/mol × 1 mol P4O10/4 mol P ×
283.9 g/mol = 103 g P4O10 would form.
If all O2 reacted: 83.0 L O2 / 22.4 L/mol × 1 mol P4O10/5 mol O2 ×
283.9 g/mol = 210 g P4O10 would form.
Thus P is the limiting reactant and the yield P4O10 would be 103 g
2. A flask contains 5.0 g He gas; a second flask contains 5.0 g Cl2. Both flasks
have the same volume and temperature. Which flask would have the greater:
(a)
pressure
(b)
(c)
average speed of gas molecules
average kinetic energy of gas molecules
(d)
rate of effusion
[ans: (a), (c) and (d): He; (b): neither]
3. In each case find the most important IMF in a sample of the molecules (note
you should draw the molecule’s Lewis structure and determine its shape first).
(a) CF4
(b) H2O
molecule is nonpolar ⇒dispersion
hydrogen bonding
(c) CH3Cl
(d) HF
molecule is polar ⇒dipole-dipole
hydrogen bonding
[ans 103 g] 3. (contd. (e) propane (CH3CH2CH3)
(f) CO2
molecule is nonpolar ⇒dispersion
dispersion
4. Rank each set of molecules by increasing strength of IMFs.
(a) CH4~BH3 < NH3
CH4 and BH3 are nonpolar with similar mass. NH3 has H-bonds.
(b) F2 < Cl2 < Br2
(c) CH3CH2CH3 < CH3CH2CH2CH3 < CH3CH2CH2CH2CH3
dispersion forces increase as mass of molecule increases
(d) CH3CH2CH3 (dispersion) < CH3OCH3 (dipole-dipole)
< CH3CH2OH (H-bonds)
5. In each case draw a hydrogen bond from the molecule shown to another
molecule of the same compound.
(a)
H
H
H
(b)
N
H
H
H
H
N
C
H
H
O
O
C
O
H
O
C
H
C
H
H
H
(the H-bond could be to the other O
on the molecule on the right also)
6. Describe how intermolecular forces affect these properties of a liquid:
(a) Boiling point
(b) Viscosity
(c) Vapor pressure
(d) Surface tension
Review questions for Exam 3:
7. Label each of these bonds as pure covalent, polar covalent or ionic
(a)
Cl–Cl
(b)
nonpolar
Zn–Cl
(c)
ionic
S–O
(d)
polar
H–F
polar
8. Give the hybridization of the central atoms in question 3, parts a, b, c and f.
(a) sp3
9.
(b) sp3
(c) sp3
(f) sp
OF2 (g) + H2O (g) → O2 (g) + 2 HF (g)
∆H = –318 kJ/mol
Use bond energies and the ∆H given for the above reaction to calculate the O–F
bond energy. [Bond energies (in kJ/mol): H–F 568; O=O 499; O–H 460)]
–318 kJ/mol = 2 BEO–F + 2 BEO–H – (BEO=O + 2 BEHF)
–318 kJ/mol = 2 × BEO–F + 2 × 460 kJ – (499 kJ + 2 × 568 kJ)
2 × BEO–F = 397 kJ/mol
[ans: +199 kJ/mol]