GENERAL CHEMISTRY 1
THIRD HOUR EXAM
NOVEMBER 26, 2008
Solutions are posted at the end of the exam.
Part 1 ______(xx points)
Part 2 ______(xx points)
Part 3 ______(xx points)
TOTAL ______(xxx points)
NA = 6.022 x 1023 °C = (5/9) (°F - 32) °F = (9/5)(°C) + 32
1 amu = 1.661 x 10-27 kg °C = K - 273.15 K = °C + 273.15
1 atm = 760 torr = 760 mm Hg DE = q + w pV = nRT pi = Xi ptotal
1 atm = 1.013 bar H = E + pV p = nRT/(V- nb) - an2/V2 vave = (3RT/M)1/2
R = 0.08206 L.atm/mol.K 1 J = 1 kg.m2/s2 DHrxn = DErxn + DngRT
R = 8.314 J/mol.K 1 L.atm = 101.3 J DH°rxn = [S DH°f(products)] - [S DH°f(reactants)]
h = 6.626 x 10-34 J.s Ephoton = hn = hc/l c = nl (Dx)(mDv) ³ (h/4p)
c = 2.998 x 108 m/s ldeBroglie = h/mv (1/l) = RH [ (1/m2) - (1/n2) ]
RH = 1.097 x 10-2 nm-1
Do all of the following problems. Show your work.
Part 1. Multiple choice. Circle the letter corresponding to the correct answer. There is one and only one correct answer per problem. [4 points each]
1) The orbital pictured below is
a) an s orbital
b) a p orbital
c) a d orbital
d) an f orbital
e) none of the above
2) Which of the following atoms is expectd to be paramagnetic?
a) Na (sodium)
b) Mg (magnesium)
c) S (sulfur)
d) both a and b
e) both a and c
3) How many valence electrons are there in one atom of arsenic (As)?
a) 2 valence electrons
b) 3 valence electrons
c) 5 valence electrons
d) 15 valence electrons
e) 18 valence electrons
4) Which of the following statements about covalent bonds between nitrogen atoms is correct?
a) The bond length for a nitrogen single bond is shorter than for a nitrogen double bond
b) The bond energy for a nitrogen single bond is smaller than for a nitrogen double bond
c) The bond energy for a nitrogen single bond is larger than for a nitrogen double bond
d) Both a and b
e) Both a and c
5) There are six electron containing regions around a particular covalently bonded atom. The electron geometry is
a) octahedral
b) square planar
c) tetrahedral
d) trigonal planar
e) trigonal pyramidal
6) Consider 1.00 moles of a real gas. This gas is expected to behave more and more like an ideal gas as
a) temperature approaches zero while pressure is held constant
b) pressure approaches zero while temperature is held constant
c) volume approaches zero while temperature is held constant
d) both b and c
e) both a and b and c
7) The average speed of a molecule of a particular pure ideal gas at some temperature Ti is vrms = 600. m/s. If the temperature of the gas is doubled to a final value Tf = 2 Ti, the average speed of a molecule of the gas will be
a) 2400. m/s
b) 1200. m/s
c) 850. m/s
d) 600. m/s
e) None of the above (to within 5 %)
Part 2. Short answer.
1) Give the electron configuration for the following atoms or ions [3 points each]
a) V ______
b) Co2+ ______
2) For each of the following questions circle the correct answer. There is one and only one correct answer per question? [3 points each]
a) Which of the following is the largest atom?
Li Na O S
b) Which of the following has the largest first ionization energy?
Li Na O S
c) Which of the following is the most metallic element?
Li Na O S
d) Which of the following is the largest ion?
Li+ Na+ O2- S2-
3) Give Lewis structures for the following molecules or ions. Include resonance structures when appropriate.
[5 points each]
a) HCFO
b) CH3CH2OCH3
c) NO3-
4) Give the electron configuration and the bond order for the N2+ cation. The appropriate molecular orbital energy diagram is given below. [5 points]
electron configuration ______bond order ______
s2p* ______
p2p* ______
s2p ______
p2p ______
s2s* ______
s2s ______
MO diagram for N2+
5) Charles' law is the statement that for a given amount of an ideal gas at a constant pressure V/T = constant. Starting with the ideal gas law (pV = nRT) derive Charles' law. [4 points]
Part 3. Problems.
1) Propane is a low molecular mass hydrocarbon whose reaction with oxygen is often used to produce heat. The equation for the reaction of propane with oxygen is
CH3CH2CH3(g) + 5 O2(g) ® 3 CO2(g) + 4 H2O(g)
Based on the table of average bond energies given below estimate the value for DH°rxn for the above reaction. [12 points]
Bond Energy Bond Energy Bond Energy Bond Energy
(kJ/mol) (kJ/mol) (kJ/mol) (kJ/mol)
C - C 347. O - O 142. C - H 414. C - O 360.
C = C 611. O = O 498. O - H 464. C = O 799.
2) The following questions concern the molecule whose Lewis structure is given below [3 points each]
a) What is the molecular geometry for the carbon atom labeled (1)? ______
b) What is the hybridization for the oxygen atom labeled (2)? ______
c) What is the formal charge for the carbon atom labeled (3)? ______
d) What is the value for the C - O - C bond angle? ______
e) What is the total number of sigma bonds present in the molecule? ______
3) The ideal gas law may be used as a way of determining the molecular mass of a gas. Consider the following experiment
The mass of a glass bulb is measured when empty (at vacuum) and found to be m = 78.231 g. The bulb is then filled with an unknown gas, and the mass of the filled bulb is measured and found to be m = 81.842 g. The pressure and temperature of the gas inside of the filled bulb were p = 774. torr and T = 18.8 °C. Finally, in a separate experiment, the volume inside the bulb is found to be V = 1.218 L. Based on that information find the molecular mass of the unknown gas. [12 points]
Solutions
Part 1 1) B 2) E 3) C 4) B 5) A 6) B 7) C
Part 2 1) a) V 1s2 2s2 2p6 3s2 3p6 4s2 3d3 or [Ar] 4s2 3d3
b) Co2+ 1s2 2s2 2p6 3s2 3p6 3d7 or [Ar] 3d7
2) a) Na b) O c) Na d) S2-
3) a) HCFO
b) CH3CH2OCH3
c) NO3-
4) electron configuration _(s2s)2 (s*2s)2 (p2p)4 (s2p)1 bond order 5/2
5) pV = nRT divide both sides by pT, to get
pV = nRT or V = nR
pT pT T p
If n and p are constant then everything on the right side of the above equation is constant, so V/T is constant. This is just Charles' law.
Part 3 1) Lewis structures, which are used to determine the number and types of bonds formed or broken, are given
DH°rxn @ [ S bond energies of reactants] - [ S bond energies of products]
= [ 8 (C-H) + 2 (C-C) + 5 (O=O) ] - [ 6 (C=O) + 8 (O-H) ]
= 8 (414) + 2 (347) + 5 (498) ] - [ 6 (799) + 8 (464) ] = - 2010. kJ/mol
2) a) trigonal planar b) sp3 c) 0 d) 109. ° e)8
3) m = 82.595 g - 78.231 g = 4.364 g
pV = nRT ; so n = pV p = 774. torr (1 atm/760 torr) = 1.0184 atm
RT T = (18.8 + 273.2) K = 292.0 K
n = (1.0184 atm) (1.218 L) = 0.05177 mol
(0.08206 L.atm/mol.K) (292.0 K)
M = m = 4.364 g 84.3 g/mol
n 0.05177 mol
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