CHAPTER 3
1.
a)
path 1 : (1 mole ideal gas, 3 bar, 450K) (1 mol, 3 bar, 250 K) (1 mol, 2 bar, 250 K)
path 2 : (1 mole ideal gas, 3 bar, 450K) (1 mol, 2 bar,450 K) (1 mol, 2 bar, 250 K)
path 1: the first step is isobaric, the second is isothermal; both rev
Practice exam on kinetics
CHM 3400, Dr. Chatfield, Fall 2011
I suggest that you study for the exam and give yourself 1 hours to take it. Making reference to
equations in the text is fine. I suggest you not allow yourself to scour the text for information
Kinetics of the mutarotation reaction of glucose
Erika Urgiles
Lab Partners:
Caron Spence
Abstract
Introduction
Optical isomers are a form of stereoisomers that possess the ability to rotate planepolarized light. A pair of optical isomers comprise the sam
CHAPTER 6
1.
a)
b)
1800 J s1m 2 3600 s hr 1 8 hr 1 m 2 0.018 kg mol1
43270 J mol1
= 21.6 kg
Vapor pressure of water at 40C is 7370 Pa
nRT [(21.6 kg) (0.018kg mol1 )](8.314 J K 1 mol1 )(313K)
=
p
(7370 Pa)
3
= 422.9 m
V=
c)
Vapor pressure of water at 20C i
Practice exam on kinetics
CHM 3400, Dr. Chatfield, Fall 2011
I suggest that you study for the exam and give yourself 1 hours to take it. Making reference to
equations in the text is fine. I suggest you not allow yourself to scour the text for information
CHAPTER 11
1.
a)
19
o (K) = / h = (2.2 eV)(1.6 10 J / eV) = 5.311014 s1
6.626 1034 J s
8 m s1
o (K) = c / o = 2.9981014 1 = 5.64 107 m = 564 nm
5.3110 s
o (Ni) = (Ni) o (K) = 5.0 o (K) =12.11014 s1
2.2
(K)
o (Ni) = (K) o (K) = 248 nm
(Ni)
b)
c)
2.
A w
The gas laws
Equations of state
The state of any sample of substance is specified by giving the values of the
following properties:
V the volume the sample occupies
p the pressure of the sample
T the temperature of the sample
n the amount of substance in
Solutions 10
1. The Nernst equation applies to half-cells as well as cells
Initially,
Ei = E - (RT/F)ln Qi = E - (25.7 mV / )ln Qi
Finally,
Ef = E - (25.7 mV / )ln Qf
The half reaction for the electrode is
2 H+(aq) + 2 e- H2(g)
E = 0
Q = pH2/[H+]2 = 1.45/
Solutions 6
1. The volume of the laboratory is
V = 5.05.03.0 = 75.0 m3
a) water: p = 24 Torr
n = pV/RT = (24 Torr) (75.0103 L) / cfw_(62.364 L Torr K-1 mol-1)
(298.15 K) = 96.855 mol
m = nM = (96.855 mol) (18.02 g mol-1) = 1745.3 g = 1.7453 kg
b) benzene
Homework 11 Solutions
1. We fit the initial rate data to the equation
log r0 = log k + a log[A]0
A = C6H12O6
_
log(r0/mol L-1 s-1)
0.69897
0.88081
1.19033
1.30103
log([A]0/mmol L-1) 0.00
0.18752
0.49415
060423
_
(a) The slope a = 0.9986 1. Therefore, the
Simple mixtures
Before dealing with chemical reactions, here we consider mixtures of substances that do not
react together. At this stage we deal mainly with binary mixtures (mixtures of two components,
A and B). We therefore often be able to simplify equ
Homework 5 Solutions
1. The substance with the lower molar Gibbs energy is the more stable; therefore, rhombic
sulfur is more stable. The application of pressure tends to favor the substance with the
smaller molar volume (higher density); therefore, rhomb
Homework 3 Solutions
1. We first determine the amount of molecules in the sample:
n = 65.0 g / 131.29 g mol-1 = 0.495 mol
Vi = nRTi/pi = (0.495 mol) (0.0820574 L atm K-1 mol-1) (298 K) / 2.00 atm =
6.053 L
(a)
For reversible adiabatic expansion,
piVi = pf
Physical transformations of pure substances
Boiling, freezing, and the conversion of graphite to diamond examples of phase
transitions changes of phase without change of chemical composition. In this chapter we
describe such processes thermodynamically, u
Homework 4 Solutions
1. U = U(0) + (3/2)nRT
U is independent of V, p
H = U + pV = U + nRT = U(0) + (5/2)nRT
(U/V)T = 0 and (U/p)T = 0
(H/V)T = 0 and (H/p)T = 0
2. The isothermal Joule-Thompson coefficient is
T = (H/p)T = -Cp = (-1.11 K atm-1) (37.11 J K-1
ngéo§¢12L5~ aggfgi/ap/L étab'/Lf
g/ZE:M i/ubH f /
W Z [7g 72
.1.__&. P02
,7/____
Ti = «W310? /< flz¥60 Tozz ZIVé/Ezz
/
1 ._ 313/4977 fsz 2 WW
.1 /1/
5131; /<' 410655-03 Jmog My my a) 3 £5: : A :
04 WA A339 ,1:
a ME ~ 1 0%
p.
{A 2 it A
Homework 1 Solutions
1. The external (atmospheric) pressure is greater than the internal pressure, hence
ex - in = gh
in = ex - gh = 758 Torr - gh
gh can be expressed in Torr as follows
gh = 3.55 cm H2O (1 cm Hg / 13.59 cm H2O) = 0.261 cm Hg = 2.61 mm Hg
Exam I
CHM 3410, Dr. Mebel, Fall 2005
1. (10 pts) The evolution of life requires the organization of a very large number of
molecules into biological cells. Does the formation of living organisms violate the
Second Law of thermodynamics? State your conclu
Solutions 7
1. For an ideal solution,
A - A* = RT lnxA = (8.3145 J K-1 mol-1)(353.25 K) ln(0.30) = -3.536 kJ mol-1
aA = pA / pA*
A xA = pA / pA*
pA = A xA pA*
At the normal boiling point, pA* = 1 atm = 760 Torr
pA = 0.93 0.30 760 Torr = 212 Torr
2.
mix S
CHAPTER 13
1.
a)
hc (6.626 10 3 4 Js)(3 108 m s1)
E=
=
= 2.92 10 1 9 J
7
6.8 10 m
19 1.602 1019 =1.82 eV
= 2.92 10
(an Einstein is a mole of photons so multiply by NA to find kJ Einstein-1)
= (2.92 1019 )(6.022 1020 ) =176 kJ mol1
b)
c)
2.
Einsteins = 485
CHAPTER 12
1.
a)
NO-
NO
NO
+
(2px)
(2px)
(2px)
*(2p)
(2p)
(2px)
(2s)
*(2p)
(2p)
(2px)
(2s)
*(2p)
(2p)
(2px)
(2s)
(2s)
(2s)
(2s)
(1s)
(1s)
(1s)
(1s)
(1s)
bond order = 2.5
bond order = 2
(1s)
bond order = 3
b)
Acetylene is H C C H . Recognize that this is a
CHM 3410 Problem Set 5
Due date: Wednesday, October 5th.
Do all of the following problems. Show your work. (NOTE: R = 0.082057 Latm/molK, and R = 8.3145 J/molK)
There is nothing to be learnt from a Professor which is not to be met with in Books. - David H
Questions
a mixture that is uniform in its properties throughout
CORRECT: homogeneous mixture
heterogeneous mixture
mixture
homogeneous; heterogeneous
a mixture that consists of physically distinct parts
element; compound
homogeneous mixture
CORRECT: hete
CHM 3410 Problem Set 4
Due date: Wednesday, September 21st. NOTE: The date of our first hour exam is Monday, September 26 th. The
exam will cover Chapters 1 (all), Chapter 2 (all), and Chapter 3 (sections A, B, and C) of Atkins, and handouts.
Do all of th