ProfDr. J Thison 9

NOTthe ahaveinforDo n

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s exam cons9 pages (21 Q

Parts

Page 3

Page 4

Page 5

Page 6

Page 7

Page 8

TOTAL

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Greenbowe ingston

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Grading Points

21 pts

18 pts

20 pts

18 pts

14 pts

12 pts

103 pts

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wers on the te

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edit on problThe final anuestions are able; the lastear away pag

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CHEM 177our Exam IItober 24, 20

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Sections 21,57 11,30,46 39,62 55 29,42 26,63 8,60 47,54 25,33 6,48 40 37,43 12,44 20,28 1,2 5,9 38,41 13,18 45 31,59 15,56 19 7,10 36,52 3,22 4,64 53,58 24 23,27

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Section____

Time 10, 12 2, 5, 3 11, 8 10 4, 1 3, 9 1, 5 4. 5 2, 9 11, 5 12 10, 1 3, 2 9, 4 8, 9 11, 1 11, 12 3, 5 2 8, 2 4, 11 8 12, 2 10, 3 10, 11 10, 12 4, 1 1 12, 3

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2Please read the following instructions carefully before proceeding! Part I &II of your exam will be computer graded. In order for the computer to identify who you are, it is important that you complete the information section properly.

You must use a #2 pencil and completely fill in the appropriate circles on the RED computer scan sheet. 1. 2.

3.

To help you code the correct circles, first write your last name, first name and middle initial in the boxes (skip a space between each). Then darken the circles that match the letters in the box above it. See the sample to the right. Write the middle nine digits of your ISU identification number in the boxes A-I. Do not skip any spaces. Below each number, darken the circle that matches this number. For example, 123456789. See the sample at bottom right.

Write your recitation section number in the special code area, boxes K-L. Do not skip any spaces. For example, if you are in section 8 of Chem 177, write 08. Again, darken the circle that matches the number above it. See the sample at bottom far right.

In Part I & II, select the one best answer for each question. Place your answer on the computer answer sheet by darkening the proper circle for that question. Your computer scan sheet will be your official answer sheet for Part I & II. All material (exam, answer sheet, scratch paper) must be returned to your TA in order for us to grade your exam. Exam #3: Chem 177, Fall 2011

PartYou

1. W

a)

2. W

a)

d

3. W

ca

a

b

c

d

e

4. Ca

5. W

6. W

7. W

t I: Multipleshould circl

Which one of

enthalpy

Which of the f

) H2O(s)

) H2O(l)

Which one o

arbonate, [H

a) Pb(s) + C

b) 2 Pb(s) +

c) Pb(s) + C

d) Pb(s) + C

e) Pb(s) + C

alculate the

For the re

the stand

a) 142 kJ

Which energy

a) n=1

Which of the f

a) 1s22s22p5

d) 1s22s22p6

Which elemen

1

a) Gold

e Choice: (3le the answe

f the followin

b)

following ph

H2O(l)

H2O(g)

of the follow

Hof (PbCO3)

C(s) + O3(g)

+ 2C(s) + 3 O

C(s) + 3/2 O2

CO(g) + O2(g

CO2(g) + O(g

standard ent

eaction, Pb(

dard enthalpy

y level can ho

b) n=

following el

3s13p4

63s33p4

nt has the fol

s22s22p63s23

b)

3 pts each). r on this she

ng is NOT a

work

hase change

wing reactio

]?

PbCO

O2(g) 2

2(g) Pb

g) PbC

g) PbC

thalpy of for

(s) + C(s) +

y of formatio

b) 142 kJ

old a maxim

=2

lectron confi

b)

e)

llowing grou

3p64s23d104p

Silver

The answereet for your o

a state functio

c) temper

is exotherm

b) H2O

e) None

on represents

3(s)

2 PbCO3(s)

bCO3(s)

CO3(s)

CO3(s)

rmation of oz

O3(g) P

on of PbCO3

J

mum of 18 el

c) n=3

iguration is n

1s22s22p63s2

1s12s22p63s2

und state ele

p65s14d10

c) Coba

r you fill in oown referenc

on?

rature

mic?

(s) H2O

e of the abov

s the equati

zone [Hof (

PbCO3(s) t

3 is 699.0 k

c) 770 kJ

lectrons?

not possible?

23p64s23d10

23p4

ctron config

alt

on your bubce.

d) energy

O(g)

ve

ion for stan

(O3)] from th

the enthalpy

kJ/mol.

d

d) n=4

?

c

guration show

d) Cadm

bble sheet is

y e

c) H2O(

dard enthalp

he following

y of reaction

d) 1540 kJ

e) non

c) 1s22s22p63

wn below?

mium

the one that

e) volume

(g) H2O

py of forma

g information

(Hrxn) is

e)

ne of the abo

3s23p4

e) Copper

3t will count.

O(l)

ation of lead

n:

841.0 kJ and

0 kJ

ove

r

3

d

d

8. W

a) [A

9. Co

a) 5,

10. W (

a) 3,

11. W

a)

b)

c)

12. W

a)

b)

c)

13) W

a)

Which of the f

Ar]3d8

orrect set of

0, 0, +1/2

Which one o(given in the

2, 2, 1/2

Which of the

4s

4s

4s[Ar]

[Ar]

[Ar]

Which electr

4s

4s

4s[Ar]

[Ar]

[Ar]

Which one o

) lithium

following is

b) [Kr

f four quantu

b) 5, 1

of the follow order n, l, m

b) 4, 0

e following o

ron orbital-fi

of the follow

b) sod

the condens

r]5s25d6

um numbers

1, 0, +1/2

wing represeml, and ms)

0, 0, 1/2

orbital-filling

3d

3d

3d

illing diagram

3d

3d

3d

wing element

dium

sed form of g

c) [Ar]4

for the valen

c) 5, 1,

ents an impo

c) 3, 3,

g diagram vi

m is correct

s will have t

c) beryl

ground state

4s24p6

nce (outer m

1, 1/2

ossible set of

3, 1/2

iolates Hund

d)

e)

for a ground

d)

e)

the smallest

llium

e electron con

d) [Kr]4

most) electron

d) 6, 0, 0

f quantum n

d) 5, 3, 0

d’s rule?

4s

4s[Ar]

[Ar]

d state Chrom

4s[Ar]

[Ar]4s

ionization e

d) potass

nfiguration f

s24p6

n of rubidium

0, +1/2

numbers for

0, 1/2

mium atom?

energy?

sium

for iron, Fe?

e) [Ar]4s2

m, Rb is ___

e) 5, 1, 1,

an electron

e) 5, 3, 2

3d

3d

?

3d

3d

e) cesium

4?

23d6

_______.

+1/2

in an atom?

2, 1/2

m

4

?

PartYou

14. an 1.0 jou

a)

15. W w

a

16. G

a)

17. F w

a)

18. C

[

a)

t II: Multipshould circl

As the reacnd the volu00 atm at coules? [1 L· a

+1940 J

What is the fwater at 20.0

a) 38.0 oC

Given the fo

2 H2(g)

Calculate

) 176 kJ

For the followould be rel

+ 960

Calculate the

[NA = 6.02 ×

) 1.66 × 104

ple Choice: le the answe

ction proceeume of the onstant tempatm = 101.3

b) +1

final temperoC. The spe

b) 50.

llowing data

) + ½ N2(g)

N2(g)

e the enthalp

NH3(g)

b) 22

owing reactieased or abs

b) +10

e energy of o

× 1023; h = 6.

48 J b)

(4 pts each)r on this she

eds in a cycylinder ch

perature of J]

590 J

ature in degrcific heat of

0 oC

a:

2 HCl(g)

+ ½ Cl2(g)

+ 3 H2(g)

py change fo

+ HCl(g)

22 kJ

on, 2NaCl(ssorbed when

000

one mole of

.63 × 1034 J

2.65 × 101

). The answeet for your o

ylinder equiphanges from398 K. Cal

c) 374

ree Celsius wf water is 4.1

c) 56.0

H2(g) +

NH4Cl

2 NH3(g

r the followi

NH4Cl(s

c) 406

s) + F2 (g) 3.0 mole of

c) +320

photon of re

J.s; c = 3.00

9 J c) 6

wer you fill inown referenc

pped with am 1.50 L tculate the c

4 J

when 60.0 m184 J/g·oC. [d

oC

+ Cl2(g)

l(s)

g)

ing reaction,

s) H =

6.8 kJ

2NaF (f Cl2 reacted

0

ed light with

0 × 108 m/s]

6.63 × 1034

n on your buce.

a moveableto 20.50 Lchange in to

d) 2280

mL of water density of w

d) 63.0 o

H = +1

H = 3

H = 9

,

= ? kJ

d) 591.

(s) + Cl2(g) with excess

d) 320

h a frequency

J d) 4.0

ubble sheet i

piston, 35. The press

otal energy

0 J

at 80.0 oC iswater = 1.00

oC

84.6 kJ

14.4 kJ

92.2 kJ

2 kJ

H = 32s of NaF to g

y of 4.00 × 1

00 × 1014 J

is the one tha

5 J of heatsure in the(E) for the

e) –37886

s mixed withg/mL]

e) 100.0 o

e) +92.2 k

20 kJ. How mgive NaCl an

e) 960

104 s1?

e) 1.60

5

at will count

t is released cylinder ie reaction in

6 J

h 40.0 mL o

C

kJ

much energynd F2?

0 × 105 J

5

t.

d s n

f

y

Partnum

19) (potaand 1.00

a) (3

b) (2

G

c) (1

(

d) (6

e) (2

e) (4

t-III: For fmber of signif

(18 pts) 85.ssium hydrothe final temg/ml and th

3 pts) Write

2 pts) Clear

Gaining Hea

pt) Is the pr

(i) exothermi

6 pts) Calcu

qrxn = q

2 pts) In the a

(0.102 m

4 pts) Calcu

Since we

qrxn corr

full credit, sficant figure

0 mL of 1.2oxide, KOH mperature ofhe specific he

a balanced c

HNO3

rly and speci

at: ________

rocess exoth

ic

ulate the hea

qsoln = ms

above reacti

mole of HNO

ulate the enth

e have only

responds to

Hrxn =0.052

The

show all yous. No work

20 M nitric at 23.1°C in

f the resultaeats of all so

chemical equ

3(aq) + KOH

ifically ident

__________

hermic or end

(ii) endo

t, qrxn, assoc

T = (120.0 g

= 601.9

on _______

O3 and 0.052

halpy change

0.0525 mol

0.0525 mole

-0.602 kJ

25 mol of ac

e solution

ur work legik shown = 0

acid, HNO3

n a calorimeant solution wolutions are 4

uation for th

H(aq)

tify what is g

______

dothermic? (

othermic

ciated with th

g) (4.18 J/g.

J = 0.602

__________

25 mole of K

e for this rea

of KOH, it

e of acid as

cid = 11.5

KOH

ibly, includepoints.

3 at 23.1°C weter with a swas 24.3°C.4.18 J/g°C.

his reaction. [

KNO3(aq)

gaining heat

Losi

(Circle one)

(iii)

his reaction,

oC) (24.3 oC

kJ

__ is the limi

KOH)

action, ∆Hrxn

will react w

well.

kJ/mol of a

e units, and

were allowesmall hole in. Assume th

[Remember

+ H2O(l)

and what is

ing Heat: __

neither endo

in units of k

C – 23.1 oC)

iting reagen

n, in units of

with only 0.0

acid

report your

ed to react wn the lid. That the dens

to show pha

losing heat.

__________

othermic nor

kJ.

nt. Justify yo

f kJ/mol of a

0525 mol of

The chemi

r answer to

with 35.0 mThe contents sities of all

ases]

.

___________

r exothermic

our answer.

acid.

f HNO3. So,

ical reaction

6the correct

L of 1.50 Mwere stirred

solutions are

____

c

the value o

n

6

M d e

f

7

20) A) (4 pts) Calculate Horxn for the following reaction of propane (C3H8) with chlorine gas to produce

carbon tetrachloride, CCl4 and HCl gas,

C3H8(g) + 10Cl2(g) 3CCl4(l) + 8 HCl(g)

Horxn = [(3 mol × 135.0 kJ/mol) + (8 mol × 92.0 kJ/mol)] – [(1 mol × 104.0 kJ/mol) + (10 mol× 0 kJ/mol)]

= [(405.0 kJ) + (736 kJ)] – [104.0 kJ]

= 1037 kJ = 1.04 × 103 kJ

B) (5 pts) Calculate the amount of heat released (enthalpy change) when 5.00 g of propane reacted completely with excess of chlorine gas. [If you don’t have answer for part-a, use Ho

rxn for the above reaction as 955 kJ]

5.00 g

1 mol C3H8

44.0 g C3H8

1.04 × 103 kJ

1 mol C3H8 = 118 kJ

C) (5 pts) Calculate the amount (in mL) of carbon tetrachloride, CCl4 produced if the reaction released only 346 kJ of heat energy. Density of CCl4 = 1.58 g/mL. [If you don’t have answer for part-a, use Ho

rxn for the above reaction as 955 kJ]

346 kJ

3 mol CCl4

1.04 × 103 kJ

153.0 g CCl4

1 mol CCl4

1 mL CCl4

1.58 g CCl4 = 96.6 mL of CCl4

Hof

(kJ/mol) Molar mass

(g/mol) C3H8 104.0 44.0 CCl4 135.0 153.8 HCl 92.0 36.45

21. A

i) ab

ii) em

iii) e

B) (5

C

D

FromE =

A) (2 pts) W

bsorption of l

mission of li

emission of l

5 pts) Calcul

E = –2.

E = –2.

= –3

C) (1 pt) Sta

D) (4 pts) Wcan obsn = 3 toSupport Circling credit!!

Red

Yello

Blu

m the energ

= h = hc/

When we watc

light when e

ight when ele

light when el

late the ener

.18 10–18 J

.18 10–18 J

.03 × 1022

ate whether e

Absor

What will beerve during

o n = 2 eneryour answan answer

d

ow

e

gy of the pho

OR

= (6.63

ch a firework

electrons cha

ectrons chan

lectrons chan

rgy (in kJ) o

J 1

nf2

1

ni2

J (0.250 – 0.

2 kJ

energy is abs

rbed

e the color og an electrrgy level? Cer with necr with NO

Orange

Green

Violet

oton that is

= hc/|E|

3 × 10-34 J.s)3.03 ×

ks display, th

ange from a h

nge from a lo

nge from a h

f a photon d

.111) = –2.1

sorbed or em

Emitte

of the lightron transitioCircle one. Ecessary calcwork will

released, w

)(3.00 × 108

10-19 J

he fireworks

higher to low

ower to high

higher to low

during an ele

E = –2.18

8 10–18 J (

mitted during

ed

that one on from Explain. culation. get NO

we can find t

m/s) 10-9n

1m

s we see are

wer energy l

her energy le

wer energy le

ectron transit

10–18 J 2

(0.139) = –3

g this electro

the wavelen

nm

m = 656 n

The colors

color  w

red  ~

orange  ~

yellow  ~

green  ~

blue  ~

violet  ~

a result of _

level.

evel.

evel.

tion from n=

2 2

1 1

2 3

.03 × 10–19 J

onic transitio

gth as show

nm

of the visib

wavelength r

~ 700–635 nm

~ 635–590 nm

~ 590–560 nm

~ 560–490 nm

~ 490–450 nm

~ 450–400 nm

1 TH

_______. (Ch

=3 to n=2 ene

J

1 kJ

1000 J

on?

wn below:

le light spec

range  frequ

m  ~ 430

m  ~ 480

m  ~ 510

m  ~ 540

m  ~ 610

m  ~ 670

Hz = 1012 Hz

8hoose one)

ergy level.

ctrum

uency range

0–480 THz

0–510 THz

0–540 THz

0–610 THz

0–670 THz

0–750 THz

z = 1012 s-1

8

e

9 q= m Cs T qsolution + qrxn = 0 qsolution = –qrxn Hrxn = q/mol of limiting reactant qv = –Cv T E = q + w w = –PV c = 3.00 108 m/s c = E = h h = 6.63 10–34 J·s

For hydrogen atoms: En = –2.18 10–18 J

1

n2 E = –2.18 10–18 J 1

nf2

1

ni2

RH = 1.096776 107 m–1 = h

mv x mv ≥ h

4 1 nm = 10–9 m

Pe riodic Table of t he Ele me nt s

1 0 3Lr

(2 6 0 )

1 0 2No

(2 5 9 )

1 0 1Md

(2 5 8 )

1 0 0Fm

(2 5 7 )

9 9Es

(2 5 2 )

9 8Cf

(2 5 1 )

9 7Bk

(2 4 7 )

9 6Cm

(2 4 7 )

9 5Am

(2 4 3 )

9 4Pu

(2 4 4 )

9 3Np

(2 3 7 )

9 2U

2 3 8

9 1Pa

2 3 1

9 0Th

2 3 2

7 1Lu

1 7 5

7 0Yb

1 7 3

6 9Tm1 6 9

6 8Er

1 6 7

6 7Ho

1 6 5

6 6Dy

1 6 2

6 5Tb

1 5 9

6 4Gd

1 5 7

6 3Eu

1 5 2

6 2Sm

1 5 0

6 1Pm

(1 4 5 )

6 0Nd

1 4 4

5 9Pr

1 4 1

5 8Ce

1 4 0

8 A1 8

7 A1 7

6 A1 6

5 A1 5

4 A1 4

3 A1 3

Lant hanide s

Act inide s

1 0 9Mt

(2 6 6 )

1 0 8Hs

(2 6 5 )

1 0 7Bh

(2 6 2 )

1 0 6Sg

(2 6 3 )

1 0 5Db

(2 6 2 )

1 0 4Rf

(2 6 1 )

8 9Ac

2 2 7

8 8Ra

2 2 6

8 7Fr

(2 2 3 )

8 3Bi

2 0 9

8 2Pb

2 0 7

8 1Tl

2 0 4

8 0Hg

2 0 1

7 9Au

1 9 7

7 8Pt

1 9 5

7 7Ir

1 9 2

7 6Os

1 9 0

7 5Re

1 8 6

7 4W

1 8 4

7 3Ta

1 8 1

7 2Hf

1 7 8

5 7La

1 3 9

5 6Ba

1 3 7

5 5Cs

1 3 3

5 1Sb

1 2 2

5 0Sn

1 1 9

4 9In

1 1 5

4 8Cd

1 1 2

4 7Ag

1 0 8

4 6Pd

1 0 6

4 5Rh

1 0 3

4 4Ru

1 0 1

4 3Tc

(9 8 )

4 2Mo

9 5 .9

4 1Nb

9 2 .9

4 0Zr

9 1 .2

3 9Y

8 8 .9

3 8Sr

8 7 .6

3 7Rb

8 5 .5

8 6Rn

(2 2 2 )

8 5At

(2 1 0 )

8 4Po

(2 0 9 )

5 2Te

1 2 8

5 3I

1 2 7

5 4Xe

1 3 1

3 6Kr

8 3 .8

3 5Br

7 9 .9

3 4Se

7 9 .0

3 3As

7 4 .9

3 2Ge

7 2 .6

3 1Ga

6 9 .7

3 0Zn

6 5 .4

2 9Cu

6 3 .5

2 8Ni

5 8 .7

2 7Co

5 8 .9

2 6Fe

5 5 .8

2 5Mn

5 4 .9

2 4Cr

5 2 .0

2 3V

5 0 .9

2 2Ti

4 7 .9

2 1Sc

4 5 .0

2 0Ca

4 0 .1

1 9K

3 9 .1

1 8Ar

3 9 .9

1 7Cl

3 5 .5

1 6S

3 2 .1

1 5P

3 1 .0

1 4Si

2 8 .1

1 3Al

2 7 .0

2He

4 .0 0

1 0Ne

2 0 .2

9F

1 9 .0

8O

1 6 .0

7N

1 4 .0

6C

1 2 .0

5B

1 0 .88 B

2 B1 2

1 B1 11 098

7 B7

6 B6

5 B5

4 B4

3 B3

1 2Mg

2 4 .3

1 1Na

2 3 .0

4Be

9 .0 1

3Li

6 .9 4

2 A2

1 A1

1H

1 .0 1

1 1 0Ds

(2 8 1 )