suggested soln to foundation physical chemistry

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Reg. No. 200710217Z

Practice Paper 7 Advo Education Group Pte Ltd

Suggested Solution to FoundationPhysical Chemistry

Stoichiometry

(a) Mole Calculation

1. Element C H Omass/g 54.54 9.09 36.37

amount/mol12.0

54.54= 4.545

1.0

9.09= 9.09

16.0

36.37

= 2.273

Simplest moleratio 2.273

4.545= 2

2.273

9.09

= 42.273

2.273

= 1

Empirical formula of W is C2H4O.

Let te molecular formula !e "2n#4n$n.M r = 2n%12.0& ' 4n%1.0& ' n%16.0& = ((.0

44 n = ((

n = 2)olecular formula of nicotine is C4H8O2.

2. (a)Ratio of te average mass of one atom of an element to

1

12of the mass of a 12 C atom.

$*

Ratio of te average mass of 1 mole of atoms of an element to 1

12of the mass of 1

mole of 12 C atoms.

(b) Ar of +! = %1.37

100, 204.0072& ' %

26.3

100, 206.0612& ' %

20.(

100, 207.0552& ' %

51.53

100 , 20(.0602&

= 2!."

". # %) - mount of "l 2 molecules in 35.5 g =35.5

2%35.5& = .$ mol

% %& - mount of #2 molecules in 3.01,1023 rogen molecules =×

×

23

23

3.01 10

6.02 10 = 0.500 mol

#2 ≡ 2e− %since eac #2 molecule contains 2 electrons i.e. #••#& mount of electrons in 3.01,1023 rogen molecules = 2 & 0.500= 1. mol

C %& - mount of #2S$4 in 1 m3 of 1 mol m 3 #2S$4%a& = 1 , 1 = 1.00 mol#2S$4 2#'

mount of #

'

ions in 1 m

3

of 1 mol m

3

#2S$4%a& = 2 & 1.00 = 2. mol' %& - mount of $2 gas molecules in 22.4 m3 =

22.4

22.4=1.00 mol

$2 2$ mount of ogen atoms %i.e. $& = 2 & 1.00 = 2. mol

#nser %

4. (a) )ass of ater present in 2.44 g of rate a"l 2. x#2$ = 2.44 2.0( = 0.36 g

A is Easy with ADVO!

Tel: (65) 6251 3359 / 8233 2753www.advoedu.com 1 Goldhill Plaza #02-43 Singapore 308899

Note: Molecular formulashould NOT ben(C2H4O and ans!ershould NOT be

*ote- Ar 8 no +nits to 1 ,.-. as re+ire, by the n

*ote1 mol of an gas occupies22.4 ,m" at s.t.p.

http://www.advoedu.com/





Reg. No. 200710217Z


: of ater of ration =0.36

100:2.44

× = 14.8/ 01m

(b) mount of anrous a"l 2 =0.208

08.2 = 0.0100 mol

mount of #2$ =

0.36

1(.0 = 0.0200 mol

Since a"l 2. x#2$ ≡ a"l 2 ≡ 2#2$"orrect formula of rate !arium clorie- %aC l 2.2H2O

OR

mount of anrous a"l 2 =0.208

08.2 = 0.0100 mol

Since a"l 2. x#2$ ≡ a"l 2 mount of rate a"l 2. x#2$ = 0.0100

M r of a"l 2. x#2$ =0.0100

2.44= 244

137.0 ' 2%35.5& ' x %2.0 ' 16.0& = 244 x = 2 01m

"orrect formula of rate !arium clorie- %aC l 2.2H2O

$. (a) O2%g& 2H2%g& → 2H2O%l & 01m

(b) ;nitial <olume / cm3 15.0 55.0

inal <olume / cm3

(i) inal <olume of reaction miture = >olume of unreacte rogen

= 2$. cm" 01m

(ii) )a. ecrease in <olume = ?otal initial <olume ?otal final <olume

= %15.0 ' 55.0& 25.0 = 4$. cm" 01m

)"@

1. C 2. ' ". # 4. # $. ' 3. C !. ' 8. C . C

(b) Volumetric Analysis



$2 is te limitingreagent

55.0 2%15.0&= 25.0

or gasesmole ratio = <ol. ratio

01m for riting correct formula

01m

01m for riting correct formula

Aater is a liui atr.t.p.

>olume of #2$%l & isnegligi!le.

nrous a"l 2 #2$ mount / mol 0.0100 0.0200

*atio 1 2






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1. (a) (*H4)2SO4 2*aOH → 2*H" *a2SO4 2H2O*aOH HCl → *aCl H2O

(b) mount of Ba$# ae = 0.(001000

250× = .2 mol

(c) Since Ba$# ≡ #"l amount of Ba$# tat reacte it #"l = 0.5001000

(5.0× = .42$ mol

(,) mount of Ba$# tat reacte it %B#4&2S$4 = 0.200 0.0425 = .1$8 mol(e) Since %B#4&2S$4 ≡ 2 Ba$#

)ass of %B#4&2S$4 = ×2

0.15( C2%14.0'4.0& ' 32.1 ' 4%16.0&D = 1.4 g

(c) Redox

1. (a) Cr 2O!25 36e2 14H → 2Cr " 36e" !H2O

(b) Since 6e2' ≡ "r 2$72

mount of e2' = 6 × 1.50 1084 = 9.00 10 4 mol

Since e2'

≡ eS$4.7#2$)ass of eS$4.7#2$ = 9.00 10 4 × %55.( ' 32.1' 4 16.0 ' 7 1(.0&= .2$ g

2. (a) $6e2 7nO4 5 8H → $6e" 7n2 4H2O

(b) mount of eS$4%B#4&2S$4.6#2$ use =9.(5

392.0= 0.0251 mol

Since eS$4%B#4&2S$4.6#2$ ≡ e2' mount of e2' in 250 cm3 = 0.0251 mol

mount of e2' in 25.0 cm3 = 0.0251 × 25.0

250= 2.$1 1 5 " mol

(c) Since 5e2'

≡ )n$4

≡ F)n$4 mount of F)n$4 reacte = 2.51 × 10 3

5

1× = $." 1 5 4 mol

(,) CF)n$4D = 5.03 × 10 4 24.75

1000÷ = .2" mol ,m 5"

CF)n$4D = 0.0203 × %39.1 ' 54.9 ' 4%16.0&& = ".21 g ,m 5"

(e) mount of F)n$4 in 1.( m3 = 0.0203 1.( = 0.0365 mol>olume of te solution to maGe 0.0200 mol m 3 = 0.0365 H 0.0200= 1.(4 m3

>olume of ater reuire = 1.(4 1.( = .4 ,m"

". mount of metallic salt reacte = ×50.0

0.1001000

= 5.00 , 10 3 mol








Reg. No. 200710217Z


mount of S$32− reacte = ×

25.00.100

1000= 2.50 , 10 3 mol

7etallic salt SO"2

#mo+nt 9 mol 5.00 , 10 3 2.50 , 10 3

7ole Ratio 2 1i.e. 2 )etallic salt ≡ S$3

2−

Since S$32− ≡ 2e %[O]: SO3

2– + H 2 O → SO42– + 2H + + 2e –&

∴ 2 )etallic salt ≡ S$32− ≡ 2e

⇒ 7etallic salt≡

e 5

1 mole of metallic salt is re,+ce, ! gaining 1 mole of electrons.

?erefore oiation num!er of metal in metallic salt ill ecrease from " to 2.

∴?e ne oiation num!er of metal is 2.

4. mount of :2$5 reacte = ×25.0

0.1001000

= 2.50 , 10 3 mol :2O$≡ 2:5

mount of :5' reacte = 2 , 2.50 , 10−3 = 5.00 10−3 mol

Let :n' !e te reuce form of :.

#mo+nt of :n forme, (after re,+ction of :2O$ by SO2) ; amo+nt of :5 reacte,

= 5.00 10−3 mol

mount of F)n$4 reacte it :n'= × 50.00.0200 1000 = 1.00 , 10 3 mol

:n <7nO4

#mo+nt 9 mol 5.00 , 10−3 1.00 , 10−3

7ole Ratio $ 1

i.e. )n$4 ≡ $In'

Since )n$4 ≡ $e %C*D- )n$4

' 5e → )n2'&

7nO4 5≡

$e 5≡

$:n

⇒ :n'

≡ e

Juring oi,ation of :n' 1 mole of :n ill lose 1 mole of e 5 to form :5'.

$iation no. of : ecreases from '5 to 4.$. (a) ?otal amount of Ba$# in 25.0 cm3 of solution H

= ×22.50

0.1001000

= 2.25 × 10 3 mol

?otal amount of Ba$# in 100 cm3 of solution H



(!) (2)






Reg. No. 200710217Z


= .

.

3 1002 25 10

25 0

−× × = . × 1 5" mol 01m

(b) mount of F)n$4 use =1000

10.00.0200 × = 2. 1 54 mol 01m

Since 27nO4

5

= $H2O2

mount of #2$2 in 25.0 cm3 of solution H

= −× × 42.00 10$

2 = 5.00 10 4 mol

mount of #2$2 in 100 cm3 of solution H

= −× × 41005.00 10

25.0 = 2. 1 5" mol 01m

(c) Since *a2O2 H2O2

mount of Ba2$2 in te original sample= 2. 1 5" mol 01m

mount of Ba$# forme from reaction of Ba2$2 = −× × 32.00 102 = 4.00 10 3 mol

mount of Ba$# forme ! !urning of Ba2$

= ?otal amount of Ba$# forme mount of Ba$# forme fromn reaction of Ba2$2

= 9.00 × 10 3 4.00 10 3 = $. 1 5" mol 01m

Since *a2O 2*aOH

mount of Ba2$ = −× × 35.00 101

2 = 2.$ 1 5" mol 01m

)"@

1. ' 2. # ". # 4. ' $. C 3. C

#tomic Str+ct+re

1. > −31 3

15 - 15 p 16 n 1( e ─ W −32 2

15 - 15 p 17 n 17 e ─

: −32 2

16 - 16 p 16 n 1( e ─ ?−31

15 - 15 p 16 n 16 e ─








Reg. No. 200710217Z


@−35

17 - 17 p 1( n 1( e ─

(a) (i) > −31 3

15 : −32 2

16 an @−35

17 are isoelectronic.

(ii) > −31 3

15 : −32 2

16 an ?−31

15 are isotonic.

(iii) 31 3

15

−V an 32 2

15

−W or 32 2

15

−W an 31

15

−Y are isotopic.

CBote- 31 3

15

−V an 31

15

−Y are not isotopes as te

a<e te same no. of protons an neutrons. ?eare ions of te same isotope it ifferent carges.D

(b) 31 3

15

−V %ue to its igest carge an smallest mass&

(c) @−35

17 %ue to its loest carge an igest mass&

2. ;on −31 2

15+ as %a& 1$ -rotons %!& 13 ne+trons %c& 1! electrons

Electronic configuration −31 2

15+ of is 1s2 2s2 2p6 3s2 3p5

3p(,) " energy levels %sells& containing electrons (n ; 1A n ; 2A n ; ")(e) $ energy s+bBlevels %su!sells& containing electrons (1sA 2sA 2-A "sA "-)(f) orbitals containing electrons (one 1sA one 2sA three 2-A one "sA three "-)(g) 8 com-letely fille, orbitals (one 1sA one 2sA three 2-A one "sA to "-)(h) 1 -artially fille, orbital %one 3p&(i) 1 +n-aire, electron %te one unpaire electron in te 3p or!ital&() ! valence electrons %total no.of e ─ from outermost uantum sell = 2 ' 5&

". (a) 7B atom- 1s2 2s2 2-"

1s 2s 2-

(b) Arite elect. config. of 20"a atom- 1s2 2s2 2p6 3s2 3p6 4s2 first ten remo<e to e ─ fromoutermost 4s or!.&

20"a2' ion- 1s2 2s2 2-3 "s2 "-3

1s 2s 2- "s "-

(c) Arite elect. config. of 16S atom- 1s2 2s2 2p6 3s2 3p4 first ten a 2 e ─ to outermost sell.

16S2

ion- 1s2

2s2

2-3

"s2

"-3

1s 2s 2- "s "-

(,) 22?i atom- 1s2 2s2 2-3 "s2 "-3 ",2 4s2

1s 2s 2- "s "- ", 4s



Recall: Extent of deflection αcharge

Recall: Isotopes are atomsof the same element with

the same no. of protons but

(Hund's rule: fill orbitals singly first before

pairing)

ote: !oxes

("ill #s orbital before filling the $d






Reg. No. 200710217Z


(e) 29"u atom- 1s2 2s2 2-3 "s2 "-3 ",1 4s1

1s 2s 2- "s "- ", 4s

4. (f) 24"r 2' ion- 1s2 2s2 2-3 "s2 "-3 ",4

1s 2s 2- "s "- ",

(g) 27"o3' ion- 1s2 2s2 2-3 "s2 "-3 ",3

1s 2s 2- "s "- ",

(h) 32Ke atom- 1s2 2s2 2-3 "s2 "-3 ",1 4s2 4-2

s 2s 2- "s "- ", 4s 4-

4. (a) *a as a larger (atomic) ra,i+s tan Li. ?is is !ecause te effects of one more -rinci-al+ant+m shell an, greater shiel,ing effect by greater n+mber of inner shell electronso+teighs the effect of a larger n+clear charge in *a as compare to Li.

(b) 7g2 as a larger (ionic) ra,i+s tan l 3'. ?is is !ecause 7g2 as a larger n+clear chargetan l 3' ile !ot a<e te same n+mber of -rinci-al +ant+m shell an, the shiel,ingeffect by inner shell electrons are relatively constant.

$. (a) Kroup D> Note: Group number must be state in roman numeral.Si9Silicon

(b) 1s2 2s2 2-3 "s2 "-2

3. (a) (i) *elati<e atomic mass is te eighte, average of te mass of te isotopes taGinginto account te relati<e a!unances of te isotopes.

(ii) ?e <alue gi<en in te +erioic ?a!le escri!es te naturall8occurring "u iccontains more tan 2 isotopes. C1mD$r naturall8occurring "u as greater a!unance of 63 "u tan tat in te sample.



C1mD






Reg. No. 200710217Z


(b) (i)

(ii) "arge species +63 2

29"u I−127

53

carge

mass

2

63

1

127

ngle of eflection 7.0oo o( . ) .

2 17 0 1 7

63 127÷ × ≈

!. (a) ?e !irst ionisation ener"# of an element is te energy re+ire, to remove one mole ofelectrons from one mole of gaseo+s atoms. C1mD

(b)

(c) (i) Ba- 1s2 2s2 2p6 "s1 F- 1s2 2s2 2p6 3s2 3p6 4s1

Ba as %one& less n +mber of -rinci-al +ant+m shell of electrons tan F resulting ina smaller atomic ra,i+s9,istance from its n+cle+s to the o+termost electron C1mDan smaller shiel,ing effect by feer inner shell electrons C1mD.

#ence 1st IE of Ba %espite its smaller nuclear carge& is iger tan tat of F.

!.(c) (ii) )g- 1s2 2s2 2p6 "s2 l - 1s2 2s2 2p6 3s2 "-1

"ompare to te 3s electron of )g te "- electron to be remove, from #l is lessable to -enetrate thro+gh the inner shells to a--roach the n+cle+s closely . ?uste 3p electron of l as higher energy than the "s electron of 7g C1mD.



(#cross -erio, 2A"A" -attern)

ote:

%eneral increase across period

&st IE (l ) &st IE (g) and

&st IE (*i) + &st IE (g)

&st IE (*) &st IE (,) and

&st IE (-l ) + &st IE (,)

&st IE () &st IE (a)

s1

s1

s2

s2p1

s2p2

s2p3 s2p4

s2p5

s2p6

11 12 13 14 15 16 17 1( 19proton no.

Note:

• /eflection should begin after the charged

particles enter the electric field0 and

• particles should continue on a straight line path

after lea1ing the field.

C1mD -u23 deflected towards 41e plate

Extent of deflectn: larger than that for I

C1mD I deflected towards 31e plate

Extent of deflectn: smaller than that for -u23

C1mDNote: "irst IE of ,eriod 2

and $ elements follow the

same pattern of 20$0$.

Suggestion: "or students

of high readiness0 tutor

can get them to s5etch 2nd

IE of ,eriod 2 elements.

C1mD

1st ionisationenerg/GM mol 1

neutron

+

I "u2'

source






Reg. No. 200710217Z


Less energ is reuire to remo<e te 3p electron from l an ence 1st IE of )g isiger tan tat of l .

(iii) Si- 1s2 2s2 2p6 3s2 "-2 +- 1s2 2s2 2p6 3s2 "-"

+ as a larger n+clear charge C1mD tan Si an te atomic ra,i+s9,istance from itsn+cle+s to the o+termost electron is smaller C1mD tan tat for Si. Sieling effect! inner sell electrons for !ot Si an + are relati<el constant since te a<e tesame num!er of inner sells.

#ence 1st IE of + is iger tan tat of Si.

(iv) +- 1s2 2s2 2p6 3s2 "-" S- 1s2 2s2 2p6 3s2 "-4

7+t+al re-+lsion beteen the -aire, "- electrons in S maGes te removal of oneof the -aire, electrons easier compare to remo<ing te unpaire 3p electron of +C1mD ic oes not eperience suc repulsion.

8. (a) (i) Shar- increase from $th to 3th IE impling 5 <alence electrons. C1mD

F is in group >. C1mD

(ii) *o. C1mD Kroup > element in +erio 2 can onl a<e 7 electrons !ut element F hasmore than ! electrons. C1mD

(b) (i) 1st I.E- F(g) → F (g) e C1mD

2n I.E- F (g) → F2 (g) e $s.s re%&' C1mD

(ii) s successi<e electrons are remo<e more energy is re+ire, to remove an electronfrom an increasingly more -ositive ion. C1mD

(c) ot te 4th an, $th electrons are remo<e from te same -rinci-al +ant+m shell C1mDile te 3th electron remo<e is from an inner +ant+m shell hich is m+ch closer to then+cle+s. C1mD

#ence more energ is reuire to remo<e an electron from te inner sell.

(,) F coul !e -hos-hor+s C1mD N

1s2 2s2 2-3 "s2 "-" %or an Kroup > element oter tan nitrogen& C1mD

. (a) "ompare to e atom 6e2 an, 6e" ions have one less -rinci-al +ant+m shellof electrons after remo<ing 4s electrons from e atom.?erefore their ionic ra,ii are smaller tan te atomic raius of e.

"ompare to S atom S25 ion has to more electrons in its o+ter shell. Jue to greater



%roup no. is in Roman numeral while

,eriod no. is in rabic numeral.

C1mD






Reg. No. 200710217Z


electronBelectron re-+lsion te electron clo+, e-an,s to gi<e a larger S25 ion. C1mD

(b)

Aile !ot e2' an e3' ions a<e te same num!er of principal uantum sells 6e"

has one less electron than 6e2. ?us tere is re,+ce, electronBelectron re-+lsion in6e" to gi<e a smaller electron clo+,. C1mD

7CF1. KaB is an ionic compoun. Since Ka is in te

form of Ka3' in KaB ten B must !e in te formof B3

7B - 1s2 2s2 2p3

7B3- 1s2 2s2 2p6 ns- '

2. ;entit of element is etermine ! atomic no.

244

94 +u ' : → 289114 Cne elementD ' 3 neutrons % 10n&

Bo. of protons in I 114 94 = 20

?e element it atomic num!er 20 is "a. ns- %

". # $ an Ie a<e ifferent atomic raii as

te a<e ifferent nos. of p..s.

% (le)tron a!!init# refers to te energreuire en one mole of gaseousatoms gain one mole of electrons.Electron affinit oes not appl in tecontet ere as !ot $2 an Ie lose e#

to form cations B$? gain e#.

C $ as ( protons Ie as 54 protons⇒ ifferent electron configurations.

' ot $2 an Ie lose an electron toform cations $2

' an Ie' respecti<el.

?us in orer to react in a similarfasion te must a<e similar first IE.

ns- '

4.

must a<e a loer carge8to8mass ratio tanBa' %to gi<e smaller eflection&

or Ba' carge/mass ratio = 1/23 = 0.0435or a2' carge/mass ratio = 2/137 = 0.0146or e2' carge/mass ratio = 2/9.0 = 0.222?us options %#& an %'& are eliminate.

Ki<en angle of eflection for Ba' = 4o

ngle of eflection of a2'

= o( . )1 2

4 0

23 137

÷ × = 1.3o

ns- %

$. 2n ;E- ) ' O ) 2' ' e#

ns- %

3. "arge = %proton no.& %e no.&

Element +roton e cargeG n n81 '1

!. Element +roton e

I2 36 2 = 34 36*!' 37 37 1 = 36



etent of eflection ∝ carge

mass

ns1 for Kroup ;; element

s1

s2

s2p1s2p2

s2p3

s2p6






Reg. No. 200710217Z


H n ' 1 n81 '2

1. ?e carge of G is alf tat of H.

2. G as loer nuclear carge %1 less proton&tan H. s G is isoelectronic it H <alence

e#

of G are less strongl attracte tonucleus gi<ing a larger ionic raius of G.

". G reuires lesser energ tan H for teremo<al of te furter electron as G as alarger ionic raius an loer nuclear carge.

ns- #

Sr 2' 3( 3( 2 = 36

1. ot I2 an *!' a<e 36 electrons.

2. I2 as a smaller nuclear carge ue tolesser no .of protons tan Sr 2'.

". Less energ is reuire to remo<e anelectron from an increasingl negati<e ionue to repulsion of te same carge %ierepulsion !eteen e# an te resulting I#&.

ns- #

Chemical %on,ing

1. (a) *a(s) is a goo, electrical con,+ctor ue to te presence of mobile electrons to carrcarges troug te lattice. *aCl (s) ,oes not con,+ct electricity !ecause of absence of mobile charge carriers since its ions Ba' an "l − are fie at teir lattice points&.

(b) luminium is mallea!le !ecause te layers in te lattice can sli,e o<er eac oter itho+tbreaing the strong metallic bon,. luminium fluorie is !rittle !ecause en a force isapplie along a particular plane layers of ions sli,e s+ch that ions of the same chargemeet an, re-el one another sattering te crstal along te fault line.

(c) 7agnesi+m chlori,e is solu!le in ater !ecause its DO*S can form ionB,i-oleinteractions ith ater molecules ile ethene cannot form ionB,i-ole interactions nor hy,rogen bon,s ith ater molecules.

2.



∙∙






Reg. No. 200710217Z


".

4. (a) ?rigonal planar ⇒ 3 !on pairs no lone pair ⇒ : as 3 <alence electrons

Kroup DDD

(b) ?rigonal pramial ⇒ 3 !on pairs 1 lonepair ⇒ ? as 5 <alence electrons

?8sape ⇒ 3 !on pairs 2 lone pairs⇒ @ as 7 <alence electrons

Kroup > Kroup >DD

$. (a) *aCl as a giant ionic str+ct+re an HCl as a sim-le molec+lar9covalent str+ct+re.7ore energy is reuire to overcome the strong electrostatic attractions9 ionic bon,s








Reg. No. 200710217Z


beteen *a an, Cl I than the ea intermolec+lar forces (OR -ermanent ,i-oleB-ermanent ,i-ole attraction) beteen HCl molec+les.#ence *aC l as higher melting -oint tan #"l .

(b) SiO2 as a giant covalent str+ct+re ile SO2 as a sim-le molec+lar9covalent str+ct+re.7ore energy is reuire to overcome the strong covalent bon,s beteen Si an, Oatoms in SiO2 tan te ea intermolec+lar forces (OR -ermanent ,i-oleB-ermanent,i-ole attraction) beteen SO2 molec+les.#ence SiO2 as higher melting -oint tan S$2.

(c) %oth a<e giant ionic str+ct+res it strong electrostatic attractions9 ionic bon,sbeteen o--ositely charge, ions. Since 7g2 an, O2 a<e higher charge an, smaller ra,i+s tan Ba' an "l − respecti<el more energy is neee to overcome stronger ionicbon,s in 7gO tan tat in Ba"l .#ence 7gO as higher melting -oint tan Ba"l .

(,) %oth a<e giant metallic str+ct+res it strong electrostatic attractions9 metallic bon,sbeteen cations an, mobile valence electrons. Since #l has more valence electronstan Ba an #l " has higher charge ,ensity tan Ba' more energy is neee to overcomethe stronger metallic bon,s in #l tan tat in Ba.

#ence #l as a higher melting -oint tan Ba.

$. (e) C as a giant covalent str+ct+re an #r as a monoatomic str+ct+re.7ore energy is reuire to overcome the strong covalent bon,s beteen C atoms thanthe ea intermolec+lar forces (OR van ,er WaalsJ forces) beteen #r atoms.#ence C as a higher boiling -oint tan r.

(f) %oth a<e sim-le molec+lar str+ct+res. 7ore energy is reuire to overcome thestronger hy,rogen bon,ing beteen H2O molec+les than the eaer van ,er WaalsJforces beteen H2S molec+les.#ence H2O as higher boiling -oint tan #2S.

(g) %oth a<e monoatomic str+ct+res. Since *e has a larger n+mber of electrons -er atomtan #e more energy is reuire to overcome the stronger van ,er WaalsJ forcesbeteen *e atoms tan tat !eteen #e atoms. #ence *e as higher boiling -oint tan#e.

(h) %oth a<e sim-le molec+lar9covalent str+ct+res. Since CCl 4 has a larger n+mber of electrons -er molec+le tan "4 more energy is reuire to overcome the stronger van,er WaalsJ forces beteen CCl 4 molec+les tan tat !eteen "4 molecules.#ence CCl 4 as higher boiling -oint tan "4.

(i) %oth a<e sim-le molec+lar9covalent str+ct+res. Jue to branching of carbon chains(CH")"CH molec+les have smaller s+rface area of contact tan "#3"#2"#2"#3 molecules.?us less energy is reuire to overcome the eaer van ,er WaalsJ forces beteen(CH")"CH molec+les compare to tat !eteen "#3"#2"#2"#3 molecules.#ence CH"CH2CH2CH" as higher boiling -oint tan %"#3&3"#.

() %oth a<e sim-le molec+lar9covalent str+ct+res. 7ore energy is reuire is overcome A is Easy with ADVO!







Reg. No. 200710217Z


the stronger hy,rogen bon,ing beteen CH"CH(OH)CH" molec+les than the eaer -ermanent ,i-oleB-ermanent ,i-ole attraction beteen CH"COCH" molec+les.#ence CH"CH(OH)CH" as higher boiling -oint tan "#3"$"#3.

() %oth a<e sim-le molec+lar9covalent str+ct+res. 7ore energy is reuire to overcomethe stronger -ermanent ,i-oleB-ermanent ,i-ole attraction beteen CH6" molec+lesthan the eaer van ,er WaalsJ forces beteen C64 molec+les.#ence CH6" as higher boiling -oint tan "4.

(l) %oth a<e giant covalent str+ct+res. Since C atom is smaller tan Si atom an tus CICbon, is shorter tan SiPSi !on more energy is reuire to overcome the stronger covalent bon,s beteen C atoms tan tat !eteen Si atoms.#ence C as a higher melting -oint tan Si.

3. (a)

(b) #l in l "l 3 is electronB,eficient %onl 6 electrons aroun l &. * in B#3 as one lone -air of electrons. ?us a ,ative covalent bon, is forme, beteen one #l Cl " molec+le an, one*H" molec+le to gi<e #3B:→ l "l 3 so tat #l can achieve a stable octet config+ration.

(c)

!. (a) S is a +erio 3 element ic as energetically accessible vacant "d orbitals to e-an,its octet so it can accommoate more tan ( electrons in te outermost sell uring teformation of S4 an S6. #oe<er O is a +erio 2 element ic cannot e-an, its octetto form $4 an $6 !ecause it ,oes not have energetically accessible vacant "d orbitals.

(b) (i) Cs6 as strong electrostatic attractions9 ionic bon,s beteen Cs an, 6 I ions icare forme troug te transfer of valence electrons from Cs to 6.

S63 as strong covalent bon,s beteen S an, 6 atoms ic share their valenceelectrons in forming S 6 covalent bon,. ?ere eist eaG <an er AaalsQ forces or intermolecular forces !eteen S6 molecules.*eason- ?e ,ifference in electronegativity !eteen Cs an 6 is large ile tat

!eteen S an 6 is small.(ii) "aesium flourie %"s&

giant ionic str+ct+reS6

sim-le molec+lar str+ct+re1. ;t as a high melting -oint %an/or

boiling -oint9 lo volatility& ue tolarge amo+nt of energy neee too<ercome te strong electrostaticattractions9 ionic bon,s beteen Cs

an, 6 5 ions.

1. ;t as a lo melting -oint %an/or boiling -oint9 high volatility& ue tosmall amo+nt of energy neee too<ercome te ea van ,er WaalsKforces beteen (S63) molec+les.

2. ;t is a goo, electrical con,+ctor in 2. ;t is a -oor electrical con,+ctor in all



Cl Al Cl

Cl

H N H

H

Al

Cl

Cl Cl

120°

H

H

N

H107°






Reg. No. 200710217Z


molten an, a+eo+s states ue topresence of mobile ions.

states ue to absence of mobilecharge carriers %OR no mobile ionsnor mobile electrons&

3. ;t is sol+ble in ater ue to teformation of ionB,i-ole interaction ith

ater molec+les %te ration energrelease from te ion8ipole interactionit ater molecules is sufficient too<ercome te strong ionic !ons in"s&.

3. ;t is insol+ble in ater as S63 is notca-able of forming hy,rogen bon,ing

nor ionB,i-ole interaction ith ater molec+les.

8. (a) (i) C1- s-" !riisation C2- s-2 !riisation(ii)

sp3 !ri or!itals aroun "1

%tetraeral arrangement&sp2 !ri or!itals aroun "2

%trigonal planar arrangement&(b) σ8!on is forme ! te hea,Bon overla- of orbitals.

π8!on is forme ! te si,eay overla- of orbitals.(c) ;n te <apour state %Rust a!o<e its !oiling point& CH"COOH ,imerises (or forms a ,imer)

troug te formation of hy,rogen bon,s beteen to CH"COOH molec+les. ?us its M r as foun to !e 120.



Check! Must include

• 2 pairs of dipole (δ$% δ& perh'drogen bond

• lone pair on O

• dotted lines bt! lone pair on






Reg. No. 200710217Z


Aen it is issol<e in ater CH"COOH molec+les form hy,rogen bon,s ith ater molecules?us its M r as 60.

)"@

1. # 2. # ". C 4. C $. ' 3. C !. C

Lhe Gaseo+s State

1. (a) >ol+me of the gas molec+les is negligible com-are, ith vol+me of the container .Gas molec+les eert *O forces of attraction on one another .

(b) "onitions- Lo pressure an ig temperature. t lo -ress+re te gas occ+-ies a larger vol+me an gas molecules are space out.?us te vol+me of the gas molec+les !ecomes insignificant9negligible com-are, to thevol+me of the container .

lso te molec+les are far a-art suc tat intermolec+lar forces of attraction tat operate!eteen tem !ecome insignificant9negligible.

t high tem-erat+re te molecules a<e iger Ginetic energ an ence a<e s+fficientenergy to overcome intermolec+lar forces. ?us te intermolecular forces of attraction!eteen te gas molecules are negligi!le.

2.r

msing p * n, ,

M

= = ÷

( ) ( ) ( )

( ) ( )4 6

0.100 (.31 305m

2.00 10 104 10r

, M

p* −= = =

× ×122

?e formula of # is <r62.

". fter te flasGs are connecte total <olume of <essel 31 2 3 m= + =

∴ 3>olume of #e >olume of Be 3 m= =

6or He1 1 2 2 p* p * =

( ) ( ) ( )2 3 2 1 p =

2 p =2

Ma"

6or *e1 1 2 2 p* p * =

( ) ( ) ( )2 3 1 2 p =

2 p =2

Ma"

sing JaltonTs La of +artial +ressure

, He Ne p p p= + =4

Ma"

; 1."" Ma



*ote -

• M r calculate by s+mmation of Ar of all atomsin te species is gi<en to 1 ,ecimal -lace.

• $terise calc+late, M r is i<en to "sf .






Reg. No. 200710217Z


4. (a) ner constant ?*? 1

> >

n p -

= = ÷

(b) ner constant >

( )*?

?>

n p - = =

(c) ner constant >

( )*?

?/ "'273>

n p - = = °

(,) ner constant ?*? 1

> >

n p -

= = ÷

(e) ner constant ?> *? p n - = =

(f) ner constant ?>

*?

p

n=

Chemical Energetics (Part I : Enthalpy change o reaction

1. (a) $

f H ∆ ( )( ))g$ s - 7g(s) NO2(g) → 7gO(s)

(b) $

f H ∆ ( ) ( )( )2)g $# a - 7g(s) O2(g) H2(g) → 7g(OH)2(a)

(c) $

cH ∆ ( )( )2 6" # g - C2H3(g) 72 O2(g) → 2CO2(g) "H2O(l )

(,) ( )LE 2# Li $∆ - 2i(g) O25(g) → i2O(s)

2. (a) enthal-y change of formation of soli, so,i+m chlori,e

(b) enthal-y change of atomisation of bromine

(c) bon, energy of %r5%r bon,

(,) enthal-y change of sol+tion of calci+m chlori,e

(e) enthal-y change of hy,ration of -otassi+m ion

". (a) en,othermic (,) eothermic (g) en,othermic(b) eothermic (e) eothermic



p

V

p

T/K 0

p

T/°C –273

p

1/V0

pV

V

nRT

pV

p

n

Check! Mustinclude states'mbols"






Reg. No. 200710217Z


(c) eothermic (f) en,othermic

4. (a) 2"%grapite& ' #2%g& → "2#2%g&

( )( ) ( ) ( )

( ) ( ) ( ) 223 HP mol

! H H H H ∆ = ∆ = ∆ − ∆

= × − + − − − =

∑ ∑$ $ $ $

2 2 rn c c" # g rtants pts

2 394 2(6 1300

(b) 3"%grapite& ' 4#2%g& ' U$2%g& → "3#7$#%l &

( )( ) ( ) ( )

( ) ( ) ( ) "13 HP mol

l H H H H ∆ = ∆ = ∆ − ∆

= × − + × − + − − =

∑ ∑$ $ $ $

f 3 7 rn c c" # $# rtants pts

3 394 4 2(6 0 2010

(c) "6#12%l & ' 9$2%g& → 6"$2%g& ' 6#2$%l &

( )( ) ( ) ( )

( ) ( ) ( ) "428 HP mol

l H H H H ∆ = ∆ = ∆ − ∆

= × − + × − − − + =

∑ ∑$ $ $ $

c 6 12 rn f f " # pts rtants

6 394 6 2(6 152 0

$. (a) "#3"#3%g& ' "l 2%g& → "#3"#2"l %g& ' #"l %g&

ons !roGen GM mol P1

ons forme GM mol P1

1 "P" !on 350 1 "P" !on 3506 "P# 6 %'410& = '2460 5 "P# 5%'410& = '20501 "l P"l '244 1 "P"l '340

1 #P"l '431?otal '3054 ?otal '3171

∆H rn = Σ%!ons !roGen& P Σ%!ons forme& = %'3054& %'3171& = I11! P mol I1

(b)

?is is !ecause te bon, energies in the Data Booklet re-resent average9mean bon,energies eri<e from a range of molecules tat contain te particular !on.

3. (a) Stanar entalp cange of com!ustion of liui etanol is te enthalpy change thatoccurs hen ! mole o" ethanol is completely burned in oxygen under standardconditions.

(b) #eat reuire to raise te temperature = 400 , 4.2 , %22812& = 16(00 M = 16.( GM $No si"n/'

∴∆H c%etanol%l && = P16.(

%0.92 H 46.0&= I84 P

(c) 1. Some heat is absorbe, by the metal calorimeter .

2. Heat loss from the calorimeter to the s+rro+n,ings (no lagging 9ins+lation).". *ot all the heat from the flame is transferre, to the calorimeter an, ater .4. oss of ethanol (hich va-orise,) from the hot ic before the s-irit lam- as

reeighe, (at the en, of the e-eriment). Lhis o+l, res+lt in QhigherQ mass of ethanol than act+al being +se, in the calc+lation.

!. (a) Ba$#%a& ' "#3"$$#%a& → "#3"$$Ba%a& ' #2$%l &








Reg. No. 200710217Z


( ) ( ) ( ))

H

= × =

+= = °

= ∆ = + −= =

∆ = −neutralisation

30.0amount of ater prouce 1.0 0.0300mol

100029.0 29.4

a<erage initial temperature of solution 29.2 "2

amount of eat e<ol<e < ? 30.0 30.0 3.2 35.( 29.2

1660M 1.66 GM

amount of eat 1$$.4 P mol = − = ÷ ÷

e<ol<e 1.66

amount of ater prouce 0.0300

(b) #"l0 a strong aci ionises completel in aueous solution ile CH"COOH is a ea aci,that ionises -artially in aueous solution.Since ionisation of "#3"$$# is an enotermic process some heat release from teneutralisation is absorbe, to f+rther ,issociate the ea aci, com-letely.?erefore te enthal-y change of ne+tralisation of CH"COOH ith *aOH is lesseothermic tan tat of #"l it Ba$#.

8. (a)

OR

#essTs La



Ca(s) + F2(g) CaF2(s)

Ca(g) + 2F(g)

Ca2+(g) + 2F – (g)

+178

(+590)+(+1150)

2(+79)

2(–328) L(CaF2(s))

Ca(s) + F2(g)

0

Ca (g) +F2(g)

Ca (g) + 2F(g)

Ca 2+(g) + 2F(g) + 2e –

Ca 2+(g) + 2F – (g)

CaF2(s)

!"#g$ /%& 'o –1

+178

2(+79)

(+590)+

(+1150)

2(–328)

–1220

L (CaF2(s))

Check!or each process%& e)uation is balanced& state s'mbol ofreactants and productsare stated correctl'

& process is labelled

!ith correct ∆H s'mbol

or *alue& arro! is pointing in the






Reg. No. 200710217Z


( )( )

( ) ( ) ( ) ( ) ( ) ( )

2

1

LE "a s

1220 17( 590 1150 2 79 2 32(

2640 GM mol−

= − − + + + + + + + + − = −

8. (b)+ −

+ −

µ+

LE

r r

Since Cl 5 as te same ionic charge an larger ionic ra,i+s9sie tan te magnit+,e of lattice energy of CaCl 2 is smaller tan tat of "a2 %OR lattice energy of CaCl 2 is lesseothermic&.

O2I as a larger ionic charge an larger ionic ra,i+s9sie tan . Since te effect of ( I)is greater than the effect of (r r I) te magnit+,e of lattice energy of CaO is larger tantat of "a2 %OR lattice energy of CaO is more eothermic&.

. (a) SiCl 4(g)→

Si(g) Cl (g)∆

# ; #verage $ (Si5Cl )(b) (i) Energ / GM mol 1

Si%g& ' 4"l %g&

4%'122& = '4((

Si%g& ' 2"l 2%g&

Si%s& ' 2"l 2%g& '33(

610

Si"l 4%g&

(ii) a<erage ( %Si8"l & = VC%'33(&'%'4((&'%'610&D = "$ P mol T1



0

" # a$erage E(Si%Cl

Check!or each process%& e)uation is balanced& state s'mbol ofreactants andproducts are statedcorrectl'

& process is labelled

!ith correct ∆H s'mbol or *alue

&






Reg. No. 200710217Z


1. (a))g"l 2%s& )g2'%a& ' 2"l %a&

)g2'%g& ' 2"l %g&

∆# soln(7gCl 2) ; 5 E(7gCl 2) ∆# hy,(7g2) 2∆# hy,(Cl 5) = %2526& ' %1(90& ' 2%3(4&

= 1"2 P mol 51

1. (a)

OR %y energy level ,iagram

Energ / GM mol 1



∆ H

$*(g2+)

= –1890

– L(gCl 2)

= – (– 252,)

∆ H

so!(gCl

2)

2∆ H

$*(Cl – )

= 2(–384)

g2+(g) + 2 Cl – (g)

gCl 2(s)

–2526

g2+(a-) + 2 Cl – (g)

–1890

g2+(a-) + 2 Cl – (a-)

2(–384)

= –768

∆ H

soln(MgCl

2)

∆ H so!(gCl 2)

= – (1890 + 7,8 – 252,)= – 132 ! "ol –1

No#e$

o NOT 'a#% 0o! " yas

C%e&'

Fo# "a6 #o6"ss

"-ao! s :aa!6"* sa" s$':o o; #"a6a!sa!* #o*6s a#" sa"*

6o##"6$

#o6"ss s a:""* <

6o##"6 ∆ H s$':o o#

a"






Reg. No. 200710217Z


(b) Ca2 has te same ionic charge !ut larger ionic ra,i+s9sie tan )g2' an tus Ca2 has aloer charge ,ensity tan tat of )g2'. ?erefore te n+merical val+e of ∆# hy, of Ca 2 issmaller tan tat of )g2' since "a2' ions form eaGer ion8ipole interactions it ater molecules.

7+lti-leBChoice F+estions

11. (%) "$%g& ' 2 #2%g& "#3$#%l &

∆# r ; Σ ∆# c (reactants) Σ ∆# c (-ro,+cts)

∆H r = ∆H c%"$& ' 2∆H c%#2& ∆H c%"#3$#&

= %2(3& ' 2%2(6& %715&= T 14 P mol 1

ORA "$%g& ' 2 #2%g& "#3$#%l & (T28"& 2(T283) %T!1$)

"$2%g& ' 2 #2$%l &

#essQ La ∆H r = %2(3& ' 2%2(6& %715&= T 14 P mol 1

12.

(C) ?e stanar entalp cange of neutralisation of an aci it an alGali is te entalp# )an"e en one mole of ater is forme, from te reaction of te aci an alGaliuner stanar )onitions.

ot reactions in<ol<e miing of a strong base an, a strong aci, an eac gi<es tomoles of ater . ?erefore te eat li!erate for !ot reactions is te same.

#2S$4%a& ' 2Ba$#%a& → Ba2S$4%a& ' 2H2O(l ) ∆H = 114 GM mol 1

2#"l %a& ' a%$#&2%a& → a2"l 2%a& ' 2H2O(l ) ∆H = 5114 GM mol 1

1".

(#) ?e entalp cange of te folloing processes/reactions is alas negati<e%eotermic&.

• "om!ustion• Beutralisation

•ormation of soli ionic compoun from its gaseous ions %1atti)e (ner"# &

• #ration of gaseous ions

?e entalp cange of te folloing processes/reactions is alas positi<e%enotermic&.

• tomisation• ;onisation• reaGing of co<alent !ons %on ener"# &

(#) !urning an element in ogen ⇒ comb+stion (alays eothermic)



∆H r

∆H r






Reg. No. 200710217Z


(%) issol<ing a compoun in ater ⇒ solution %∆H can !e <e or '<e&(C) forming an ion from an atom ⇒ ;E alas enotermic %to form cation from atom&

!ut E can !e <e or '<e %to form anion from atom&(') sntesising a compoun from its elements ⇒ formation %∆H can !e <e or '<e&

14. (C) 0P39M"9F!A *19M"9F








Reg. No. 200710217Z


6oo5 out for statement that indicates -a-l 2 being more stable than -a-l .

#. 6alse "a%g& → "a'%g& ' eB 1st IE = ' 590 GM mol71

"a%g& → "a2'%g& ' 2eB 1st IE ' 2n IE = ' 590 ' 1145 = '1735 GM mol71

%. Statement is tr+e %UL ,oes not e-lain hy CaCl 2 is forme, rather than CaCl .or "a"l %s&- U "l 2%g& → "l %g& → "l %g& ∆H 1

∆H 1 = U ( %"l 7"l & ' 1st E %"l & = U %'244& ' %−364& = −242 GM mol71

or "a"l 2%s&- "l 2%g& → 2"l %g& → 2"l %g& ∆H 2

∆H 2= ( %"l 7"l & ' 2 1st E %"l & = '244 ' 2%−364& = −4(4 GM mol71

⇒ more energ is release in forming 2 "l from "l 2 in te formation of "a"l 2

C. attice energy of CaCl is less eothermic than that of CaCl 2

⇒ eaer ionic bon,s in CaCl ⇒ CaCl is less stable.

'. 6alse ;f more energ is release en "a"l is forme⇒ energ le<el of "a"l is loer tan tat of "a"l 2 ⇒ "a"l is energeticall more sta!le tan "a"l 2 %6alse&

1$. (C) 0*89M"9F8

Sinceq q

LE r r

+ −

+ −

×µ

+

)agnitue of Lattice energ- )2'W2− X M'I− X L'Y−



0 elements in stanar states

"a"l %s&

"a"l 2%s&

energ

ou!l8carge

cation an anion

singl8carge cations an anions

MI as a smaller interionic istance %r ' ' r 8 &tan LY






Reg. No. 200710217Z


13. (') 0*$9M19!

#. Since entalp cange of solution for a2 is a small positi<e <alue a2 migt still!e solu!le/sparingl solu!le in ater. %Statement is inconcl+sive&

%. ot a2'

an )g2'

ions a<e te same ionic carge !ut a2'

as a larger ionicraius tan )g2'. #ence te ration energ for a2' soul !e numericall smaller %less eotermic& tan tat of )g2'. %6alse&

C. + −

+ −

×µ

+q q

LEr r

Since a2' as a larger ionic raius % !ot compouns a<e te same prouct of ionic carges& lattice energ for a2 soul !e numericall smaller %lesseotermic& tan tat of )g2. %6alse&

'. ∆# soln(%a62); E(%a62) ∆# hy,(%a2) 2 Hhy,(6 5)

1!. (#) 0*23 M1 F""

1 ?e ∆H at%iamon& is smaller tan ∆H at%grapite&. (Lr+e)

2 Since it reuires more energ to atomise grapite te ( %""& in grapite isgreater tan tat in iamon. (Lr+e)

" ?e ∆H c%iamon& is numeri)all# greater tan ∆H c%grapite&. (Lr+e)



or ∆H soln to !e positi<e tis term must !e more positi<e tan te negati<e term ue to

ration processes. ?erefore te numerical <alue of lattice energ must !e greater tan te sum of ration energies. (Lr+e)

Energ/ GM mol 1

C(gra-hite)

C(,iamon,)

'3

C(g)

∆H at%grapite&

∆H at%iamon&

CO2(g)

∆H c%grapite&∆H c%iamon&






Reg. No. 200710217Z


Chemical Energetics Mart 2 Entro-y

18. (a) Entrop is a meas+re of the ,isor,er in a system. ?e more ,isor,ere, a system te larger its entro-y.

(b)

(i)Ba%s& → Ba%g&∆%

V ⇒ increase in ,isor,er from a more or,ere, soli, state to a more ,isor,ere,gaseo+s state.

(ii) 2+!%B$3&2%s& → 2+!$%s& ' 4B$2%g& ' $2%g&∆%

V ⇒ increase in ,isor,er !ecause te reaction -rocee,s ith an increase in then+mber of gas molec+les.

(iii) "l 2%g& ' 13 ;2%s& → 23 ;"l 3%l &

∆%

⇒ ,ecrease in ,isor,er !ecause te reaction -rocee,s ith a ,ecrease in then+mber of gas molec+les.

(iv) 1 mol of "l 2%g& is ae to 1 mol of B2%g&∆%

V ⇒ increase in ,isor,er !ecause miing of to ifferent gas molecules ill

result in a more ,isor,ere, arrangement of the molec+les.(v) 1 mol of "l 2%g& at 29( F is eate to 373 F

∆%

V ⇒ increase in ,isor,er !ecause an increase in tem-erat+re increases theinetic energy of the molec+les.?is causes a broa,ening of the %oltmann energy,istrib+tion ic ill result in more ays of arranging energy +anta in te otter gas.

1. (a) l 2$3%s& ' 6 #%g& → 2 l 3%s& ' 3 #2$%g&

∆H = X∆# $f (-ro,+cts) X∆# $f (reactants)

= 2 ∆H $

f %l 3& ' 3 ∆H $

f %#2$& ∆H $

f %l 2 $3& 6 ∆H $

f %#&

= 2%−1504& ' 3%−242& %−1676& 6%−271&

= 4"2 P mol 1

(b)

negative ∆% $f inicates tat te reaction procees it a ,ecrease in ,isor,er as te

reaction -rocee,s ith a ,ecrease in the n+mber of gas molec+les.(c) ∆G = ∆H , ∆S

= %432& 28 %−1000

390& = 5"13 P mol 1

(,)∆& ; ∆# 5 ' ∆%

Since ∆% A 5' ∆% V .

When tem-erat+re increases LY%

becomes more -ositive.#t high tem-erat+re ∆&

V since #

5'∆%

.?erefore te reaction is not s-ontaneo+s at high tem-erat+re.

#t lo tem-erat+re ∆&

since #

V 5'∆%

.?erefore te reaction is s-ontaneo+s at lo tem-erat+re.

#oe<er te reaction ill be too slo at lo tem-erat+re. #ence te reaction sho+l, becarrie, o+t at mo,erate tem-erat+re an not at <er ig temperature to pre<ent te reactionfrom !eing non8spontaneous.



– " –"

+"






Reg. No. 200710217Z


2. (a) (i) ∆Goppt = 2.303 *? log sp

∆Goppt = 2.303 × (.31 × 29( × log%2.0 10 10&

= 55300 M mol 1

= $$." P mol −1

(ii)∆& o--t ; ∆# o--t 5 L∆% o--t

55.3 = 66.0 29(%∆So &

∆Soppt = 0.035( GM mol 1 F 1

= "$.8 P mol 1 < 1

(iii)∆%o

--t is negative inicating tere is a ,ecrease in ,isor,er in te sstem,+ring -reci-itation. ?is is !ecause te free a+eo+s ions combinetogether to give a more or,ere, soli, #gCl .

(b) ∆Goppt = 2.303 × (.31 × 29( × log%1.006&

= 14.8 P mol −1

Since ∆&o--t is -ositiveA -reci-itation of #g6 ill not occ+r at 29(F.

?erefore #g6 is sol+ble in ater at 28 <.

g is solu!le in ater !ecause te hy,ration energy release, ! te formation of teion8ipole interaction !eteen g' an te ater molecules an tat !eteen anater molecules is s+fficient to overcome the hy,rogen bon,s beteen ater molec+les an, the electrostatic attraction (or ionic bon,) beteen #g an, 6ions in #g6.

Chemical E+ilibri+m

1. "$2%g& ' #2%g& + "$%g& ' #2$%g&Euili!rium partial pressure/ atm 20 9 15 20

(a) 2

2 2

CO H O

-

CO H

M M; 1.3!

M M =

%15&%20&=

%20&%9&

(b)

% since te euili!rium position sifts rigt to remo<e S$)E of te #2.

2. (a) c

0: ?;

0: 0?

2

2

22

= 5 mol1 m3

÷

4 2

2 2c 4 2 2

2 2

CI YD CI YDQ = =

CI D CYD CI D CYD= 52 = 2$ nit = mol I2 ,m3

(b) 3

−

− ÷

11 2

21

2

2

2 2c 2

2 2

CI DCYD CI YDQQ = = = %5&

CI YD CI D CYD = .44! nit = mol 192 ,m I"92

". (a) Le "atelierTs +rinciple states tat if a system is s+becte, to a change hich ,ist+rbsthe e+ilibri+m the system res-on,s in s+ch a ay to co+nteract some of the effect








Reg. No. 200710217Z


of the change.(b) (i) B2%g& ' 3#2%g& + 2B#3%g& 888%Z& [H = 92.6 GMmol 1

Aen <olume ecreases -ress+re of te sstem increases.?e e+ilibri+m -osition of (Z) shifts to te right to reuce te pressure ! re,+cingthe amo+nt of gas molec+les.

E+ilibri+m constant remains +nchange, as te tem-erat+re is constant.(ii) Aen temperature increases te e+ilibri+m -osition of (Z) shifts to te left tofavo+r en,othermic reaction so as to remove some heat energ.?e e+ilibri+m constant ,ecreases since te !acGar reaction is fa<oure.

(iii)

Aen iron catalst is ae the e+ilibri+m -osition remains the same since terate of bacar, reaction an, forar, reaction are increase, by the same etent .an, e+ilibri+m constant remain the same since tem-erat+re remains constant.

4. ?emperature increases ⇒ c increases.Aen c increases ith increasing tem-erat+re it implies tat te e+ilibri+m -osition of reaction shifts right to favo+r en,othermic reaction ! removing9absorbing some heatenerg. ?us te forar reaction is en,othermic.

$. (a) (i)( )

( ) ( )2

2

2

3

2

S$

$ S$

p

= unit - atm 51

(ii) 2S$2%g& ' $2%g& 2S$3%g&Em partial pressure /atm 2- - 4.!

( )

( ) ( )

( )

( ) ( )

1 atm

.1 atm.2 atm .1 atm

$$2 p

=+ + = ⇒ =

= =

= = =

2 2

2

2

2

3

2

S$ $

2

2

S$

$ S$

?otal pressure at em 5 atm 2p p 4.7 5 atm p

at em at em

4.7

0.1 0.2

(b) (i)

2

2

total S$

$

?otal initial pressure 3 atm

2 initial 3 2 atm

3

1 initial 3 1 atm

3

A A

total

n

n

÷ ÷ ÷

÷

=

= × ⇒ = × =

⇒ = × =

2S$2%g& ' $2%g& 2S$3%g&;nitial partial pressure /atm 2 1

Em partial pressure /atm .1 .$ 1..1 atm .$ atm = =

2 2S$ $ at em at em

(ii) 2.$ atm= + + =ne total pressure 0.10 0.05 1.9

Since S$3 ≡ S$2 teoretical pressure of S$3 prouce = 2.00 atm

$. / ÷

= × =2 3

1.90: con<ersion of S$ into S$ 100:

2.00



+2(095) –095 –(2095)






Reg. No. 200710217Z


(iii)

( )

( ) ( )

( )

( ) ( )

1 atm!!2 p

= = =

2

2

2

3

2

2

2

S$

$ S$

1.9

0.05 0.1

$. (c) 2 S$2%g& ' $2%g& 2 S$3%g& 88%Z& p at 500 \" = 5520 atm 1 p at 430 \" = 7720 atm 1

Aen - increases ith, ,ecreasing tem-erat+re tis implies tat te e+ilibri+m-osition of (Z) shifts to te right to favo+r eothermic reaction ! releasing some heatenerg. #ence te forar, reaction is eothermic.

3. (a)

(b) (i)2 3

2 4

2

> ?2

> ? >?

(@ )

(@ )(@ ) p K = nit = atm 51 01m 5 - e-ression[ 01m 5 +nits

(ii) - ;

2

2

)50.1)(75.0(

)25.0( ; ."! atm 51 01m

(c) (i) Aen temperature is increase E+ilibri+m -osition shifts to the left 01m to favo+r the en,othermic reaction

so as to remove some heat 01m to ecrease temperature.(ii) Aen pressure is increase

E+ilibri+m -osition shifts to the right 01m to ,ecrease the n+mber of gaseo+smolec+les 01m to reuce te sstem pressure.

()$ *o#% ,es &o--e&#l. l/elle

()$ e&-esng &-e #%# leel o

o- o-- -e&#on

(

)$ n&-esng &-e #%# leel o

o- /&- -e&#on

()$ *o#% g-%s leel o # #%e s"e

&oo-n#es

()$ &o--e&#l. l/el #%e #o g-%s s

o-- n /&- -e&#ons

5(

) – 3"

34(

) – 2"

–

suggested soln to foundation physical chemistry

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