!!"#$%&'()*+#,-#.)/#0#.)1)2+0$(-#+)345&6+6$%7)89*)6()9:#;6&<%73$;=.#)2&&#&&;#(<)>$<#%6$.&)?)@&&4#)A)?)B#5%4$%7)CDAE)F)"#$%&'()*+4-$<6'()/6;6<#+)CDAE

CH

EMIS

TRY

AS

PA

PER

2 M

AR

KS

CH

EME

Qu

esti

on

Nu

mb

erA

nsw

erA

dd

itio

nal

Gu

idan

ceM

ark

1(a

)

(

1)

Ever

y at

om a

nd e

very

bon

d m

ust

be

show

n1

1(b

)C

2H5

(1)

1

1(c

)B

1

1(d

)C

1

1(e

)•

re-a

rran

gem

ent

of p

V=

nRT

equa

tion

(

1)

•co

nver

sion

of

volu

me

and

subs

titu

tion

(

1)

•ev

alua

tion

of

num

ber

of m

oles

of

alka

ne

(1

)

•ca

lcul

atio

n of

Mr,

usin

g nu

mbe

r of

mol

es a

nd m

ass

of

alka

ne t

o id

entify

it

(1)

Cor

rect

ans

wer

, no

wor

king

sco

res

4 m

arks

Exam

ple

of c

alcu

lation

n=

pV

R

T

=10

0 00

0 x

(83

x 10

-6)

8.31

x 4

25

= 0

.002

35…

(m

oles

)

Mr

= 0

.235

/ 0

.002

35…

= 9

9.99

or

100

i.e.

this

is h

epta

ne

4

(To

tal

for

Qu

esti

on

1 =

8 m

arks

)

PMT

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!G

Qu

esti

on

Nu

mb

erA

nsw

erA

dd

itio

nal

Gu

idan

ceM

ark

2(a

)An

answ

er t

hat

mak

es r

efer

ence

to

the

follo

win

g po

ints

:

•(a

ll eq

uation

s) s

how

oxy

gen/

O2

as a

rea

ctan

t/on

the

left

(1)

•(a

ll) Δ

rHo

valu

es a

re n

egat

ive

/(al

l) r

eact

ions

are

ex

othe

rmic

(1)

2

2(b

)An

answ

er t

hat

mak

es r

efer

ence

to

the

follo

win

g po

ints

:

•re

action

2 is

not

, be

caus

eth

ere

is m

ore

than

1 p

rodu

ct

(1

)

•re

action

3 is

not

, be

caus

e 2

mol

es o

f pr

oduc

t ar

e fo

rmed

(1)

Acc

ept

beca

use

met

hane

/CH

4is

not

an

elem

ent

2

2(c

)B

1

2(d

)D

1

(T

ota

l fo

r Q

ues

tio

n 2

= 6

mar

ks)

PMT

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!H

Qu

esti

on

Nu

mb

erA

nsw

erA

dd

itio

nal

Gu

idan

ceM

ark

*3

(a)

This

que

stio

n as

sess

es a

stu

dent

’s a

bilit

y to

sho

w a

coh

eren

t an

d lo

gica

lly s

truc

ture

d an

swer

with

linka

ges

and

fully

-su

stai

ned

reas

onin

g.

Mar

ks a

re a

war

ded

for

indi

cative

con

tent

and

for

how

the

an

swer

is s

truc

ture

d an

d sh

ows

lines

of

reas

onin

g.

The

follo

win

g ta

ble

show

s ho

w t

he m

arks

sho

uld

be a

war

ded

for

indi

cative

con

tent

.N

umbe

r of

in

dica

tive

m

arki

ng

poin

ts s

een

in a

nsw

er

Num

ber

of

mar

ks

awar

ded

for

indi

cative

m

arki

ng

poin

ts6

4 5–

4 3

3–2

2 1

1 0

0

Gui

danc

e on

how

the

mar

k sc

hem

e sh

ould

be

app

lied:

The

mar

k fo

r in

dica

tive

con

tent

sho

uld

be

adde

d to

the

mar

k fo

r lin

es o

f re

ason

ing.

Fo

r ex

ampl

e, a

n an

swer

with

five

indi

cative

mar

king

poi

nts

that

is p

artial

ly

stru

ctur

ed w

ith

som

e lin

kage

s an

d lin

es o

f re

ason

ing,

sco

res

4 m

arks

(3

mar

ks f

or

indi

cative

con

tent

and

1 m

ark

for

part

ial

stru

ctur

e an

d so

me

linka

ges

and

lines

of

reas

onin

g).

If t

here

are

no

linka

ges

betw

een

poin

ts,

the

sam

e fiv

e in

dica

tive

mar

king

poi

nts

wou

ld y

ield

an

over

all s

core

of

3 m

arks

(3

mar

ks f

or in

dica

tive

con

tent

and

no

mar

ks

for

linka

ges)

.

6

PMT

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GD

Qu

esti

on

Nu

mb

erA

nsw

erA

dd

itio

nal

Gu

idan

ceM

ark

*3

(a)

con

t.Th

e fo

llow

ing

tabl

e sh

ows

how

the

mar

ks s

houl

d be

aw

arde

d fo

r st

ruct

ure

and

lines

of

reas

onin

g.

Num

ber

of m

arks

aw

arde

d fo

r st

ruct

ure

of a

nsw

er a

nd

sust

aine

d lin

e of

re

ason

ing

Ans

wer

sho

ws

a co

here

nt a

nd

logi

cal s

truc

ture

with

linka

ges

and

fully

sus

tain

ed li

nes

of

reas

onin

g de

mon

stra

ted

thro

ugho

ut

2

Ans

wer

is p

artial

ly s

truc

ture

d w

ith

som

e lin

kage

s an

d lin

es o

f re

ason

ing

1

Ans

wer

has

no

linka

ges

betw

een

poin

ts a

nd is

uns

truc

ture

d0

Ind

icat

ive

con

ten

t:

•te

mpe

ratu

re d

ecre

ase

low

ers

the

rate

of

the

reac

tion

•

beca

use

ther

e ar

e fe

wer

mol

ecul

es/p

articl

es w

ith

EE a

•an

dth

eref

ore

ther

e ar

e fe

wer

suc

cess

fulc

ollis

ions

per

se

cond

•

tem

pera

ture

dec

reas

e in

crea

ses

the

yiel

d (o

f th

e pr

oduc

t)•

beca

use

the

(for

war

d) r

eact

ion

is e

xoth

erm

ic•

low

er r

ate

and

incr

ease

d yi

eld

are

oppo

sing

fac

tors

and

it

is n

ot p

ossi

ble

to t

ell w

hich

has

gre

ater

eff

ect

on o

vera

ll yi

eld

in a

giv

en t

ime.

Acc

ept

slow

s do

wn

the

reac

tion

Acc

ept

shi

fts

the

posi

tion

of

equi

libri

um

to t

he r

ight

/in

forw

ard

dire

ctio

n

PMT

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GA

3(b

)(ii

)

•Δ

rHsh

own

as t

he a

ppro

xim

atel

y ve

rtic

al d

ista

nce

betw

een

reac

tant

s an

d pr

oduc

ts

(1

)

•E a

show

n as

the

app

roxi

mat

ely

vert

ical

dis

tanc

e be

twee

n re

acta

nts

and

peak

of

draw

n cu

rve

for

cata

lyse

d re

action

(1)

ec

f fr

om c

andi

date

’s c

urve

2

3(c

)K

c =

[!"!]![!

!!(!)]!

[!! !

!]![!

!(!)]!

(

1)Sta

te s

ymbo

ls n

ot e

ssen

tial

1

(To

tal

for

Qu

esti

on

3 =

10

mar

ks)

Qu

esti

on

Nu

mb

erA

nsw

erA

dd

itio

nal

Gu

idan

ceM

ark

3(b

)(i)

•cu

rve

star

ting

at

reac

tant

s le

vel,

endi

ng a

t pr

oduc

ts le

vel,

and

peak

ing

low

er t

han

orig

inal

cur

ve

(1

)

1

PMT

Qu

esti

on

Nu

mb

erA

nsw

erA

dd

itio

nal

Gu

idan

ceM

ark

4(a

)(i)

(

1)

Line

nee

ds t

o be

to

the

left

of

E a(n

o ca

taly

st)

but

to t

he r

ight

of

the

hum

p in

the

cu

rve

Sha

ding

not

req

uire

d bu

t ca

n be

incl

uded

an

d us

ed in

(a)

(ii)

1

4(a

)(ii

)An

expl

anat

ion

that

mak

es r

efer

ence

to

the

follo

win

g po

ints

:

•th

ear

ea u

nder

the

cur

ve t

o th

e ri

ght

of t

he a

ctiv

atio

n en

ergy

rep

rese

nts

the

frac

tion

of

mol

ecul

es t

hat

have

E≥

E a

(1

)

•th

e ar

ea un

der

the

curv

e to

th

e ri

ght

of

E a(w

ith

cata

lyst

) is

gre

ater

tha

n th

e ar

ea u

nder

the

cur

ve t

o th

e ri

ght

of E

a(n

o ca

taly

st)

(

1)

•(t

here

fore

)th

ere

are

mor

e su

cces

sful

colli

sion

s (p

er

seco

ndpe

r un

it v

ol.)

(1

)

Acc

ept

desc

ript

ion

that

invo

lves

sha

ded

area

s un

der

the

curv

e

Acc

ept

mor

e m

olec

ules

hav

e en

ough

ene

rgy

to r

eact

3

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GC

PMT

"#$%&'()*+#,-#.)/#0#.)1)2+0$(-#+)345&6+6$%7)89*)6()9:#;6&<%73$;=.#)2&&#&&;#(<)>$<#%6$.&)?)@&&4#)A)?)B#5%4$%7)CDAE)F)"#$%&'()*+4-$<6'()/6;6<#+)CDAE

G1

Qu

esti

on

Nu

mb

erA

nsw

erA

dd

itio

nal

Gu

idan

ceM

ark

4(b

)•

calc

ulat

ion

of t

otal

bon

ds f

orm

ed

(1)

•re

-arr

ange

men

t of

ΔrH

expr

essi

on

(1)

•

eval

uation

of

final

ans

wer

(1)

Cor

rect

ans

wer

, no

wor

king

sco

res

3 m

arks

Exam

ple

of c

alcu

lation

(bon

ds f

orm

ed)

= E

(N≡N

) +

3E(

H–H

)=

944

+ (

3 x

436)

=

225

2 (k

J m

ol1 )

ΔrH

o =

bond

s br

oken

–

(bon

ds f

orm

ed),

so

(bon

ds f

orm

ed)

=

bond

sbr

oken

– Δ

rHo

6E(N

–H)

= 2

252

– (-

92)

E(N

–H)

= (

+)

391

(kJ

mol

1 )

Acc

ept

390.

7

3

(To

tal

for

Qu

esti

on

4 =

7 m

arks

)

PMT

Qu

esti

on

Nu

mb

erA

nsw

erA

dd

itio

nal

Gu

idan

ceM

ark

5(a

)•

calc

ulat

ion

of

tem

pera

ture

cha

nge

and

subs

titu

tion

into

Q =

mcΔ

T

(

1)

•

conv

ersi

on t

o kJ

(1)

•ca

lcul

atio

n of

am

ount

of

copp

er(I

I) s

ulfa

te

(1)

•us

e of

ΔH

=

–  ! !

(1)

•co

rrec

t an

swer

with

sign

and

to

3. s

.f

(

1)

Allo

w e

cffo

r st

eps

in c

alcu

lation

; in

clud

ing

for

final

ans

wer

dep

ende

nt o

n ro

undi

ng in

st

eps

of t

he c

alcu

lation

.

Cor

rect

ans

wer

with

sign

, to

3.s

.f a

nd n

o w

orki

ng s

core

s 5

mar

ks

Mar

k co

nseq

uent

ially

on

inco

rrec

t te

mpe

ratu

re c

hang

e

Exam

ple

of c

alcu

lation

Q =

50

.0 x

4.1

8 x

7.2

(J)

=

1504

.8 (

J)

Q =

 !"#$.!

!"""

= 1

.504

8 (k

J)

mol

esof

cop

per(

II)

sulfa

te

=

4.7

0 ÷

159

.6

= 0

.029

449

mol

ΔH

=

–  ! !

=

– 1.

5048

÷ 0

.029

449

= (

– 51

.099

) =

–

51.1

(kJ

mol

–1)

5

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GI

PMT

"#$%&'()*+#,-#.)/#0#.)1)2+0$(-#+)345&6+6$%7)89*)6()9:#;6&<%73$;=.#)2&&#&&;#(<)>$<#%6$.&)?)@&&4#)A)?)B#5%4$%7)CDAE)F)"#$%&'()*+4-$<6'()/6;6<#+)CDAE

GE

Qu

esti

on

Nu

mb

erA

nsw

erA

dd

itio

nal

Gu

idan

ceM

ark

5(b

)A d

escr

iption

tha

t m

akes

ref

eren

ce t

o th

e fo

llow

ing

poin

ts:

•re

cord

tem

pera

ture

at

tim

ed in

terv

als

(aft

er a

ddin

g th

e so

lid)

(1)

•pl

ot t

hese

tem

pera

ture

val

ues

(on

a gr

aph)

(1)

•ex

trap

olat

e th

e lin

e of

bes

t fit

to

the

tim

e of

add

ing

the

solid

(1)

3

5(c

)•

use

of H

ess

cycl

e or

exp

ress

ion

e.g.

Δ

rH=Δ

rH2

– Δ

rH1

or

ΔrH

=Δ

Hso

lutio

n(an

hydr

ous)

– Δ

Hso

lutio

n(hy

drat

ed)

(1)

•co

rrec

t fin

al a

nsw

er w

ith

sign

(–

49.5

kJ

mol

–1)

(1

)

Cor

rect

ans

wer

with

sign

and

no

wor

king

sc

ores

2 m

arks

2

5(d

)An

expl

anat

ion

that

mak

es r

efer

ence

to

the

follo

win

g po

ints

:

•(t

he p

ale

blue

col

our

show

s th

at)

anhy

drou

s co

pper

(II

) su

lfate

is p

artial

ly h

ydra

ted

(1

)

•th

eref

ore

less

anh

ydro

us c

oppe

r (I

I) s

ulfa

te r

eact

s w

ith

wat

er /

som

e of

the

(bl

ue)

solid

rea

cts

endo

ther

mic

ally

(1)

•th

eref

ore

less

the

rmal

ene

rgy

tran

sfer

red

/he

at e

nerg

y re

leas

ed d

urin

g ex

peri

men

t

or

(exp

erim

enta

l val

ue)

has

a sm

alle

r ne

gative

val

ue

(1)

3

(To

tal

for

Qu

esti

on

5 =

13

mar

ks)

PMT

Qu

esti

on

Nu

mb

erA

nsw

erA

dd

itio

nal

Gu

idan

ceM

ark

6(a

)(i)

•(p

ale)

yel

low

(1)

1

6(a

)(ii

)

•co

nver

sion

to

seco

nds

and

use

of

!!"#

$  !"  !"#

(

1)

•

divi

de b

y lo

wes

t to

get

who

le n

umbe

r ra

tio

(

1)

Exam

ple

of c

alcu

lation

!!"#$

:

! !!!

:! !

or 8

.00

x 10

4 :

1.

80 x

10

3 :

0.

20

1 :

2.25

: 2

50

2

6(a

)(ii

i)An

expl

anat

ion

that

mak

es r

efer

ence

to

the

follo

win

gpo

ints

:

•th

e C

–Cl b

ond

is s

hort

erth

an t

he C

–I b

ond

/ th

e C

l ato

m is

sm

alle

rth

an t

he I

ato

m

(1)

•th

eref

ore,

the

C–C

l bon

d is

str

onge

rth

an t

he C

–Ibo

nd

(1)

•th

eref

ore,

the

C–C

l bon

d ne

eds

mor

een

ergy

to

brea

k, a

nd s

o th

e re

action

is s

low

er

(1

)

Allo

w r

ever

se a

rgum

ent

thro

ugho

ut

3

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GJ

PMT

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G!

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ceM

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6(b

)(i)

D

1

6(b

)(ii

)•

H/

(CH

3)3

CBr

(1)

•be

caus

e C

of

C–B

r is

joi

ned

to 3

alk

yl

grou

ps/j

oine

dto

3 c

arbo

n at

oms

(1

)

2

6(b

)(ii

i)•

num

ber

pref

ix u

sed

shou

ld b

e as

low

as

poss

ible

/

brom

ine

atom

is o

n th

e fir

st c

arbo

n at

om o

f th

e ch

ain

(1)

1

(To

tal

for

Qu

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on

6 =

10

mar

ks)

PMT

Qu

esti

on

Nu

mb

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nsw

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dd

itio

nal

Gu

idan

ceM

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7(a

)(r

ed-)

brow

n to

col

ourl

ess

(1)

Allo

w o

rang

e fo

r re

d-br

own

Do

not

allo

w c

lear

for

col

ourl

ess

1

7(b

)(i)

elec

trop

hilic

(ad

dition

)

(

1)1

7(b

)(ii

)

•pa

rtia

l cha

rges

on

Br

atom

s

(

1)

•cu

rly

arro

w f

rom

C=

C b

ond

to (

+)

Br

atom

AN

D

curl

y ar

row

fro

m B

r–Br

bond

to

(–)

Br

atom

(1)

•st

ruct

ure

of c

arbo

cation

with

+ c

harg

e

(

1)

•cu

rly

arro

w f

rom

Br

to C

with

+ c

harg

e AN

D

stru

ctur

e of

1,

2-di

brom

obut

ane

prod

uct

(1

)

Acc

ept

cycl

ic b

rom

oniu

m io

n w

ith

+ c

harg

e sh

own

in c

entr

e of

rin

g

4

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GG

PMT

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GH

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nsw

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itio

nal

Gu

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ceM

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7(c

)(i)

•ca

lcul

atio

n of

% b

y m

ass

of B

r

(1)

•ev

alua

tion

of

num

ber

of m

oles

of

C,

H,

and

Br

(1)

•fin

al e

mpi

rica

l for

mul

a ba

sed

onw

hole

num

ber

ratio

(1)

Exam

ple

of c

alcu

lation

%Br

= 1

00 –

34.

9 –

6.60

= 5

8.5%

C!".! !"

= 2

.91

H

!.!

" !=

6.6

0

Br

!".! !".! =

0.7

32

2.91

=

3

.98

6.

60

=

9.02

0

.732

=

1

0.73

2

0.73

2

0.73

2

i.e.

C4H

9Br

3

7(c

)(ii

)•

2–br

omob

utan

e

(1

)1

PMT

Qu

esti

on

Nu

mb

erA

nsw

erA

dd

itio

nal

Gu

idan

ceM

ark

7(d

)An

answ

er t

hat

incl

udes

at

leas

t on

e si

mila

rity

and

one

di

ffer

ence

to

gain

max

imum

mar

ks:

Sim

ilari

ties

:•

both

invo

lve

the

over

lap

of t

wo

(ato

mic

) or

bita

ls

(eac

h co

ntai

ning

a s

ingl

e el

ectr

on)

(1)

•bo

th in

volv

e an

ele

ctro

stat

ic a

ttra

ctio

n be

twee

n a

bond

ing

pair

of

elec

tron

s an

d tw

o nu

clei

(1

)

Diff

eren

ces:

•th

e b

ond

repr

esen

ts e

lect

ron

dens

ity/

elec

tron

s be

twee

n th

e tw

o ca

rbon

ato

ms/

nucl

ei/e

nd-o

nov

erla

ppin

g of

ato

mic

orb

ital

s

(1

)

•th

e b

ond

repr

esen

ts e

lect

ron

dens

ity/

elec

tron

s ab

ove

and

belo

w t

he

bon

d/ s

idew

ays

over

lapp

ing

of o

rbital

s

(

1)

•th

e 𝜎𝜎

bond

is f

orm

ed b

y th

e ov

erla

p of

a s

ingl

e lo

be f

rom

one

(at

omic

) or

bita

l with

a si

ngle

lobe

fr

om t

he a

noth

er

(ato

mic

) or

bita

l

(1

)

•th

e 𝜋𝜋

bond

is f

orm

ed b

y th

e ov

erla

p of

tw

o lo

bes

from

one

(at

omic

) or

bita

l with

two

lobe

s of

an

othe

r (a

tom

ic)

orbi

tal

(1)

Acc

ept

elec

tros

tatic

attr

action

bet

wee

n a

shar

edpa

ir o

f el

ectr

ons

and

two

nucl

ei

Acc

ept

end-

to-e

nd/h

ead-

on/a

xial

ove

rlap

ping

3

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HD

PMT

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HA

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dd

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ceM

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7(e

)

(1

)

Do

not

acce

pt a

str

uctu

ral o

r di

spla

yed

form

ula.

1

(T

ota

l fo

r Q

ues

tio

n 7

= 1

4 m

arks

)

PMT

Qu

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Nu

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ceM

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8(a

)•

PC

H3C

H2C

H2C

H2C

H2O

H

(1)

•Q

(CH

3)2C

HC

H(O

H)C

H3

(1

)

•R

(CH

3)2C

(OH

)CH

2CH

3

(1)

Acc

ept

disp

laye

d fo

rmul

ae3

8(b

)•

m/z

= 4

5 is

due

to

CH

3CH

(OH

)+(1

)

•m

/z=

43

is d

ue t

o C

H3C

H2C

H2+

(1

)

•th

eref

ore

the

stru

ctur

e is

CH

3CH

2CH

2CH

(OH

)CH

3

(1

)

Pena

lise

the

lack

of

a pl

us s

ign

once

onl

y

(CH

3)2C

HC

H(O

H)C

H3

not

allo

wed

as

it is

Q

3

8(c

)(i)

An

answ

er t

hat

mak

es r

efer

ence

to

the

follo

win

g po

ints

:

•Y

is a

ket

one

beca

use

the

spec

trum

con

tain

s an

ab

sorp

tion

for

C=

O (

carb

onyl

gro

up)

1700

– 1

720

cm-1

(1

)

•Z

is a

car

boxy

lic a

cid

beca

use

the

spec

trum

con

tain

s ab

sorp

tion

s fo

r C

=O

172

5-17

00 c

m-1

AN

D O

–H (

acid

s) 3

300

– 25

00 c

m-1

(1)

Abs

ence

of

bond

s or

abs

ence

of

wav

enum

ber

rang

es m

axim

um 1

Allo

w v

alue

s w

ithi

n th

e ra

nges

2

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HC

PMT

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H1

Qu

esti

on

Nu

mb

erA

nsw

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dd

itio

nal

Gu

idan

ceM

ark

8(c

)(ii

)•

calc

ulat

ion

of m

ass

of c

arbo

n fr

om m

ass

of C

O2

(1)

•ca

lcul

atio

n of

mas

s of

hyd

roge

n fr

om m

ass

of H

2O(1

)

•su

btra

ctio

n to

fin

d m

ass

of O

, an

d ev

alua

tion

of

num

ber

of m

oles

of

C,

H a

nd O

(1

)

•co

nfir

m w

hole

num

ber

ratio

(1

)

Exam

ple

of c

alcu

lation

Mas

s of

car

bon

= (!" !!

x 5.

89)

= 1

.606

g

Mas

s of

hyd

roge

n =

(! !"

x 2.

44)

= 0

.271

g

Mas

s of

oxy

gen

= 2

.74

– 1.

606

– 0.

271

= 0

.863

g

C!.!

"! !"=

0.1

34

H!.!

"# !=

0.2

71

O!.!

"# !"=

0.0

54

Rat

io =

2.4

8 :

5.02

: 1

= 5

: 1

0 :

2

4

(To

tal

for

Qu

esti

on

8 =

12

mar

ks)

PMT

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PMT