chemistry 30 - alberta education · 2015-11-26 · alberta education, assessment sector 5 chemistry...

Chemistry 302013 Classroom Assessment Materials and Examples

This document was written primarily for:

Students

Teachers of Chemistry 30

Administrators

Parents

General Audience

Others

For further information, contact Tim Coates ([email protected]) or Jack Edwards ([email protected]) at the Assessment Sector, or call (780) 427-0010. To call toll-free from outside Edmonton, dial 310-0000.

Our Internet address is www.education.alberta.ca.

Copyright 2013, the Crown in Right of Alberta, as represented by the Minister of Education, Alberta Education, Assessment Sector, 44 Capital Boulevard, 10044 108 Street NW, Edmonton, Alberta T5J 5E6, and its licensors. All rights reserved.

Special permission is granted to Alberta educators only to reproduce, for educational purposes and on a non-profit basis, parts of this document that do not contain excerpted material.

Excerpted material in this document shall not be reproduced without the written permission of the original publisher (see credits, where applicable).

ContentsIntroduction ......................................................................................................................................1

Assessment of Communication Skills in the Classroom ..................................................................2

Communication Guidelines for Classroom Assessment ...................................................................3

Significant Digits ..............................................................................................................................4

Analytic Communication Scoring Guide .........................................................................................6

Explanation of the Holistic Scoring Guide .......................................................................................6

Holistic Scoring Rubrics ...................................................................................................................7

Sample Written-Response Questions and Responses for Classroom Assessment ............................8

Sample Student Responses and Rationales .....................................................................................13

Sample Written-Response Questions for Classroom Assessment ...................................................20

Alberta Education, Assessment Sector 1 Chemistry 30

Introduction

The items here are intended to promote the use of extended-response items in high-quality classroom assessments. Teachers may wish to use these items in a variety of ways to improve the degree to which students develop and demonstrate an understanding of the concepts described in the Chemistry 30 Program of Studies.

http://education.alberta.ca/teachers/program/science/programs.aspx


Assessment of Communication Skills in the ClassroomAlthough written-response questions are no longer part of the Chemistry 30 Diploma Examination, we encourage the use of written-response questions in classroom assessment. In this way, there can be a greater assessment of the outcomes included in the program of studies.

The following pages give examples of written-response questions, together with scoring rubrics and examples of student work.

Chemistry is a discipline in which there is a stringent set of rules for proper scientific communication. Communication skills are most evident in, and can be directly assessed on, the written-response questions.

In previous years, we have scored analytic-style written-response questions out of 6, with 5 marks for chemistry content and 1 mark for communication. The communication mark is partially determined by the extent to which the question has been attempted. Communication is marked based on organization, clarity, use of correct scientific conventions, and use of proper language conventions.

Proper scientific conventions include:

• labelling of graphs and diagrams

• mathematical formulas and equations

• significant digits, units of measurement, and unit conversion

• states of matter

• abbreviations

In previous years, we have scored holistic written-response questions using the holistic scoring rubrics, which integrate the assessment of communication skills into the marking matrix that is used to assess the overall response.

Therefore, on the analytic questions, communication skills can be assessed more independently; whereas on the holistic question, communication can be assessed as part of the total response.

The intent of evaluating communication is to reward students for creating responses that are on topic, clear, concise, and well written using the conventions of scientific language.


Communication Guidelines for Classroom AssessmentThe following list is a set of guidelines that were used during the marking of the communication scale for any of the written-response questions.

• Do not score work that the student has indicated should not be scored—this includes partially erased or clearly crossed-out work.

• If a student’s response contains contradictory information, then score the work as either ambiguous or incorrect.

• Do not score any irrelevant and extra information that is not incorrect but that does not contribute to the correct response.

• The omission of leading zeros is not a scientific error and therefore will not be scored as such.

• States, units, significant digits, and ion charges must be included within the response. The student must be consistent in their use. (The exception to this is equilibrium expressions, which do not require units.) Units used should respect the conventions of the International System of Units (SI).

• Significant digits in the final recorded answer must be correct. It is not necessary to carry extra significant digits in intermediate steps, but it is a preferred practice to carry at least one extra digit throughout intermediate steps. If the number of significant digits in intermediate steps has been truncated (is less than the required number), then this will be considered an error.

• If spelling and grammatical errors limit the understanding of the response and cause ambiguity, then this will be considered a communication error.

• Graphs should include an appropriate title, labelled axes with units, and an appropriate scale. The manipulated variable should always be on the x-axis.

• When the student is asked to draw a diagram of a cell, the diagram should include labels for the anode, cathode (or the specific substances), reagent solutions (electrolytes), a salt bridge or porous cup, voltmeter or power source, and a connecting wire to the electrodes. Students are not required, unless specifically asked, to label the migration of ions, the solution in the salt bridge (although the diagram or procedure should indicate that there is a solution present), or the electron flow. If a student chooses to include these labels, then they will also be marked as part of the response.

• The y-axis on an energy diagram can be labelled Ep, H, or ∆H with appropriate units. However, on diploma examinations and field tests the y-axis will be labelled Ep (kJ).

• Portions of a response not assessed for chemistry marks will not be assessed for communication.


Significant DigitsThe examples below illustrate the proper use of significant digits for the Chemistry 30 Diploma Examination in a response.

Example 1

A 10.0 mL sample of a Fe2+(aq) solution of unknown concentration is titrated with a standardized 0.120 mol/L KMnO4(aq) solution. The following data are recorded.

Trial I II III

Final burette reading (mL) 10.10 19.22 28.33

Initial burette reading (mL) 1.00 10.10 19.22

Titrant added (mL) 9.10* 9.12 9.11

The concentration of the Fe2+(aq) is __________.

Reaction Equation

MnO4–(aq) + 8 H+(aq) + 5 Fe2+(aq) → Mn2+(aq) + 4 H2O(l) + 5 Fe3+(aq)

Average volume of titrant added is 9.11 mL

[Fe2+(aq)] = 9.11 mL MnO4–(aq) × 0.120 mol/L MnO4

–(aq) × 5 mol Fe2+(aq)

×1

1 mol MnO4–(aq) 10.0 mL Fe2+(aq)

[Fe2+(aq)] = 0.547 mol/L

Example 2

The pH of a 0.100 mol/L solution of ethanoic acid is __________.

Ka = 1.8 × 10–5 =

xx

(0.100 mol/L )

2

-

The value of x can be ignored when compared to 0.100 mol/L in the case of such a weak acid.

Ka is approximately x0.100 mol/L

2

x = [H3O+(aq)] = 0.001342pH = –log(0.001342 mol/L) = 2.87

2 decimal places (10.10–1.00)

according to the addition/subtraction

rules leaves 3 significant digits

* Final answer has 3 significant digits (least number present according to the multiplication/division rule)

Additional digits carried through on an interim basis

Final answer has 2 significant digits

Exact number, therefore does not change the final number of significant digits

Ka value has 2 significant digits


Example 3

A student conducts a calorimetry experiment to determine the energy transferred when Solution A is mixed with Solution B. The data collected are shown below. Assume the specific heat capacity for each solution is the same as that of water.

Mass of Solution A (g) 100.0

Mass of Solution B (g) 100.0

Mass of final solution mixture (g) 200.0

Initial temperature of solutions A and B (°C) 20.0

Final temperature of the solution mixture (°C) 23.0

DH = mcDt

DH = (200.0 g) (4.19 J/g.°C) (3.0 °C)

DH = 2.5 kJ

The original data are limited to 3 significant digits.

The final answer should be rounded to the same number

of significant digits contained in the input data for the

calculation DH = mcDt that has the fewest number of

significant digits.

The final answer has 2 significant digits because the input data for the DH = mcDt calculation is limited by the temperature difference of 3.0 °C, which has 2 significant digits.

The resulting temperature has 2 significant digits.


Analytic Communication Scoring Guide

Communication Scoring Guide for Closed-Response (Analytic) Questions

Score Criteria

1

The teacher does not have to interpret any part of the response, and no reference to the question is needed to understand the response. The response is clear, concise, and presented in a logical manner. Scientific conventions have been followed. The response may contain a minor error.

0Ambiguous

or >1 scientific

error

The teacher has to interpret the response (ambiguous) or the response is so poorly organized that the marker has to refer to the question in order to understand the response. The response may be ambiguous, incomprehensible, and/or disorganized, and/or contains errors (more than one) in scientific conventions.

0 50% or less of the question has been attempted. There is not enough of a response present to be able to score for communication.

NR No response given.

Scientific conventions to be followed:

• Correct, appropriate units are used throughout the response.

• States are given throughout the response except in calculation labels and when a formula replaces a word in a sentence.

• Significant digits are used throughout the response.

• Branch names in organic molecules are in alphabetical order, using lowest possible numbers to indicate the position of branches.

*Note: Content and communication are scored on separate scales for the analytic question.

Explanation of the Holistic Scoring Guide

Holistic questions are designed so that students can demonstrate their understanding of science from more than one valid approach or perspective and are assessed in a holistic fashion. The holistic question will be scored on two rubrics that combine to form a 5-point scale.

Teachers must read the student response in its entirety in order to decide whether it contains the key component(s) for the particular question. The teacher then looks for the necessary support details. These two aspects are used to assess the quality of students’ responses. Communication skills and scientific conventions are considered in the determination of the overall quality of the key component(s) and support.


Holistic Scoring Rubrics

Score Key–no Key Criteria

1 Key

(weight of 2)

The student has addressed the key component(s) of the question asked. The key component(s) of the question can be found in the stem of the question.

0 No Key The student has not addressed the key component(s) of the question asked.

Score Support

3 Very Good to

Excellent

The student has provided good support for all of the bullets. The support may include a minor error/weakness in one of the support bullets.

2 Satisfactory

to Good

The student has provided support for the majority of the bullets but not necessarily all of the bullets. The support provided may contain minor errors/weaknesses. There is more correct than incorrect support.

1 Minimal

The student has provided minimal support for one or more of the bullets, but there are many errors throughout. There is more incorrect than correct support.

0 Limited to No support

The student has not provided enough support to demonstrate more than a limited understanding. The support is either off topic or contains major errors throughout all of the bullets.


Sample Written-Response Questions and Responses for Classroom Assessment

Use the following information to answer the first question.

Sour gas contains a significant amount of hydrogen sulfide gas mixed with methane gas. Hydrogen sulfide gas is a colourless, toxic gas that smells like rotten eggs. Hydrogen sulfide gas can be converted to sulfur dioxide gas in a process called flaring, as represented by the equation below.

2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g)

1. a. Determine the enthalpy change for the flaring process represented by the equation above.

(3 marks)

b. Sketch and label a potential energy diagram that represents the enthalpy change for the flaring process.

(2 marks)

Use the following information to answer the next question.

Large amounts of ammonia for the production of fertilizers and other consumer goods are made by the Haber process. During the Haber process, hydrogen gas combines with nitrogen gas to produce ammonia gas. This process is carried out in the presence of a catalyst.

2. a. Write a balanced equilibrium equation for the Haber process. Include the enthalpy change as an energy term in the balanced equation.

(3 marks)

b. Describe what happens to the equilibrium position and the value of the equilibrium constant when the temperature of the system is increased from 200 °C to 500 °C.

(2 marks)



The copper covering on the hull of a ship, which is the main body of the ship that is in contact with water, corrodes when it is exposed to water and oxygen. To protect against such corrosion on British naval ships, Sir Humphry Davy was the first to use blocks of either zinc, tin, or iron as sacrificial anodes, which were attached to the ship’s hull.

Explain how a block of zinc, tin, or iron would prevent the corrosion of the copper on a ship’s hull.

Your response should include

• an explanation of the corrosion of copper

• an explanation of how a block of zinc, tin, or iron protects the copper from corrosion

• relevant balanced equations and E°cell calculations to support each of your explanations

3.


Ana

lyti

c-St

yle

Wri

tten

-Res

pons

e Q

uest

ion

Sam

ple

Res

pons

e

Que

stio

nM

arks

Sam

ple

Res

pons

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omm

ents

1.a.

32

H2S

(g)

+ 3

O2(

g) →

2SO

2(g)

+ 2

H2O

(g)

∆H°

= ∑

n∆fH

° (pro

duct

s) –

∑n∆

fH° (r

eact

ants

)

= [(

2 m

ol)(

–296

.8 k

J/m

ol)

+ (2

mol

)(–2

41.8

kJ/

mol

)]

– [(2

mol

)(–2

0.6

kJ/m

ol)

+ (3

mol

)(0

kJ/m

ol)]

= (–

1 07

7.2

kJ)

– (–

41.2

kJ)

= –

1 03

6.0

kJ

• 1 m

ark

for

corr

ect m

etho

d

• 1 m

ark

for

subs

titut

ion

co

nsis

tent

with

met

hod

• 1 m

ark

for

corr

ect a

nsw

er

1.b.

2C

ombu

stio

n of

H2S

(g)

Not

e: C

an a

lso

be la

belle

d re

acta

nts

and

prod

ucts

.

• 1 m

ark

for

corr

ect l

abel

s

• 1 m

ark

for

shap

e of

gra

ph

cons

iste

nt w

ith c

alcu

latio

n

1C

omm

unic

atio

n—Se

e G

uide

Use

Ana

lytic

Sco

ring

Gui

de

Tota

l pos

sibl

e m

arks

= 6


Ana

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c-St

yle

Wri

tten

-Res

pons

e Q

uest

ion

Sam

ple

Res

pons

e

Que

stio

nM

arks

Sam

ple

Res

pons

eC

omm

ents

2.a.

33

H2(

g) +

N2(

g) ⇌

2N

H3(

g) +

91.

8 kJ

• 1 m

ark

for

bala

nced

equ

atio

n

• 1 m

ark

for

the

corr

ect e

ntha

lpy

valu

e

• 1 m

ark

for

the

incl

usio

n of

the

enth

alpy

term

on

the

corr

ect

side

2.b.

2T

he e

quil

ibri

um p

ositi

on w

ould

shi

ft to

war

d th

e re

acta

nts

beca

use

the

forw

ard

reac

tion

is e

xoth

erm

ic, a

nd th

e K

c va

lue

wou

ld d

ecre

ase.

• 1 m

ark

for

corr

ect s

hift

in

equi

libri

um c

onsi

sten

t with

en

thal

py te

rm

• 1 m

ark

for

a ch

ange

in K

c co

nsis

tent

with

the

shif

t

1C

omm

unic

atio

n—Se

e G

uide

Use

Ana

lytic

Sco

ring

Gui

de

Tota

l pos

sibl

e m

arks

= 6


Hol

isti

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Wri

tten

-Res

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Res

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Que

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arks

Sam

ple

Res

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omm

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3.

Cor

rosi

on E

xpla

nati

on

The

cor

rosi

on o

f co

pper

is th

e sp

onta

neou

s ox

idat

ion

reac

tion

that

occ

urs

whe

n co

pper

re

acts

with

wat

er a

nd o

xyge

n. S

olid

cop

per

is o

xidi

zed

to C

u2+(a

q).

O

2(g)

+ 2

H2O

(l)

+ 4

e– → 4

OH

– (aq)

E

° =

+0.

40 V

C

u(s)

→ C

u2+(a

q) +

2e–

E°

= +

0.34

V

O2(

g) +

2H

2O(l

) +

2C

u(s)

→ 4

OH

– (aq)

+ 2

Cu2+

(aq)

E

° cel

l = +

0.06

V

O

R →

2C

u(O

H) 2

(s)

Sacr

ifici

al A

node

Exp

lana

tion

The

met

al fo

und

in th

e sa

crifi

cial

ano

de p

reve

nts

the

corr

osio

n of

cop

per

beca

use

it (Z

n, S

n, o

r Fe

) is

a s

tron

ger

redu

cing

age

nt th

an c

oppe

r an

d th

e m

etal

und

ergo

es

oxid

atio

n be

fore

the

copp

er.

If b

oth

iron

and

cop

per

are

pres

ent w

ith w

ater

and

oxy

gen,

the

reac

tion

that

occ

urs

is th

e fo

llow

ing.

O2(

g) +

2H

2O(l

) +

2Fe

(s) →

4O

H– (a

q) +

2Fe

2+(a

q)

E°

= +

0.85

V

O

R →

2Fe

(OH

) 2(s

)

Key

Com

pone

nt

• exp

lana

tion

that

Fe

(s),

Sn(s

)

or Z

n(s)

rea

cts

spon

tane

ousl

y w

ith th

e ox

idiz

ing

agen

t be

fore

Cu(

s)

Supp

ort

• exp

lana

tion

of

the

corr

osio

n

of c

oppe

r

• exp

lana

tion

of

the

sacr

ifici

al

anod

e

• rel

evan

t eq

uatio

ns

and

pote

ntia

l di

ffer

ence

ca

lcul

atio

n


Sample Student Responses and Rationales

A number of sample responses to written-response questions are given on the following pages. For each, the score awarded and a rationale for that score are also provided.

Written-Response Questions—Student Response 1

Score—6 out of 6 (5 for content and 1 for communication)

RationaleThis example meets the requirements of the standard of excellence.

Rationale The correct method for determining ∆H is used, but the final answer does not have the correct number of significant digits. The enthalpy diagram has the correct shape and is labelled.

The response received 1 mark for communication because the only error in communication was in part a with the significant digits in the final answer.




RationaleThis example meets the requirements of the acceptable standard.

The correct method for determining ∆H is used, but the units included are incorrect, as are the significant digits. The shape of the enthalpy diagram is correct but the diagram is missing a title.

The response received 0 marks for communication because of the errors in significant digits and in units.




RationaleThis example meets the minimum requirements of the acceptable standard.

The equation was correctly balanced with an equilibrium arrow. The enthalpy calculation appears to have been done correctly, but an incorrect value was substituted into the equation, and was on the wrong side of the equation. The shift and the change in the Kc were consistent with the student’s equation.

The response received 0 for communication, as units were incorrect more than once.




RationaleThis example does not meet the requirements of the acceptable standard.

The equation was not correctly balanced and did not have an equilibrium arrow. The value for the enthalpy change was correct. The enthalpy change was on the wrong side of the equation. The change in the value of the equilibrium constant was incorrect and no shift was indicated.

The response received 1 mark for communication, as states and units were correct.



Score—5 out of 5(Holistically Scored)

RationaleThis example meets the requirements of the standard of excellence.

The response addresses the question asked by making reference to the sacrificial anode reacting instead of copper (key component). The response explains why the sacrificial anode reacts instead of copper and it is supported with correct equations and potential difference calculations. Scientific conventions are followed and the information is clearly communicated.

A total score of 5 was awarded (2 for key component and 3 for support).




RationaleThis example just meets the minimum requirements of the acceptable standard.

The student has addressed the question by making reference to the sacrificial anode in a comparison to copper in a minimal way (key component). The student does not provide correct equations, or an explanation of why the sacrificial anode reacts instead of copper. The student does provide a potential difference calculation for the equation the student provided. The student has provided minimal support for one of the bullets, but there is more incorrect than correct support.





RationaleThis example does not meet the requirements of the acceptable standard.

The student has not addressed the question as the student has not made a correct reference to Zn as a sacrificial anode in comparison to copper. The student does not provide correct equations or an explanation of why zinc reacts instead of copper. The student does provide a potential difference calculation for the equation they provided. The student has provided an explanation of the corrosion of copper, but there is more incorrect than correct support.



Sample Written-Response Questions for Classroom Assessment


A student titrated 10.0 mL of a 0.10 mol/L solution of sodium hypochlorite, NaOCl(aq), with 0.10 mol/L HCl(aq) and recorded the pH. The data are shown in the table below.

Volume of HCl(aq) (mL) pH

0.0 10.2

2.0 8.0

4.0 7.6

6.0 7.2

8.0 6.8

10.0 4.2

12.0 2.0

14.0 1.8

16.0 1.6

18.0 1.5

20.0 1.5

Written Response—10%

a. On the grid below, plot and label the data given above. Identify a buffering region. (3 marks)

1.



Decorative copper coatings can be applied to objects, such as a vase, using an electrolytic process.


a. Write the two half-reaction equations that occur in this cell, and identify one observation that could be made during the operation of the cell.

(3 marks)

2.

b. Describe the role of a buffer, and write the net ionic equation that represents the buffer reaction that occurs in the buffer region that you identified on your graph.

(2 marks)

b. Calculate the mass of copper that could be produced in the electrolytic cell when a current of 3.00 A is applied for 20.0 min.

(2 marks)



A cutting torch can be used to cut metal in places such as construction sites, shipyards, and rail yards. The flame used for cutting is produced by a combustion reaction in which a gaseous fuel is mixed with oxygen and forced through a nozzle in the cutting torch. Gaseous products are formed.

Two gases that can be used as fuel in a cutting torch are hydrogen gas, H2(g), and ethyne gas, C2H2(g), which is commonly known as acetylene.

Chem30_0522 Cutting torch


Compare, in terms of enthalpy, the use of hydrogen gas and ethyne gas as fuels, and identify which of the two is the better fuel to use in a cutting torch.


• a balanced chemical equation representing the combustion reaction for each fuel

• enthalpy calculations for the combustion of 1.00 g of each fuel in kJ/g

• identification of the better fuel in terms of energy per gram of fuel and one other reason for choosing the fuel

3.


Mar

ker

ID N

umbe

r ___

____

_

Que

stio

n 1

– H

olis

tic

Scor

ing

Cri

teri

a (J

une

2009

)

Que

stio

nM

arks

Sam

ple

Res

pons

eC

omm

ents

Bub

ble

#

1.a.

3T

itra

tion

of

NaO

Cl(a

q) w

ith

HC

l(aq)

0 2

4 6

8 10

12

14

16

18

20

0 2 4 6 8 10

12

Not

e: th

e bu

ffer

reg

ion

as s

how

n on

thei

r gr

aph,

and

not

a s

ingl

e po

int

• 1

mar

k fo

r tit

le a

nd

axis

labe

ls (w

ith

appr

opri

ate

scal

e)

• 1

mar

k fo

r pl

otte

d po

ints

(or

corr

ect

curv

e)

• 1

mar

k fo

r la

bell

ing

the

buff

er r

egio

n,

appr

ox. 2

-8 m

L,

(con

sist

ent w

ith g

raph

be

fore

equ

ival

ence

po

int)

1

1.b.

2A

buf

fer

mai

ntai

ns a

rel

ativ

ely

cons

tant

pH

(re

sist

s ch

ange

in p

H)

Or

rea

cts

with

a s

tron

g ac

id o

r ba

se to

for

m it

s co

njug

ate

wea

k ac

id/b

ase

pair.

OC

l– (aq)

+ H

3O+

(aq)

→ H

OC

l(aq

) +

H2O

(l)

OR

H

OC

l(aq

) +

H2O

(l)

⇌ O

Cl– (a

q) +

H3O

+(a

q)

Not

e: e

quili

briu

m e

quat

ion

is a

ccep

tabl

e be

caus

e it

can

indi

cate

eith

er d

irec

tion,

an

d H

+(a

q) io

n ve

rsus

H3O

+(a

q) is

acc

epta

ble

• 1

mar

k fo

r de

scri

ptio

n

• 1

mar

k fo

r ne

t io

nic

equa

tion

cons

iste

nt

with

labe

l on

grap

h

2

1C

omm

unic

atio

n—Se

e G

uide

Use

Ana

lytic

Sco

ring

G

uide

3

Tota

l pos

sibl

e m

arks

= 6


Mar

ker

ID N

umbe

r ___

____

_

Que

stio

n 2

– H

olis

tic

Scor

ing

Cri

teri

a (J

une

2009

)

Que

stio

nM

arks

Sam

ple

Res

pons

eC

omm

ents

Bub

ble

#

2.a.

3C

atho

de

Cu2+

(aq)

+ 2

e– →

C

u(s)

Ano

de

2H

2O(l

) →

O

2(g)

+ 4

H+(a

q) +

4e–

Obs

erva

tion

s

• B

lue

colo

ur in

sol

utio

n di

sapp

ears

.

• C

oppe

r-co

lour

ed e

lect

ropl

ate

form

s at

the

cath

ode

(vas

e).

• M

ass

of th

e ca

thod

e (v

ase)

incr

ease

s.

• G

as b

ubbl

es f

orm

at t

he a

node

(in

ert C

(s)

elec

trod

e). R

elig

ht g

low

ing

splin

t

• S

olut

ion

beco

mes

mor

e ac

idic

ove

r tim

e (p

H d

ecre

ases

).

• 1

mar

k fo

r ox

idat

ion

half

-rea

ctio

n eq

uatio

n

• 1

mar

k fo

r re

duct

ion

half

-rea

ctio

n eq

uatio

n

• 1

mar

k fo

r on

e ob

serv

atio

n co

nsis

tent

w

ith r

eact

ions

1

2.b.

2C

u2+(a

q) +

2e– →

Cu(

s)

mC

u =

9.

6510

/

(3.0

0C

/s)(

20.0

/)

min

min

Cm

ol

s60

4#

# ×

21 ×

63.

55g/

mol

(0

.037

3 m

oles

of

elec

tron

s)

(0.

0187

mol

es o

f C

u(s)

)

=

1.19

g

• 1

mar

k fo

r m

etho

d

• 1

mar

k fo

r co

rrec

t an

swer

2

1C

omm

unic

atio

n—Se

e G

uide

Use

Ana

lytic

Sco

ring

G

uide

3

Tota

l pos

sibl

e m

arks

= 6


Mar

ker

ID N

umbe

r ___

____

_

Que

stio

n 3

– H

olis

tic

Scor

ing

Cri

teri

a (J

une

2009

)

Mar

ksSa

mpl

e R

espo

nse

Com

men

tsB

ubbl

e #

52

H2(

g) +

O2(

g) →

2H

2O(g

)

∆H°

= (

2 m

ol)(

–24

1.8

kJ/m

ol)

= –

483.

6 kJ

(–

241.

8 kJ

/mol

)

∆H° H

2 =

n∆

H° H

2 =

2m

ol.kJ

483

6-

×

2.02

gm

ol1

= –

120

kJ/g

2C

2H2(

g) +

5O

2(g)

→ 4

CO

2(g)

+ 2

H2O

(g)

∆H

°C2H

2 =

∑n∆

fH° p

rodu

cts −

∑n∆

fH° r

eact

ants

=

.

.m

olm

olkJ

mol

mol

kJ4

393

52

241

8-

+-

__

ii

<F

.

.m

olm

olkJ

mol

molkJ

222

74

50

-+

+_

_i

i<

F =

–2

512.

4 kJ

(–

1 25

6.2

kJ/m

ol)

∆H° C

2H2 =

n∆

H°

=

2m

ol.kJ

2512

4-

×

26.0

4g

mol

1 =

–48

.2 k

J/g

Sam

ple

Con

side

rati

ons

to S

uppo

rt t

he F

uel C

hose

n

H2(

g) –

m

ore

ener

gy p

rodu

ced

per

gram

of

fuel

–

envi

ronm

enta

lly s

afe

fuel

bec

ause

onl

y w

ater

vap

our

is p

rodu

ced

–

H2(

g) is

pro

duce

d fr

om f

ossi

l fue

ls, u

sing

a p

roce

ss th

at c

reat

es

gree

nhou

se g

ases

–

the

light

est o

f al

l the

gas

es

C2H

2(g)

–

mor

e en

ergy

pro

duce

d pe

r m

ole

of f

uel,

or to

tal e

nerg

y re

leas

ed

–

prod

uces

CO

2(g)

, a g

reen

hous

e ga

s th

at c

ontr

ibut

es to

clim

ate

chan

ge

–

read

ily a

vaila

ble,

or

wid

ely

used

, or

burn

s ho

tter

Key

Com

pone

nt

• an

ent

halp

y co

mpa

riso

n of

the

com

bust

ion

of th

e tw

o fu

els

(in

term

s of

H

ess’

s L

aw, k

J/m

ol o

r kJ

/g)

Supp

ort C

ompo

nent

s

• co

rrec

t bal

ance

d eq

uatio

n

• ∆H

° ca

lcul

atio

n in

kJ/g

• id

entifi

catio

n of

the

bette

r fu

el b

ased

on

ene

rgy

and

one

othe

r co

nsid

erat

ion

to s

uppo

rt th

e fu

el

chos

en

1 2


June 2009 Holistic Split Scoring Guide


1 Key

(weighted 2)

The student has addressed the question by comparing the two fuels, hydrogen and ethyne gas, in terms of an enthalpy (enthalpy of combustion) comparison. This may be in terms of kJ/g, kJ/mol, or total kJ produced using the results obtained through their calculations. Note that there may be errors in their calculations: e.g., reversing Hess’s law or missing coefficients, correct calculation of Hess’s Law but incorrect kJ/mol or kJ/g.

0 No Key

The student has not addressed the question asked, or only one fuel has been addressed and there is no comparison.

Score Support

3 Very Good to Excellent


2 Satisfactory

to Good

The student has provided support for the majority of the bullets but not necessarily all of the bullets; the support provided may contain minor errors/weaknesses. There is more correct than incorrect support.

1 Minimal




Minor errors: Unbalanced equation

Method for calculating kJ/g is correct, but the student has made a minor calculation error (the comparison is g/kJ versus kJ/g)

Identifying the best fuel in terms of energy, and then a positive reason for the other fuel

Using H2O(l) instead of H2O(g)

Major errors: Incorrect method for calculating Hess’s Law, or kJ/mol or kJ/g

Incorrect products in the equation, or a fuel other than hydrogen or ethyne is used

Not making a choice for the best fuel in terms of energy



Methane gas can be produced in a laboratory by reacting carbon disulfide, CS2(g), and hydrogen gas, as represented by the equation below.

CS2(g) + 4 H2(g) ⇌ CH4(g) + 2 H2S(g)

The initial concentrations that are placed into an empty flask at a temperature of 90.0 °C are 0.175 mol/L CS2(g) and 0.310 mol/L H2(g). When equilibrium is established, 0.125 mol/L CS2(g) is present.

Written Response—10% (5 marks for content, 1 mark for overall communication)

a. Determine the concentration of hydrogen gas present in the flask at equilibrium.

(3 marks)

b. Write the equilibrium law expression for the reaction, and determine the value of the equilibrium constant for the reaction at 90.0 °C.

(2 marks)

1.


Que

stio

n 1

– A

naly

tic

Scor

ing

Cri

teri

a*

Ple

ase

note

that

thes

e ar

e on

ly s

ampl

e re

spon

ses,

and

oth

er v

aria

tion

s of

the

resp

onse

may

als

o ha

ve r

ecei

ved

full

mar

ks.

Que

stio

nM

arks

Sam

ple

Res

pons

eC

omm

ents

1.a.

3

CS 2

(g)

+

4H

2(g)

⇌

C

H4(

g)

+

2H

2S(g

)

initi

al

0.1

75 m

ol/L

0.

310

mol

/L

0 0

chan

ge

–0.

050

mol

/L –

4 1(0

.050

mol

/L)

+ 1 1

(0.0

50 m

ol/L

) +

2 1(0

.050

mol

/L)

equi

libri

um

0.1

25 m

ol/L

0.

110

mol

/L

0.05

0 m

ol/L

0.

100

mol

/L

The

con

cent

ratio

n of

H2(

g) =

0.1

10 m

ol/L

(3

sign

ifica

nt d

igits

con

tain

ed in

the

orig

inal

dat

a)

Not

e: A

lso

allo

w 0

.11

mol

/L (

0.05

0 m

ay b

e co

nsid

ered

ori

gina

l dat

a)

• 1

mar

k fo

r

corr

ect c

hang

e

in c

once

ntra

tion

(can

be

impl

ied)

• 1

mar

k fo

r co

rrec

t m

ole

ratio

for

H2(

g)

• 1

mar

k fo

r co

rrec

t an

swer

(co

nsis

tent

w

ith c

hang

e

and

mol

e ra

tio)

fo

r H

2(g)

1.b.

2

K

c =

Que

stio

n M

arks

Sa

mpl

e R

espo

nse

Com

men

ts

1.a.

3

C

S 2(g

) +

4H

2(g)

CH

4(g)

+

2H

2S(g

) in

itial

0.17

5 m

ol/L

0.

310

mol

/L

0

0

chan

ge

–0.

050

mol

/L –

4 1(0

.050

mol

/L) +

1 1(0.0

50 m

ol/L

) + 2 1

(0.0

50 m

ol/L

)

equi

libriu

m

0.12

5 m

ol/L

0.

110

mol

/L

0.

050

mol

/L

0.

100

mol

/L

The

conc

entra

tion

of H

2(g)

= 0

.110

mol

/L (3

sign

ifica

nt d

igits

con

tain

ed in

the

orig

inal

dat

a)

Not

e: A

lso

allo

w 0

.11

mol

/L (0

.050

may

be

cons

ider

ed o

rigin

al d

ata)

• 1

mar

k fo

r co

rrec

t cha

nge

in

con

cent

ratio

n (c

an b

e im

plie

d)

• 1

mar

k fo

r cor

rect

m

ole

ratio

for H

2(g)

• 1

mar

k fo

r cor

rect

an

swer

(con

sist

ent

with

cha

nge

an

d m

ole

ratio

) fo

r H2(

g)

1.b.

2

Kc

= 2

42

42

2

[CH

(g)]

[HS(

g)]

[CS

(g)]

[H(g

)]

=

2 4

(0.0

50 m

ol/L

)(0.1

00 m

ol/L

)(0

.125

mol

/L)(0

.110

mol

/L)

=

27.

3 (3

sign

ifica

nt d

igits

con

tain

ed in

the

orig

inal

dat

a)

Not

e: A

lso

allo

w 2

7 (0

.050

may

be

cons

ider

ed o

rigin

al d

ata)

• 1

mar

k fo

r eq

uilib

rium

law

ex

pres

sion

• 1

mar

k fo

r cor

rect

an

swer

con

sist

ent

with

the

conc

entra

tions

ob

tain

ed in

par

t 1a

1

Com

mun

icat

ion—

See

Gui

de

Use

Ana

lytic

Sco

ring

Gui

de

Tota

l pos

sibl

e m

arks

= 6

=

Que

stio

n M

arks

Sa

mpl

e R

espo

nse

Com

men

ts

1.a.

3

C

S 2(g

) +

4H

2(g)

CH

4(g)

+

2H

2S(g

) in

itial

0.17

5 m

ol/L

0.

310

mol

/L

0

0

chan

ge

–0.

050

mol

/L –

4 1(0

.050

mol

/L) +

1 1(0.0

50 m

ol/L

) + 2 1

(0.0

50 m

ol/L

)

equi

libriu

m

0.12

5 m

ol/L

0.

110

mol

/L

0.

050

mol

/L

0.

100

mol

/L

The

conc

entra

tion

of H

2(g)

= 0

.110

mol

/L (3

sign

ifica

nt d

igits

con

tain

ed in

the

orig

inal

dat

a)

Not

e: A

lso

allo

w 0

.11

mol

/L (0

.050

may

be

cons

ider

ed o

rigin

al d

ata)

• 1

mar

k fo

r co

rrec

t cha

nge

in

con

cent

ratio

n (c

an b

e im

plie

d)

• 1

mar

k fo

r cor

rect

m

ole

ratio

for H

2(g)

• 1

mar

k fo

r cor

rect

an

swer

(con

sist

ent

with

cha

nge

an

d m

ole

ratio

) fo

r H2(

g)

1.b.

2

Kc

= 2

42

42

2

[CH

(g)]

[HS(

g)]

[CS

(g)]

[H(g

)]

=

2 4

(0.0

50 m

ol/L

)(0.1

00 m

ol/L

)(0

.125

mol

/L)(0

.110

mol

/L)

=

27.

3 (3

sign

ifica

nt d

igits

con

tain

ed in

the

orig

inal

dat

a)

Not

e: A

lso

allo

w 2

7 (0

.050

may

be

cons

ider

ed o

rigin

al d

ata)

• 1

mar

k fo

r eq

uilib

rium

law

ex

pres

sion

• 1

mar

k fo

r cor

rect

an

swer

con

sist

ent

with

the

conc

entra

tions

ob

tain

ed in

par

t 1a

1

Com

mun

icat

ion—

See

Gui

de

Use

Ana

lytic

Sco

ring

Gui

de

Tota

l pos

sibl

e m

arks

= 6

=

27.3

(3

sign

ifica

nt d

igits

con

tain

ed in

the

orig

inal

dat

a)

Not

e: A

lso

allo

w 2

7 (0

.050

may

be

cons

ider

ed o

rigi

nal d

ata)

• 1

mar

k fo

r eq

uilib

rium

law

ex

pres

sion

• 1

mar

k fo

r co

rrec

t ans

wer

co

nsis

tent

with

th

e co

ncen

trat

ions

ob

tain

ed in

par

t 1.a

.

1C

omm

unic

atio

n—Se

e G

uide

Use

Ana

lytic

Sco

ring

G

uide

Tota

l pos

sibl

e m

arks

= 6



Potassium permanganate in an acidic solution is commonly used in redox titrations.


a. Choose an aqueous solution that could be titrated with aqueous potassium permanganate in an acidic solution. Write the balanced net ionic equation for the titration reaction and determine the potential difference for the reaction. (3 marks)

Use the following additional information to answer the next part of the question.

A student places a strip of solid iron metal into a beaker that contains aqueous potassium permanganate in an acidic solution.

b. Describe what happens to the iron metal strip when it is added to the solution in the beaker, and identify the oxidizing agent and the reducing agent.

(2 marks)

2.


Que

stio

n 2

– A

naly

tic

Scor

ing

Cri

teri

a*

Ple

ase

note

that

thes

e ar

e on

ly s

ampl

e re

spon

ses,

and

oth

er v

aria

tion

s of

the

resp

onse

may

als

o ha

ve r

ecei

ved

full

mar

ks.

Que

stio

nM

arks

Sam

ple

Res

pons

eC

omm

ents

2.a.

3R

educ

tion

Hal

f-R

eact

ions

:

M

nO4– (a

q) +

8H

+(a

q) +

5e– (

aq)

⇌

Mn2+

(aq)

+ 4

H2O

(l)

E°

= +

1.51

V

Fe

3+(a

q) +

e– ⇌

Fe

2+(a

q)

E°

= –

0.77

V

net

MnO

4– (aq)

+ 8

H+(a

q) +

5Fe

2+(a

q) →

5

Fe3+

(aq)

+ M

n2+(a

q) +

4H

2O(l

)

E° c

ell

= +

1.51

V –

0.7

7 V

=

+0.

74 V

• 1

mar

k fo

r su

itabl

e re

duci

ng a

gent

• 1

mar

k fo

r ba

lanc

ed

net i

onic

equ

atio

n

• 1

mar

k fo

r po

tent

ial

diff

eren

ce c

onsi

sten

t w

ith n

et io

nic

equa

tion

2.b.

2O

xidi

zing

age

nt is

aci

difie

d K

MnO

4(aq

) (o

r M

nO4– (a

q)).

Red

ucin

g ag

ent i

s Fe

(s).

Ele

ctro

ns a

re tr

ansf

erre

d fr

om th

e ir

on s

trip

to th

e ox

idiz

ing

agen

t,

MnO

4– (aq)

(or

MnO

4– (aq)

+ H

+(a

q)).

or The

iron

str

ip is

oxi

dize

d (m

ass

decr

ease

s) b

y th

e M

nO4– (a

q) s

olut

ion.

(The

iron

met

al r

eact

ed.)

• 1

mar

k fo

r id

entif

ying

M

nO4– (a

q) +

H+(a

q) a

s O

A, a

nd F

e(s)

as

RA

• 1

mar

k fo

r de

scri

ptio

n of

wha

t hap

pens

to th

e ir

on s

trip

1C

omm

unic

atio

n—Se

e G

uide

Use

Ana

lytic

Sco

ring

G

uide

Tota

l pos

sibl

e m

arks

= 6



A student is investigating the amount of carbon dioxide that is produced during the combustion of fossil fuels commonly used in automobiles. In order to identify the best fuel to use in automobiles, the student determines the enthalpy change needed to produce gaseous products from the combustion of each of the following three fuels: propane gas, liquid octane, and liquid ethanol.


Compare the three fuels listed above in terms of energy and the production of carbon dioxide gas, and identify the best fuel to burn in automobiles.

To support your choice, your response should include

• balanced chemical equations for the combustion of each fuel and an enthalpy calculation for each reaction

• the number of moles of carbon dioxide produced per kilojoule of energy produced by each fuel

• an explanation of the rationale that supports the fuel you have identified as the best one to use in automobiles

3.


Que

stio

n 3

– H

olis

tic

Scor

ing

Cri

teri

a*

Ple

ase

note

that

thes

e ar

e on

ly s

ampl

e re

spon

ses,

and

oth

er v

aria

tion

s of

the

resp

onse

may

als

o ha

ve r

ecei

ved

full

mar

ks.

Mar

ksSa

mpl

e R

espo

nse

Com

men

ts

5

C3H

8(g)

+ 5

O2(

g) →

3

CO

2(g)

+ 4

H2O

(g)

∆H

= [

4 m

ol(–

241.

8 kJ

/mol

) +

3 m

ol(–

393.

5 kJ

/mol

)] –

[1

mol

(–10

3.8

kJ/m

ol)

+ 0

kJ]

∆H

= –

2 04

3.9

kJ

C8H

18(l

) +

25 2

O2(

g) →

8

CO

2(g)

+ 9

H2O

(g)

∆H

= [

9 m

ol(–

241.

8 kJ

/mol

) +

8 m

ol(–

393.

5 kJ

/mol

)] –

[1

mol

(–25

0.1

kJ/m

ol)

+ 0

kJ]

∆H

= –

5 07

4.1

kJ

C2H

5OH

(l)

+ 3

O2(

g) →

2

CO

2(g)

+ 3

H2O

(g)

∆H

= [

3 m

ol(–

241.

8 kJ

/mol

) +

2 m

ol(–

393.

5 kJ

/mol

)] –

[1

mol

(–27

7.6

kJ/m

ol)

+ 0

kJ]

∆H

= –

1 23

4.8

kJ

C3H

8(g)

2

3 m

ol C

O(g

)2

043.

9 kJ

= 1

.47

× 1

0-3

32

mol

CO

(g)

1.47

10kJ

-´

C8H

18(g

)

28

mol

CO

(g)

5 07

4.1

kJ =

1.5

8 ×

10-3

3

2m

ol C

O(g

)1.

5810

kJ-

´

C2H

5OH

(l)

22

mol

CO

(g)

1 23

4.8

kJ =

1.6

2 ×

10-3

3

2m

ol C

O(g

)1.

6210

kJ-

´

C3H

8(g)

pro

duce

s th

e le

ast a

mou

nt o

f C

O2(

g) p

er 1

kJ

of e

nerg

y.

Not

e: A

lso

acce

ptab

le is

kJ/

mol

CO

2(g)

com

pari

son.

Key

Com

pone

nt

• a

com

pari

son

in te

rms

of e

ntha

lpy

and

CO

2(g)

for

eac

h fu

el

Supp

orts

• ba

lanc

ed c

ombu

stio

n eq

uatio

ns in

clud

ing

enth

alpy

cal

cula

tions

• ca

lcul

atio

n fo

r m

oles

of

CO

2(g)

per

kJ

of

ener

gy f

or e

ach

fuel

• id

entif

y an

d pr

ovid

e

a ra

tiona

le to

sup

port

ch

oice

of

fuel


Question 3 – Holistic Split Scoring Guide


2 Key

The student has addressed the question asked by making a comparison between at least two of the equations in terms of energy (reasonable approach to enthalpy change or molar enthalpy change) and amount of CO2(g) (mol CO2(g)/kJ, kJ/mol CO2(g), mol of CO2(g)).

0 No Key

The student has not addressed the question asked.

Score Support



2 Satisfactory

to Good


1 Minimal





Calculator error, but substitution is correct

Solving for kJ/mol versus mol/kJ

Major errors: Substitution error in enthalpy calculation

Incorrect approach to enthalpy change

Incorrect fuel used in the comparison (i.e., methanol)



An electrolytic cell is set up that uses a copper anode. The products of the electrolysis are hydrogen gas and aqueous copper(II) ions. The half-reaction involving hydrogen gas is represented by the following equation.

2 H+(aq) + 2 e– → H2(g)


a. Write the half-reaction equation that represents the reaction that occurs at the anode, and write the balanced equation that represents the net redox reaction that occurs during the electrolysis.

(2 marks)

b. If a current of 3.00 A is applied to a cell containing a 400 g copper anode, determine the final mass of the copper anode after the cell runs for 48.0 h.

(3 marks)

4.


Que

stio

n 4

– A

naly

tic

Scor

ing

Cri

teri

a*

Ple

ase

note

that

thes

e ar

e on

ly s

ampl

e re

spon

ses,

and

oth

er v

aria

tion

s of

the

resp

onse

may

als

o ha

ve r

ecei

ved

full

mar

ks.

Que

stio

nM

arks

Sam

ple

Res

pons

eC

omm

ents

4.a.

2

Cu(

s) →

C

u2+(a

q) +

2e–

2

H+ (a

q) +

2e– →

H

2(g)

C

u(s)

+ 2

H+ (a

q) →

C

u2+(a

q) +

H2(

g)

1 m

ark

each

for

•co

rrec

t oxi

datio

n ha

lf-r

eact

ion

•va

lid n

et r

edox

re

actio

n (c

onsi

sten

t)

4.b.

3n e

– =

=

5.3

7 m

ol

n Cu

=

e1 2

n-

n e–

= 2

.69

mol

mC

u =

2.

69 m

ol(6

3.55

g/m

ol)

= 1

71 g

∆m

anod

e =

40

0 g

– 17

1 g

= 2

29 g

1 m

ark

each

for

•va

lid m

etho

d to

de

term

ine

mas

s co

nsum

ed

•co

rrec

t fina

l mas

s of

C

u(s)

con

sum

ed

•co

rrec

t fina

l mas

s

of a

node

(con

sist

ent

with

thei

r ca

lcul

atio

ns)

1C

omm

unic

atio

n—Se

e G

uide

Use

Ana

lytic

Sco

ring

G

uide

Tota

l pos

sibl

e m

arks

= 6



A student obtained a 25.0 mL sample of a 0.176 mol/L unknown acid solution, HA(aq). The pH of the sample was 2.25.


a. Write the balanced chemical equation that represents the ionization of the unknown acid and write the equilibrium law expression.

(2 marks)

b. Determine the value of Ka and identify one property, other than the value of Ka, of the unknown acid.

(3 marks)

5.


Que

stio

n 5

– A

naly

tic

Scor

ing

Cri

teri

a*

Ple

ase

note

that

thes

e ar

e on

ly s

ampl

e re

spon

ses,

and

oth

er v

aria

tion

s of

the

resp

onse

may

als

o ha

ve r

ecei

ved

full

mar

ks.

Que

stio

nM

arks

Acc

epta

ble

Res

pons

eC

omm

ents

5.a.

2H

A(a

q) +

H2O

(l)

⇌ A

– (aq)

+ H

3O+(a

q) o

r H

A(a

q) ⇌

H+(a

q) +

A– (a

q)

Ka

=

[HA

(aq)

][H

O(a

q)][

A(a

q)]

3+

-

•1

mar

k fo

r ba

lanc

ed

equa

tion

•1

mar

k fo

r K

a ex

pres

sion

5.b.

3In

itial

pH

=

2.2

5

[H3O

+(a

q)]

= [

A– (a

q)]

= 1

0–pH

=

5.6

× 1

0–3 m

ol/L

Ka

=

(0.1

76m

ol/L

5.6

10m

ol/L

)(5

.610

mol

/L) 3

32

#

#

--

-

= 1

.9 ×

10–

4

Pro

pert

ies

•w

eak

acid

•on

ly p

artia

lly io

nize

s in

wat

er

•m

ost o

f th

e ac

id d

oes

not

diss

ocia

te

•pr

oton

don

or

•m

onop

rotic

•[H

3O+(a

q)]

is le

ss th

an [

HA

(aq)

]

•w

hen

titra

ted

with

a s

tron

g ba

se,

pH a

t equ

ival

ence

poi

nt is

abo

ve 7

•lo

w c

ondu

ctiv

ity

•so

ur ta

ste

•1

mar

k fo

r su

bstit

utio

n co

nsis

tent

with

m

etho

d

•1

mar

k fo

r co

rrec

t an

swer

•1

mar

k fo

r th

e va

lid

prop

erty

of

the

acid

1C

omm

unic

atio

n—Se

e G

uide

Use

Ana

lytic

Sco

ring

G

uide

Tota

l pos

sibl

e m

arks

= 6



A two-layer catalytic converter has been designed for diesel engines in automobiles. The first layer in the catalytic converter removes some nitrogen monoxide gas from the exhaust gas and converts it to ammonia gas. In the second layer, ammonia gas is removed as it reacts with the remaining nitrogen monoxide gas. The reactions in the two layers occur at 300 °C and are represented by the following equations.

Equation I 2 NO(g) + 3 H2(g) catalyst

Chem30_06 Arrow catalyst

2 NH3(g) + O2(g)

Equation II 4 NH3(g) + 6 NO(g) catalyst

Chem30_06 Arrow catalyst

5 N2(g) + 6 H2O(g)


Compare the two reactions that occur in the catalytic converter. Identify one similarity and one difference between the reactions in terms of enthalpy.


• enthalpy calculations per mole of nitrogen monoxide gas for each reaction

• enthalpy diagrams that represent the enthalpy change that occurs in each reaction

• explanations of the similarity and difference you have identified

6.


Que

stio

n 6

– H

olis

tic

Scor

ing

Cri

teri

a*

Ple

ase

note

that

thes

e ar

e on

ly s

ampl

e re

spon

ses,

and

oth

er v

aria

tion

s of

the

resp

onse

may

als

o ha

ve r

ecei

ved

full

mar

ks.

Mar

ksSa

mpl

e R

espo

nse

Com

men

ts

5C

alcu

lati

ons

∆H

= ∑

n∆fH

°(pr

oduc

ts)

− ∑

n∆fH

°(re

acta

nts)

Equ

atio

n 1

2 N

O(g

) +

3 H

2(g)

2

NH

3(g)

+ O

2(g)

∆H

= [(

2 m

ol)(

– 45.

9 kJ

/mol

) +

0 k

J)]

– [(2

mol

)(+9

1.3

kJ/m

ol)

+ 0

kJ)

]

=

–27

4.4

kJ/2

mol

= –

137.

2 kJ

/mol

of

NO

(g)

Equ

atio

n 2

4

NH

3(g)

+ 6

NO

(g)

5 N

2(g)

+ 6

H2O

(g)

∆H

= [0

kJ

+ (6

mol

)( –2

41.8

kJ/

mol

)] –

[(4

mol

)(–

45.9

kJ/

mol

) +

(6 m

ol)(

+91.

3 kJ

/mol

)]

=

–1

815.

0 kJ

/6 m

ol =

–30

2.5

kJ/m

ol o

f N

O(g

)

E

quat

ion

1 E

quat

ion

2

Not

e: E

p, ∆

H, o

r E

nerg

y ar

e ac

cept

able

for

the

y-ax

is la

bel,

and

reac

tion

pro

gres

s, r

eact

ion

coor

dina

te, r

eact

ion,

or

tim

e ar

e ac

cept

able

for

the

x-ax

is.

Sim

ilari

ties

— B

oth

are

exot

herm

ic b

ecau

se th

e ca

lcul

ated

∆H

is n

egat

ive

—

Bot

h ar

e ex

othe

rmic

bec

ause

the

reac

tant

s ha

ve m

ore

pote

ntia

l ene

rgy

than

the

prod

ucts

Dif

fere

nces

— E

quat

ion

2 re

leas

es m

ore

ener

gy p

er m

ol o

f N

O(g

) th

an E

quat

ion

1 (c

onsi

sten

t with

cal

cula

tions

)

— T

he p

rodu

cts

of E

quat

ion

2 ha

ve a

hig

her

pote

ntia

l ene

rgy

than

the

prod

ucts

of

Equ

atio

n 1

Key

Com

pone

nt:

• a

com

pari

son

betw

een

each

re

actio

n is

m

ade

in te

rms

of e

nerg

y (k

J or

kJ

/mol

)

Supp

ort:

• en

thal

py c

hang

e ca

lcul

atio

ns

(kJ/

mol

of

NO

(g))

• en

thal

py

diag

ram

s co

nsis

tent

with

ca

lcul

atio

ns

• ex

plan

atio

n an

d id

enti

ficat

ion

of

a si

mila

rity

and

a

diff

eren

ce

sandra.bit

Inserted Text

period

sandra.bit

Inserted Text

period

sandra.bit

Inserted Text

period

sandra.bit

Inserted Text

period

sandra.bit

Inserted Text

period

sandra.bit

Inserted Text

period

sandra.bit

Inserted Text

period



Score Key–No Key Criteria

2 Key

The student has addressed the question by making a comparison between the two reactions in terms of enthalpy (reasonable approach to enthalpy change or molar enthalpy change).

Note: There may be errors in their calculations (e.g., reversing Hess’s Law) or missing coefficients or correct Hess’s Law but incorrect kJ/mol calculation.

0 No Key


Score Support



2 Satisfactory

to Good

The student has provided support for the majority of the bullets but not necessarily all of the bullets. The support provided may contain minor errors/weaknesses or a major weakness in one of the bullets. There is more correct than incorrect support.

1 Minimal




Minor errors: • Unbalanced equation

• Method for kJ/mol is correct but has a minor calculation error

Major errors: • Incorrect method for kJ/mol

• No similarities or differences identified



Nitrogen monoxide is a commonly used gas in the chemical industry, and is also a toxic air pollutant produced by automobiles. Two reactions involving nitrogen monoxide gas are represented below.

I 2 NO2(g) → 2 NO(g) + O2(g)

II N2(g) + O2(g) → 2 NO(g) ∆H° = +182.6 kJ


a. Determine the enthalpy change for reaction I, and sketch an enthalpy diagram that represents reaction I.

(3 marks)

b. In terms of energy, identify two similarities or two differences between the two reactions involving nitrogen monoxide gas represented above.

(2 marks)

7.


Que

stio

n 7

– A

naly

tic

Scor

ing

Cri

teri

a*

Ple

ase

note

that

thes

e ar

e on

ly s

ampl

e re

spon

ses,

and

oth

er v

aria

tion

s of

the

resp

onse

may

als

o ha

ve r

ecei

ved

full

mar

ks.

Que

stio

nM

arks

Acc

epta

ble

Res

pons

eC

omm

ents

7.a.

3∆

H°

= ∑

n∆fH

°(pr

oduc

ts)

− ∑

n∆fH

°(re

acta

nts)

∆H

° 1

=

[(2

mol

)(+

91.3

kJ/

mol

) +

0 k

J] –

[(2

mol

)(+

33.2

kJ/

mol

)]

=

+

116.

2 kJ

Ent

halp

y D

iagr

am

Not

e:

•T

he a

ctiv

atio

n en

ergy

bar

rier

doe

s no

t nee

d to

be

incl

uded

on

the

ener

gy d

iagr

am.

•T

he te

rms

reac

tion

pro

gres

s, r

eact

ion

coor

dina

te, r

eact

ion,

or

tim

e m

ay b

e us

ed

to la

bel t

he x

-axi

s.

•A

title

is n

ot r

equi

red.

• 1

mar

k fo

r co

rrec

t m

etho

d to

cal

cula

te

enth

alpy

cha

nge

• 1

mar

k fo

r co

rrec

t an

swer

, inc

ludi

ng th

e si

gn (

cons

iste

nt w

ith

met

hod)

• 1

mar

k fo

r en

thal

py

diag

ram

(co

rrec

t or

cons

iste

nt w

ith a

nsw

er)


Que

stio

n 7

– A

naly

tic

Scor

ing

Cri

teri

a*

Ple

ase

note

that

thes

e ar

e on

ly s

ampl

e re

spon

ses,

and

oth

er v

aria

tion

s of

the

resp

onse

may

als

o ha

ve r

ecei

ved

full

mar

ks.

7.b.

2Si

mila

riti

es:

(Not

e: T

his

is th

e co

rrec

t res

pons

e if

the

resp

onse

to q

uest

ion

1.a.

is c

orre

ct.)

•B

oth

abso

rb e

nerg

y fr

om th

e su

rrou

ndin

gs d

urin

g th

e re

actio

n.

•B

oth

are

endo

ther

mic

.

•T

he p

rodu

cts

of b

oth

reac

tions

hav

e a

high

er p

oten

tial e

nerg

y th

an th

e re

acta

nts.

•T

he e

nerg

y te

rm w

ould

be

incl

uded

as

a re

acta

nt in

bot

h ch

emic

al e

quat

ions

.

Dif

fere

nces

:

•R

eact

ion

I ab

sorb

s le

ss e

nerg

y pe

r m

ole

of N

O(g

) th

an r

eact

ion

II.

•R

eact

ants

in r

eact

ion

I ha

ve h

ighe

r po

tent

ial e

nerg

y th

an th

e re

acta

nts

in

reac

tion

II.

(If

in p

art

(a)

stud

ent

show

s an

exo

ther

mic

rea

ctio

n)D

iffe

renc

es:

•R

eact

ion

II a

bsor

bs e

nerg

y; r

eact

ion

I re

leas

es e

nerg

y (c

onsi

sten

t with

dia

gram

).

•R

eact

ion

I is

exo

ther

mic

; rea

ctio

n II

is e

ndot

herm

ic (c

onsi

sten

t with

∆H

°).

•T

he p

rodu

cts

of r

eact

ion

II h

ave

a hi

gher

pot

entia

l ene

rgy

than

the

reac

tant

s; th

e op

posi

te is

true

for

reac

tion

I (c

onsi

sten

t with

dia

gram

).

•In

rea

ctio

n I

the

ener

gy te

rm w

ould

be

incl

uded

as

a pr

oduc

t; in

rea

ctio

n II

the

ener

gy te

rm w

ould

be

incl

uded

as

a re

acta

nt (c

onsi

sten

t with

∆H

°).

• 1

mar

k fo

r ea

ch

sim

ilari

ty o

r di

ffer

ence

th

at is

con

sist

ent w

ith th

e in

form

atio

n pr

ovid

ed in

pa

rt a

1C

omm

unic

atio

n—Se

e G

uide

Use

Ana

lytic

Sco

ring

Gui

de

6To

tal p

ossi

ble

mar

ks =

6



A student predicts that the value of Ka of ethanoic acid, a weak monoprotic acid, may be affected by the temperature of the acid.


Design an experiment to test this prediction using commonly available laboratory apparatus, and a 0.100 mol/L solution of ethanoic acid.


• a detailed procedure

• a definition for the term Ka

• a relevant equilibrium law expression and balanced chemical equation, and the formula(s) necessary for the calculation(s) to solve for Ka

8.


Que

stio

n 8

– H

olis

tic

Scor

ing

Cri

teri

a*

Ple

ase

note

that

thes

e ar

e on

ly s

ampl

e re

spon

ses,

and

oth

er v

aria

tion

s of

the

resp

onse

may

als

o ha

ve r

ecei

ved

full

mar

ks.

Que

stio

nM

arks

Acc

epta

ble

Res

pons

eC

omm

ents

82

for

key

3 for

supp

ort

Pro

cedu

re:

1. O

btai

n th

ree

100.

00 m

L b

eake

rs, a

nd fi

ll ea

ch b

eake

r w

ith 5

0.0

mL

of

0.1

00 m

ol/L

eth

anoi

c ac

id s

olut

ion.

2. U

sing

thre

e w

ater

bat

hs a

t dif

fere

nt te

mpe

ratu

res,

war

m o

r co

ol,

plac

e on

e be

aker

of

etha

noic

aci

d in

eac

h w

ater

bat

h.

3. M

easu

re a

nd r

ecor

d th

e te

mpe

ratu

re o

f ea

ch e

than

oic

acid

sol

utio

n.

4. U

sing

a p

H m

eter

, mea

sure

and

rec

ord

the

pH o

f ea

ch a

cid

sam

ple

Not

e: T

his

can

also

be

done

by

titra

tion

to d

eter

min

e hy

dron

ium

ion

conc

entr

atio

n at

the

half

equ

ival

ence

poi

nt.

Var

iabl

es:

(m

ay b

e im

plie

d)

cont

rolle

d –

sam

e ac

id, c

once

ntra

tion

of a

cid,

am

ount

of

acid

man

ipul

ated

–

tem

pera

ture

of

etha

noic

aci

d so

lutio

n

resp

ondi

ng

– pH

of

the

etha

noic

aci

d so

lutio

n

Ka

is a

mea

sure

of

the

exte

nt to

whi

ch a

n ac

id io

nize

s.

Exp

ress

ions

and

For

mul

as:

CH

3CO

OH

(aq)

+ H

2O(l

) →

H3O

+(a

q) +

CH

3CO

O– (a

q)

Ka

=

[HA

(aq)

][H

O(a

q)][

A(a

q)]

3+

-

(T

he io

niza

tion

equa

tion

may

be

impl

ied

if

the

equi

libri

um e

xpre

ssio

n is

cor

rect

.)

[H3O

+(a

q)]

= [

CH

3CO

O –

(aq)

] =

10–p

H

Key

Com

pone

nt:

• an

exp

erim

ent o

r de

sign

for

an

expe

rim

ent t

hat c

ould

allo

w th

e st

uden

t to

mea

sure

the

pH o

f th

e ac

id a

t dif

fere

nt te

mpe

ratu

res

(pH

pro

be, m

eter

, titr

atio

n,

univ

ersa

l ind

icat

ors,

indi

cato

rs)

Supp

ort:

• de

finiti

on o

f K

a

• de

tails

of

proc

edur

e (i

dent

ifyi

ng

the

man

ipul

ated

, res

pond

ing

and

cont

rolle

d va

riab

les,

may

be

impl

ied)

• re

leva

nt e

quat

ion(

s) a

nd

form

ula(

s)

Tota

l pos

sibl

e m

arks

= 5

Use

Hol

istic

Sco

ring

Gui

de

sandra.bit

Inserted Text

period



The smelting of iron occurs in a blast furnace, as represented by the following overall equation.

Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)

In a blast furnace, the smelting process occurs in three steps, as represented by the following equations.

Step I 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ∆H° = – 47.2 kJ

Step II Fe3O4(s) + CO(g) → 3 FeO(s) + CO2(g) ∆H° = +19.4 kJ

Step III FeO(s) + CO(g) → Fe(s) + CO2(g) ∆H° = –11.0 kJ


a. Determine the enthalpy change for the overall reaction. Write the equation representing the overall reaction, including the enthalpy change as a term in the equation. (3 marks)

b. Sketch and label an enthalpy diagram to represent the energy change(s) that occur in the overall reaction. (2 marks)

9.


Que

stio

n 9

– A

naly

tic

Scor

ing

Cri

teri

a*

Ple

ase

note

that

thes

e ar

e on

ly s

ampl

e re

spon

ses,

and

oth

er v

aria

tion

s of

the

resp

onse

may

als

o ha

ve r

ecei

ved

full

mar

ks.

Que

stio

nM

arks

Acc

epta

ble

Res

pons

eC

omm

ents

9.a.

3

Fe2O

3(s)

+

1 3C

O(g

) →

2 3Fe

3O4(

s) +

1 3

CO

2(g)

∆H

º =

1 3(–

47.2

kJ)

2 3

Fe3O

4(s)

+

2 3C

O(g

) →

2Fe

O(s

) +

2 3

CO

2(g)

∆H

º =

2 3(+

19.4

kJ)

2

FeO

(s)

+ 2

CO

(g) →

2Fe

(s)

+ 2

CO

2(g)

∆H

º =

2(–

11.0

kJ)

Fe

2O3(

s) +

3C

O(g

) →

2Fe

(s)

+ 3

CO

2(g)

∆H

º =

–24

.8 k

J

or

∆Hº

= ∑

n∆f H

° (pr

oduc

ts)

– ∑

n∆f H

° (re

acta

nts)

=

[(2

mol

)(0

kJ/m

ol)

+ (

3 m

ol)(

–393

.5 k

J/m

ol)]

– [(

1 m

ol)(

–82

4.2

kJ/m

ol)

+ (

3 m

ol)(

–11

0.5

kJ/m

ol)]

=

–24

.8 k

J

Fe2O

3(s)

+ 3

CO

(g)

→ 2

Fe(s

) +

3C

O2(

g) +

24.

8 kJ

• 1

mar

k fo

r co

rrec

t met

hod

• 1

mar

k fo

r co

rrec

t ans

wer

• 1

mar

k fo

r en

ergy

term

on

side

of

equa

tion

cons

iste

nt

with

cal

cula

tion

9.b.

2N

ote:

V

ertic

al a

xis

can

be

labe

lled ∆H

°, E

p, o

r en

ergy

. H

oriz

onta

l axi

s ca

n be

labe

lled

Rea

ctio

n pr

ogre

ss, R

eact

ion

coor

dina

te, R

eact

ion,

or

Tim

e.

Not

e:

A ti

tle is

not

req

uire

d,

and

neith

er is

the

enth

alpy

ch

ange

(∆H

°).

Smel

ting

of

Iron

• 1

mar

k fo

r la

bels

(m

ust

incl

ude

x- a

nd y

-axe

s an

d id

entifi

catio

n of

rea

ctan

ts

and

prod

ucts

)

• 1

mar

k fo

r sh

ape

of

enth

alpy

dia

gram

co

nsis

tent

with

ent

halp

y ca

lcul

atio

n

1C

omm

unic

atio

n—Se

e G

uide

Use

Ana

lytic

Sco

ring

Gui

de

Tota

l pos

sibl

e m

arks

= 6



Dinitrogen tetraoxide was used as one of the rocket fuels on the lunar landers of the Apollo missions. In the gaseous phase, it decomposes according to the following equation. N2O4(g) ⇌ 2 NO2(g) ∆H° = +55.3 kJ

A technician placed a sample of N2O4(g) in a 2.00 L flask and allowed the system to reach equilibrium. At 25 °C, the Kc was 5.40 × 10–3, and the concentration of N2O4(g) at equilibrium was 3.00 × 10–1 mol/L.


a. Determine the amount, in moles, of NO2(g) present in the 2.00 L flask at equilibrium. (3 marks)

b. Identify whether the value of Kc will increase, decrease, or remain the same when heat is added to the system. Support your answer with an explanation.

(2 marks)

10.


Que

stio

n 10

– A

naly

tic

Scor

ing

Cri

teri

a*

Ple

ase

note

that

thes

e ar

e on

ly s

ampl

e re

spon

ses,

and

oth

er v

aria

tion

s of

the

resp

onse

may

als

o ha

ve r

ecei

ved

full

mar

ks.

Que

stio

nM

arks

Acc

epta

ble

Res

pons

eC

omm

ents

10.a

.3

N2O

4(g)

⇌ 2

NO

2(g)

Kc

=

Que

stio

n M

arks

Sa

mpl

e R

espo

nse

Com

men

ts

10.a

. 3

N2O

4(g)

2

NO

2(g)

Kc =

2

22

4

[NO

(g)]

[NO

(g)]

[N

O2(

g)]

= 2

4[N

O(g

)]c

K

=

3

1(5

.40

10)(3

.00

10 m

ol/L

)-

-´

´

=

4.0

2 ×

10–2

mol

/L

m

ol o

f NO

2(g)

= (

4.02

× 1

0–2 m

ol/L

)(2.

00 L

) = 8

.05

× 10

–2 m

ol

• 1

mar

k fo

r cor

rect

Kc e

xpre

ssio

n/

met

hod

• 1

mar

k fo

r cor

rect

con

cent

ratio

n of

N

O2(

g)

• 1

mar

k fo

r ans

wer

in m

oles

co

nsis

tent

with

con

cent

ratio

n

10.b

. 2

Kc w

ill in

crea

se

beca

use

the

reac

tion

will

shift

to th

e rig

ht to

con

sum

e th

e ad

ded

heat

(e

ndot

herm

ic re

actio

n)

or

beca

use

the

conc

entra

tion

of N

O2(

g) in

crea

ses,

and

the

conc

entra

tion

of N

2O4(

g) d

ecre

ases

, and

Kc =

[pro

duct

s][re

acta

nts]

.

• 1

mar

k fo

r Kc i

ncre

ases

• 1

mar

k fo

r exp

lana

tion

of th

e sh

ift

cons

iste

nt w

ith a

nsw

er o

f in

crea

se/d

ecre

ase.

1

Com

mun

icat

ion—

See

Gui

de

Use

Ana

lytic

Sco

ring

Gui

de

Tota

l pos

sibl

e m

arks

= 6

[N

O2(

g)]

=

[NO

(g)]

K2

4c

=

(5

.40

10)(

3.00

10m

ol/L

)3

1#

#-

-

=

4.0

2 ×

10

–2 m

ol/L

m

ol o

f N

O2(

g) =

(4.

02 ×

10

–2 m

ol/L

)(2.

00 L

) =

8.0

5 ×

10

–2 m

ol

• 1

mar

k fo

r co

rrec

t Kc

expr

essi

on/

met

hod

• 1

mar

k fo

r co

rrec

t con

cent

ratio

n of

N

O2(

g)

• 1

mar

k fo

r an

swer

in m

oles

co

nsis

tent

with

con

cent

ratio

n

10.b

.2

Kc

wil

l inc

reas

e

beca

use

the

reac

tion

wil

l shi

ft to

the

righ

t to

cons

ume

the

adde

d he

at

(end

othe

rmic

rea

ctio

n)

or beca

use

the

conc

entr

atio

n of

NO

2(g)

incr

ease

s, a

nd th

e co

ncen

trat

ion

of N

2O4(

g) d

ecre

ases

, and

Kc

= [p

rodu

cts]

[rea

ctan

ts].

• 1

mar

k fo

r K

c in

crea

ses

• 1

mar

k fo

r ex

plan

atio

n of

the

shif

t co

nsis

tent

with

ans

wer

of

in

crea

se/d

ecre

ase.

1C

omm

unic

atio

n—Se

e G

uide

Use

Ana

lytic

Sco

ring

Gui

de

Tota

l pos

sibl

e m

arks

= 6



A student hypothesizes that if the mass of the iron electrode is increased in a lead–iron voltaic cell, the cell potential will also increase.


Design an experiment that would allow you to test the student’s hypothesis using a lead–iron voltaic cell.



• a labelled diagram of the lead–iron voltaic cell

• relevant balanced equation(s) and a potential difference calculation for your voltaic cell

11.


Que

stio

n 11

– H

olis

tic

Scor

ing

Cri

teri

a*

Ple

ase

note

that

thes

e ar

e on

ly s

ampl

e re

spon

ses,

and

oth

er v

aria

tion

s of

the

resp

onse

may

als

o ha

ve r

ecei

ved

full

mar

ks.

Mar

ksSa

mpl

e R

espo

nse

Com

men

ts

5P

roce

dure

:1.

C

onst

ruct

a v

olta

ic c

ell,

usin

g th

e re

agen

ts g

iven

in th

e di

agra

m.

2.

Plac

e 10

0.0

mL

of

1.0

mol

/L P

b(N

O3)

2(aq

) in

the

beak

er c

onta

inin

g th

e Pb

(s)

elec

trod

e

and

100.

0 m

L o

f 1.

0 m

ol/L

Fe(

NO

3)2(

aq)

in th

e be

aker

con

tain

ing

the

Fe(s

) el

ectr

ode.

3.

Mea

sure

and

rec

ord

the

cell

pote

ntia

l usi

ng a

vol

tmet

er a

fter

the

cell

oper

ates

for

5 m

inut

es.

4.

Rep

eat s

teps

1 to

3 u

sing

dif

fere

nt m

asse

s of

Fe(

s) f

or th

e ir

on e

lect

rode

.

Var

iabl

es:

Man

ipul

ated

–

mas

s of

iron

ele

ctro

de

R

espo

ndin

g –

the

cell

pote

ntia

l (or

cel

l vol

tage

)

Con

trol

led

– te

mpe

ratu

re, c

once

ntra

tion

of e

lect

roly

tes,

tim

e th

e ce

ll ru

ns, c

ell s

etup

, etc

.

Not

e: T

he v

aria

bles

can

be

impl

ied

in th

e pr

oced

ure.

Dia

gram

:

*

Or

any

othe

r ve

rsio

n of

a w

orki

ng le

ad–i

ron

volta

ic c

ell

Equ

atio

n an

d C

alcu

lati

on

Pb2+

(aq)

+ F

e(s)

→ P

b(s)

+ F

e2+(a

q)

Eº c

ell =

–0.

13 V

– (

–0.

45 V

) =

+0.

32 V

Key

Com

pone

nt

• ex

peri

men

tal

desi

gn o

r pr

oced

ure

that

atte

mpt

s to

m

anip

ulat

e th

e m

ass

of F

e(s)

in

an e

lect

roch

emic

al

cell,

and

mea

sure

po

tent

ial d

iffe

renc

e

Supp

ort C

ompo

nent

s

• de

taile

d pr

oced

ure

• la

belle

d le

ad–i

ron

volta

ic c

ell d

iagr

am

• re

leva

nt b

alan

ced

equa

tion(

s) a

nd

pote

ntia

l dif

fere

nce

calc

ulat

ion

(hal

f re

actio

ns o

r ne

t re

actio

n)




2 Key

The student has addressed the question by designing an experiment (procedure or design) involving a cell (voltaic or electrolytic, and not necessarily lead–iron) and has manipulated the mass of one of the electrodes (or in the student’s understanding manipulated the mass), and the student’s responding variable relates to measuring the cell potential.

0 No Key


Score Support



2 Satisfactory

to Good


1 Minimal





Method for calculating potential difference is correct, but the student has made a calculation error

Using a solution in the salt bridge that would precipitate

Major errors: Electrolytic cell

A voltaic cell other than a lead–iron cell

Missing electrolyte or other major error in the cell diagram

Experimental design or procedure provided



A student performs a calorimetry experiment to determine the molar enthalpy of combustion of ethanol. The student uses an aluminium can as a calorimeter. The combustion of liquid ethanol produces gaseous products, as represented by the equation below.

C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g)

The data collected during the experiment are recorded below.

Calorimetry Experiment Data

Mass of aluminium can 10.65 g

Mass of water 250.00 g

Initial mass of ethanol burner 256.34 g

Final mass of ethanol burner 252.45 g

Initial temperature of water and aluminium can 21.40 °C

Final temperature of water and aluminium can 82.30 °C



a. Calculate the student’s experimental molar enthalpy of combustion of ethanol.(3 marks)

b. The theoretical value of the molar enthalpy of combustion of ethanol is higher than that obtained in the student’s experiment. Identify an improvement that could be made to the experimental design and explain how your suggestion would reduce experimental error in the calorimetry experiment.

(2 marks)

12.


Que

stio

n 12

– A

naly

tic

Scor

ing

Cri

teri

a*

Ple

ase

note

that

thes

e ar

e on

ly s

ampl

e re

spon

ses,

and

oth

er v

aria

tion

s of

the

resp

onse

may

als

o ha

ve r

ecei

ved

full

mar

ks.

Que

stio

nM

arks

Acc

epta

ble

Res

pons

eC

omm

ents

12.a

.3

∆H° l

ost

= H

gain

ed

∆H°

=

mc∆

t (H2O

) + m

c∆t (A

l)

=

(0.

250

kg)(

4.19

kJ/

kg∙°C

)(82

.30–

21.4

0)°C

+ (0

.010

65

kg)(

0.89

7 kJ

/kg∙

°C)

(82

.30–

21.4

0)°C

=

–64

.374

5 k

J

n =

Que

stio

n M

arks

Sa

mpl

e R

espo

nse

Com

men

ts

12.a

. 3

∆H° lo

st

= H

gain

ed

∆H°

= m

c∆t (H

2O) +

mc∆

t (Al)

=

(0.2

50 k

g)(4

.19

kJ/k

g•°C

) (82.

30–2

1.40

) °C

+ (0

.010

65

kg)(

0.89

7 kJ

/kg•

°C)

(82

.30–

21.4

0)°C

=

–64

.374

5 k

J

3.89

g =

=

=

0.0

84 4

mol

46.0

8 g/

mol

mn

M

∆H°

= –

64.3

7 kJ

0.08

4 4

mol

=

–76

3 k

J/m

ol

• 1 m

ark

for

met

hod

• 1 m

ark

for

subs

titut

ion

• 1 m

ark

for

corr

ect a

nsw

er

in k

J/m

ol

12.b

. 2

Impr

ovem

ent

• a

bom

b ca

lorim

eter

wou

ld li

mit

heat

loss

to th

e su

rrou

ndin

gs

• us

ing

a sh

ield

aro

und

the

calo

rimet

er a

nd b

urne

r wou

ld li

mit

heat

loss

to th

e su

rrou

ndin

gs

• us

ing

a m

ore

effic

ient

bur

ner w

ith b

ette

r air

flow

wou

ld re

duce

inco

mpl

ete

com

bust

ion

•

susp

endi

ng th

e ca

n di

rect

ly a

bove

the

flam

e w

ould

elim

inat

e he

at lo

ss to

the

iron

stan

d an

d rin

g

• 1 m

ark

for

impr

ovem

ent

• 1 m

ark

for

expl

anat

ion

1

Com

mun

icat

ion—

See

Gui

de

Use

Ana

lytic

Sc

orin

g G

uide

Tota

l pos

sibl

e m

arks

= 6

=

Que

stio

n M

arks

Sa

mpl

e R

espo

nse

Com

men

ts

12.a

. 3

∆H° lo

st

= H

gain

ed

∆H°

= m

c∆t (H

2O) +

mc∆

t (Al)

=

(0.2

50 k

g)(4

.19

kJ/k

g•°C

) (82.

30–2

1.40

) °C

+ (0

.010

65

kg)(

0.89

7 kJ

/kg•

°C)

(82

.30–

21.4

0)°C

=

–64

.374

5 k

J

3.89

g =

=

=

0.0

84 4

mol

46.0

8 g/

mol

mn

M

∆H°

= –

64.3

7 kJ

0.08

4 4

mol

=

–76

3 k

J/m

ol

• 1 m

ark

for

met

hod

• 1 m

ark

for

subs

titut

ion

• 1 m

ark

for

corr

ect a

nsw

er

in k

J/m

ol

12.b

. 2

Impr

ovem

ent

• a

bom

b ca

lorim

eter

wou

ld li

mit

heat

loss

to th

e su

rrou

ndin

gs

• us

ing

a sh

ield

aro

und

the

calo

rimet

er a

nd b

urne

r wou

ld li

mit

heat

loss

to th

e su

rrou

ndin

gs

• us

ing

a m

ore

effic

ient

bur

ner w

ith b

ette

r air

flow

wou

ld re

duce

inco

mpl

ete

com

bust

ion

•

susp

endi

ng th

e ca

n di

rect

ly a

bove

the

flam

e w

ould

elim

inat

e he

at lo

ss to

the

iron

stan

d an

d rin

g

• 1 m

ark

for

impr

ovem

ent

• 1 m

ark

for

expl

anat

ion

1

Com

mun

icat

ion—

See

Gui

de

Use

Ana

lytic

Sc

orin

g G

uide

Tota

l pos

sibl

e m

arks

= 6

= 0

.084

4 m

ol

∆H°

=

Que

stio

n M

arks

Sa

mpl

e R

espo

nse

Com

men

ts

12.a

. 3

∆H° lo

st

= H

gain

ed

∆H°

= m

c∆t (H

2O) +

mc∆

t (Al)

=

(0.2

50 k

g)(4

.19

kJ/k

g•°C

) (82.

30–2

1.40

) °C

+ (0

.010

65

kg)(

0.89

7 kJ

/kg•

°C)

(82

.30–

21.4

0)°C

=

–64

.374

5 k

J

3.89

g =

=

=

0.0

84 4

mol

46.0

8 g/

mol

mn

M

∆H°

= –

64.3

7 kJ

0.08

4 4

mol

=

–76

3 k

J/m

ol

• 1 m

ark

for

met

hod

• 1 m

ark

for

subs

titut

ion

• 1 m

ark

for

corr

ect a

nsw

er

in k

J/m

ol

12.b

. 2

Impr

ovem

ent

• a

bom

b ca

lorim

eter

wou

ld li

mit

heat

loss

to th

e su

rrou

ndin

gs

• us

ing

a sh

ield

aro

und

the

calo

rimet

er a

nd b

urne

r wou

ld li

mit

heat

loss

to th

e su

rrou

ndin

gs

• us

ing

a m

ore

effic

ient

bur

ner w

ith b

ette

r air

flow

wou

ld re

duce

inco

mpl

ete

com

bust

ion

•

susp

endi

ng th

e ca

n di

rect

ly a

bove

the

flam

e w

ould

elim

inat

e he

at lo

ss to

the

iron

stan

d an

d rin

g

• 1 m

ark

for

impr

ovem

ent

• 1 m

ark

for

expl

anat

ion

1

Com

mun

icat

ion—

See

Gui

de

Use

Ana

lytic

Sc

orin

g G

uide

Tota

l pos

sibl

e m

arks

= 6

=

–76

3 k

J/m

ol

• 1

mar

k fo

r m

etho

d

• 1

mar

k fo

r su

bstit

utio

n

• 1

mar

k fo

r co

rrec

t ans

wer

in k

J/m

ol

12.b

.2

Impr

ovem

ent

• a

bom

b ca

lori

met

er w

ould

lim

it he

at lo

ss to

the

surr

ound

ings

• us

ing

a sh

ield

aro

und

the

calo

rim

eter

and

bur

ner

wou

ld li

mit

heat

loss

to th

e su

rrou

ndin

gs

• us

ing

a m

ore

effic

ient

bur

ner

with

bet

ter

air

flow

wou

ld r

educ

e in

com

plet

e co

mbu

stio

n

• su

spen

ding

the

can

dire

ctly

abo

ve th

e fla

me

wou

ld e

lim

inat

e he

at lo

ss to

the

iron

sta

nd a

nd r

ing

• 1

mar

k fo

r im

prov

emen

t

• 1

mar

k fo

r ex

plan

atio

n

1C

omm

unic

atio

n—Se

e G

uide

Use

Ana

lytic

Sco

ring

G

uide

Tota

l pos

sibl

e m

arks

= 6



Carbon monoxide gas and oxygen gas react to form carbon dioxide gas as represented by the equilibrium equation below.

2 CO(g) + O2(g) ⇌ 2 CO2(g)

A technician added carbon monoxide gas and oxygen gas to an empty flask. The technician then recorded the concentrations of each gas present in the flask until the reaction reached equilibrium at 500 K, as shown in the graph below.


a. Write the equilibrium law expression, and calculate the value for the equilibrium constant at 500 K. (3 marks)

b. Indicate whether the products or reactants are favoured in this reaction and explain your answer. (2 marks)

13.


Que

stio

n 13

– A

naly

tic

Scor

ing

Cri

teri

a*

Ple

ase

note

that

thes

e ar

e on

ly s

ampl

e re

spon

ses,

and

oth

er v

aria

tion

s of

the

resp

onse

may

als

o ha

ve r

ecei

ved

full

mar

ks.

Que

stio

nM

arks

Acc

epta

ble

Res

pons

eC

omm

ents

13.a

.3

Kc

=

Que

stio

n M

arks

Sa

mpl

e R

espo

nse

Com

men

ts

13.a

. 3

Kc =

2

2 22

[CO

(g)]

[CO

(g)]

[O(g

)]

Kc =

2

2(1

.1)

(2.0

)(4

.5) =

2

6.7

10-

´

=

26.

710

-´

• 1

mar

k fo

r equ

ilibr

ium

exp

ress

ion

• 1

mar

k fo

r sel

ectin

g co

rrec

t eq

uilib

rium

equ

atio

ns

• 1

mar

k fo

r cor

rect

ans

wer

13.b

. 2

The

equi

libriu

m fa

vour

s the

reac

tant

s bec

ause

• th

e K

c va

lue,

2

6.7

10-

´, i

s les

s tha

n 1

• th

ere

is a

hig

her c

once

ntra

tion

of re

acta

nts t

han

prod

ucts

pre

sent

• at

equ

ilibr

ium

, the

re is

mor

e th

an 5

0% o

f the

initi

al c

once

ntra

tions

N

ote:

The

Kc

valu

e do

es n

ot h

ave

to b

e re

stat

ed h

ere.

• 1

mar

k fo

r cor

rect

iden

tific

atio

n of

sh

ift c

onsi

sten

t with

Kc

in 2

.a.

• 1

mar

k fo

r rat

iona

le c

onsi

sten

t w

ith c

hoic

e

1

Com

mun

icat

ion—

See

Gui

de

Use

Ana

lytic

Sco

ring

Gui

de

Tota

l pos

sibl

e m

arks

= 6

Kc

=

Que

stio

n M

arks

Sa

mpl

e R

espo

nse

Com

men

ts

13.a

. 3

Kc =

2

2 22

[CO

(g)]

[CO

(g)]

[O(g

)]

Kc =

2

2(1

.1)

(2.0

)(4

.5) =

2

6.7

10-

´

=

26.

710

-´

• 1

mar

k fo

r equ

ilibr

ium

exp

ress

ion

• 1

mar

k fo

r sel

ectin

g co

rrec

t eq

uilib

rium

equ

atio

ns

• 1

mar

k fo

r cor

rect

ans

wer

13.b

. 2

The

equi

libriu

m fa

vour

s the

reac

tant

s bec

ause

• th

e K

c va

lue,

2

6.7

10-

´, i

s les

s tha

n 1

• th

ere

is a

hig

her c

once

ntra

tion

of re

acta

nts t

han

prod

ucts

pre

sent

• at

equ

ilibr

ium

, the

re is

mor

e th

an 5

0% o

f the

initi

al c

once

ntra

tions

N

ote:

The

Kc

valu

e do

es n

ot h

ave

to b

e re

stat

ed h

ere.

• 1

mar

k fo

r cor

rect

iden

tific

atio

n of

sh

ift c

onsi

sten

t with

Kc

in 2

.a.

• 1

mar

k fo

r rat

iona

le c

onsi

sten

t w

ith c

hoic

e

1

Com

mun

icat

ion—

See

Gui

de

Use

Ana

lytic

Sco

ring

Gui

de

Tota

l pos

sibl

e m

arks

= 6

= 6

.7 ×

10−

2

=

6.7

× 1

0−2

• 1

mar

k fo

r eq

uilib

rium

ex

pres

sion

• 1

mar

k fo

r se

lect

ing

corr

ect e

quili

briu

m

equa

tions

• 1

mar

k fo

r co

rrec

t an

swer

13.b

.2

The

equ

ilibr

ium

fav

ours

the

reac

tant

s be

caus

e

•th

e K

c va

lue,

6.7

× 1

0−2

, is

less

than

1

•th

ere

is a

hig

her

conc

entr

atio

n of

rea

ctan

ts th

an p

rodu

cts

pres

ent

•at

equ

ilibr

ium

, the

re is

mor

e th

an 5

0% o

f th

e in

itial

con

cent

ratio

ns

Not

e: T

he K

c va

lue

does

not

hav

e to

be

rest

ated

her

e.

• 1

mar

k fo

r co

rrec

t id

entifi

catio

n of

shi

ft

cons

iste

nt w

ith K

c in

2.a

.

• 1

mar

k fo

r ra

tiona

le

cons

iste

nt w

ith c

hoic

e

1C

omm

unic

atio

n—Se

e G

uide

Use

Ana

lytic

Sco

ring

G

uide

Tota

l pos

sibl

e m

arks

= 6



A student hypothesizes that if the concentration of nickel(II) ions is increased in a nickel–zinc voltaic cell, the potential difference of the cell will also increase.


Design an experiment that would allow you to test the student’s hypothesis using a nickel–zinc voltaic cell.



• a labelled diagram of the nickel–zinc voltaic cell

• relevant balanced equation(s) and a potential difference calculation for your voltaic cell

14.


Que

stio

n 14

– H

olis

tic

Scor

ing

Cri

teri

a*

Ple

ase

note

that

thes

e ar

e on

ly s

ampl

e re

spon

ses,

and

oth

er v

aria

tion

s of

the

resp

onse

may

als

o ha

ve r

ecei

ved

full

mar

ks.

Mar

ksSa

mpl

e R

espo

nse

Com

men

ts

5V

aria

bles

: M

anip

ulat

ed

– co

ncen

trat

ion

of N

i2+(a

q)

Res

pond

ing

– th

e ce

ll p

oten

tial (

or c

ell v

olta

ge)

C

ontr

olle

d –

tem

pera

ture

, oxi

dizi

ng a

nd r

educ

ing

agen

t use

d, c

ell s

et-u

p,

and

tim

e th

e ce

ll r

uns

prio

r to

mea

suri

ng v

olta

ge

Ni2+

(aq)

+ Z

n(s)

→ N

i(s)

+ Z

n2+(a

q)

E° c

ell

= –

0.26

V –

(–

0.76

V)

=

+0.

50 V

*Or

any

othe

r ve

rsio

n of

a w

orki

ng n

icke

l-zi

nc v

olta

ic c

ell.

Pro

cedu

re:

1. C

onst

ruct

a v

olta

ic c

ell,

usin

g th

e re

agen

ts g

iven

in th

e di

agra

m.

2. P

lace

a 1

00.0

mL

sam

ple

of 1

.0 m

ol/L

Ni(

NO

3)2(

aq)

in th

e be

aker

con

tain

ing

the

Ni(s

) el

ectr

ode.

3. M

easu

re a

nd r

ecor

d th

e ce

ll p

oten

tial u

sing

a v

oltm

eter

.

4. R

epea

t ste

ps 1

to 3

usi

ng d

iffe

rent

con

cent

ratio

ns o

f N

i(N

O3)

2(aq

) so

lutio

n.

Key

Com

pone

nt

• va

lid e

xper

imen

tal

desi

gn o

r pr

oced

ure

that

atte

mpt

s to

m

anip

ulat

e th

e co

ncen

trat

ion

of

Ni2+

(aq)

in a

n el

ectr

oche

mic

al c

ell

Supp

ort C

ompo

nent

s

• de

taile

d pr

oced

ure

• la

belle

d ni

ckel

-zin

c ce

ll di

agra

m

• re

leva

nt b

alan

ced

equa

tion(

s) a

nd

pote

ntia

l dif

fere

nce

ca

lcul

atio

n




2 Key

The student has addressed the question asked by designing an experiment (procedure or design) involving a cell (voltaic or electrolytic, and not necessarily nickel-zinc) and has attempted to manipulate the concentration of nickel solution (or in the student’s understanding manipulated the concentration), and the student’s responding variable relates to measuring the cell potential.

0 No Key


Score Support



2 Satisfactory

to Good


1 Minimal





Method for calculating potential difference is correct, but the student has made a calculation error

Using a solution in the salt bridge that would precipitate

Major errors: Electrolytic cell, or a voltaic cell other than nickel-zinc

Missing electrolyte or other major error in the cell diagram

Flawed experimental design, or procedure not provided

chemistry 30 - alberta education · 2015-11-26 · alberta education, assessment sector 5 chemistry...

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