Name CT Group 07 Subject Tutor Index No.

NANYANG JUNIOR COLLEGE Promotional Examination

H2 CHEMISTRY Paper 2 JC 1/2007

9746/02

26 September 2007

1 hour

Mark Scheme

2

For Examiners

use:

1 Sodium halides such as NaF and NaCl, while having similar appearances and physical properties, have completely different practical uses in real life. NaCl is the main ingredient in table salt while NaF is commonly used in toothpaste.

The table below shows the melting points and lattice energy data for NaF,

NaCl and MgCl2:

Table 1

Compound Melting Point / C

Lattice Energy / kJ mol-1 Experimental

value Calculated

value Percentage discrepancy

NaF 996 923 910 1.43

NaCl 801 786 769 2.21

MgCl2 714 2526 2326 8.60

(a) With reference to their structure and bonding, compare the melting points of NaF and NaCl and account for the difference. [3]

o Both have giant ionic structures o with strong electrostatic attraction between oppositely charged

ions. o Cl- is larger than F-, o hence interionic distance is larger for NaCl

o Ionic bonds are weaker in NaCl, o hence melting point is lower.

Every pt mark. Round down to nearest whole number

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For Examiners

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The lattice energy of an ionic compound can be determined in two ways, by using experimental data in a Born-Haber cycle, or from theoretical calculations. However, the two methods yield lattice energy data that do not necessarily agree. This is due to the fact that theoretical calculations are based on the assumption that the compounds are purely ionic. Table 1 shows the percentage discrepancy between experimental lattice energy and calculated lattice energy for NaF, NaCl and MgCl2.

(b) Explain why MgCl2 has a larger percentage discrepancy between its

experimental and calculated lattice energy, compared to NaCl. [2]

Mg2+ is smaller and more highly charged than Na+ (higher charge density)

Cl- is large and easily polarised Mg2+ is able to polarize the electron cloud of Cl- to a greater extent Hence MgCl2 possess greater covalent character Every pt mark. Round down to nearest whole number

NaCl is also the raw material used to produce chlorine. Industrially, chlorine gas is produced by the electrolysis of concentrated sodium chloride solution (commonly known as brine). The industrial process is carried out in a particular diaphragm cell, as shown below:

Figure 1

Concentrated NaCl(aq)

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For Examiners

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In this particular industrial diaphragm cell (Figure 1), the cost of electricity is $22 / dm3 of chlorine gas produced at room temperature and pressure. The operating current of the cell is 1.50 x 103 A. (c) Calculate the hourly chlorine gas production (in moles) in this industrial

diaphragm cell. [2]

Q = I t = 1.50 x 103 x 3600 = 5.4 x 106 C - 1 mk ne = 5.4 x 106 / 96500 = 55.95 mol Since 2Cl- Cl2 + 2e nCl2 = 55.95 / 2 = 27.97 mol - 1 mk (no ecf)

(d) Hence, deduce the hourly electricity cost incurred during the operation of

this industrial diaphragm cell. [1]

VCl2 = 27.97 x 24.0 = 671.28 dm3 Hourly electricity cost = 671.28 x 22 = $14768 - 1 mk (ecf)

(e) Using relevant data from the Data Booklet, explain the need to use

concentrated sodium chloride solution, rather than dilute sodium chloride solution in the production of chlorine. [2]

E Cl2 + 2e 2Cl- +1.36 O2 + 2H2O + 4e 4OH- + 0.40

As E (O2/OH-) is less positive, OH- will be oxidized preferentially if dilute sodium chloride is used. Hence concentrated sodium chloride is required to ensure oxidation of Cl-.

1 mk for explanation

1 mk

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For Examiners

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Using relevant data from the Data Booklet, explain why the same diaphragm cell cannot be used to produce fluorine from either dilute or concentrated aqueous NaF. [1]

F2 + 2e 2F- +2.87

E (F2/F-) is too positive. Hence F- will not be oxidized even when concentrated NaF is used.

(f) The chlorine gas produced is an important raw material in the production

of chlorine dioxide (ClO2) which is used as a disinfectant during water treatment. The oxidising action of ClO2 is as follows:

ClO2(aq) + 4H+(aq) + 5e Cl (aq) + 2H2O(l) E = + 1.50 V

Use relevant data from the Data Booklet to predict the reaction, if any, when acidified chlorine dioxide is added to aqueous tin(II) sulphate. Calculate the E cell for the reaction and write a balanced equation. [2]

E ClO2(aq) + 4H+(aq) + 5e Cl (aq) + 2H2O(l) + 1.50 Sn4+ + 2e Sn2+ +0.15

E cell = 1.50 0.15 = +1.35 V (> 0, feasible) - 1 mk

2ClO2(aq) + 8H+(aq) + 5Sn2+ 2Cl (aq) + 4H2O(l) + 5Sn4+ 1 mk (no ecf)

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For Examiners

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2 It is important for industrial chemists to find methods that give maximum yield while minimising costs. One of the very important factors to consider is the reaction conditions, namely, temperature and pressure.

The diagram below shows how the percentage (%) yield of a particular

reaction can vary with pressure and temperature.

Figure 2

(a) (i) From Figure 2, predict with reasons, whether the reaction is exothermic or endothermic. [2]

Exothermic Increasing the temperature favours the backward reaction more /

results in lesser yield. Endothermic reaction is the backward reaction.

ALL 3 points 2 m, Exo + 1 point 1 m, Exo only 1 m

(ii) Use Figure 2 to suggest conditions which favour a high yield of the product. [1]

Low temp, high pressure

1 or 0 marks

% yield

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For Examiners

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(b) The Haber Process is a very important industrial process which produces ammonia, and feedstock for the production of nitrogen fertilizer and explosives,

N2 (g) + 3H2 (g) 2NH3 (g)

A plant operates at around 700 K and 250 atm. The molar ratio of nitrogen and hydrogen employed is 1:3.

The percentage of ammonia in the equilibrium mixture was found to be 20%. (i) Find the equilibrium partial pressures of nitrogen, hydrogen and

ammonia gases. [2] Method 1:

N2 (g) + 3H2 (g) 2NH3 (g) Initial amt 1 3 0 Change in amt - x - 3x + 2x Eqm amt 1-x 3-3x 2x n (total) = 1-x+3-3x+2x = (4 2x) mol 2x / 4 2x = 0.2 x = 1/3 PN2 = 1/5 (250) = 50 atm PH2 = 3/5 (250) = 150 atm PNH3 = 1/5 (250) = 50 atm

Method 2:

PN2 = 0.20 x 250 = 50 atm PH2 = (250 50) = 150 atm PNH3 = (250 50) = 50 atm

mole fraction 1 mark Partial Pressures- 1 mark

(ii) Hence, calculate the value of Kp at 700 K. [1]

Kp = PNH32 / (PN2 x PH23) = 1.48 x 10-5 atm-2

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For Examiners

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(iii) A method to increase yield is to cool the reaction mixture such that NH3 condenses into a liquid and is drained away as it forms. How does this method increase the yield? [1]

As NH3 is removed, by LCP, POE will be shifted to the right to form more NH3 to minimize the effect in the drop in the amt of NH3.

(c) Apart from studying the reaction conditions, it is also very important to

know the energetics of the reaction. (i) Given:

2NH3(g) + 72 O2(g) 2NO2(g) + 3H2O(l) H = 585.2 kJ mol1

N2(g) + 2O2(g) 2NO2(g) H = + 180.8 kJ mol1 and Hc (H2(g)) = 286.0 kJ mol1

Draw an energy cycle to determine the enthalpy change of reaction in the Haber Process, N2 (g) + 3H2 (g) 2NH3 (g) [2]

H N2 (g) + 3H2 (g) 2NH3 (g) +5/2 O2 +5/2 O2 2NO2 (g) + 3H2O (l) - 1 mark for missing state symbols / not balanced equation

(ii) Hence, state the value of enthalpy change of reaction for the Haber Process: [1] H = +180.8 + 3(-286.0) (-585.2) = - 92 kJ mol-1

(iii) State whether entropy increases or decreases in the Haber Process.

Explain your answer. [2]

Entropy decreases, no. of gaseous particles on the LHS > RHS, less no. of ways to arrange the gaseous particles.

1 mark for stating decrease, 1 mark for explain

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For Examiners

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3 Nitrosyl chloride is a yellow gas obtained as a decomposition product of aqua regia, a mixture of hydrochloric acid and nitric acid. The resulting mixture was historically used to dissolve gold. In modern times, it can be produced by the direct combination of nitrogen monoxide and chlorine gas according to the equation:

2NO(g) + Cl2(g) 2NOCl(g) H = +138 kJ mol1

(a)(i) On the axes below, sets of initial reactant concentrations and hypothetical rate equations are given. For each set of conditions, sketch the corresponding graph for the given axes. [4]

rate

[NO] / mol dm3

rate

Set 2: rate = k [Cl2] [NO] = 1.00 mol dm3 [Cl2] = 0.01 mol dm3

Set 3: rate = k [Cl2] [NO] = 0.01 mol dm3 [Cl2] = 1.00 mol dm3

time

[NO] [Cl2]

time

Set 4: rate = k [NO] [Cl2] [NO] = 0.01 mol dm3 [Cl2] = 0.01 mol dm3

Set 1: rate = k [NO] [Cl2] [NO] = 0.01 mol dm3 [Cl2] = 1.00 mol dm3

[NO] / mol dm3

1 mark for each graph

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For Examiners

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(b) Investigation of the actual reaction gives the following results.

Initial concentrations of reactants / mol dm3 Relative rate of

formation of NOCl [NO] [Cl2]

0.10 0.10 1

0.20 0.10 4

0.30 0.20 18 (i) Deduce the order of reaction with respect to NO. [1] Comparing experiments 1 and 3, When [NO] increases by 2 times, the relative rate increases by 4 times. Hence, the order of reaction with respect to NO = 2 1 mark for correct order of rxn with correct explanation (ii) Deduce the order of reaction with respect to Cl2. [1] rate = k[NO]2[Cl2]x Comparing experiments 1 and 3,

2

218 k(0.30) (0.2)1 k(0.10) (0.1)

x

x= 18 = 9 (2)x x = 1 Hence, the order of reaction with respect to Cl2 = 1 1 mark for correct order of rxn with correct explanation (iii) Write the rate equation for the reaction. [1] rate = k[NO]2[Cl2] ECF: rate equation based on answers to (i) & (ii)

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For Examiners

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(iv) The activation energy for the reaction is +223 kJ mol1, sketch and label a reaction pathway diagram for the reaction in the axis provided. [2]

Half-mark for each of the following; total round down Correct Ea = +223 correctly labeled Reactants Correct H = +138 correctly labeled Products 4 Compounds with the molecular formula C5H11Br show two types of isomerism,

structural and optical.

(a) The full structural formula of one of the isomers is given below. Draw its structural formula (condensed). [1]

Structural formula (condensed): .

2NO + Cl2

+223

CH

H

H

C

Br

C

C

H

HH

H

H

C

H

H

H

Energy / kJ mol-1

Reaction pathway

+138

2NOCl

One of the following: (do not penalize if Br is bracket) CH3C(CH3)BrCH2CH3 or (--BrC2H5) (CH3)2CBrCH2CH3 or (--BrC2H5) CH3CH2CBr(CH3)2 or (C2H5C--) CH3CH2C(CH3)Br(CH3) or (C2H5C--)

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For Examiners

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(b) In the boxes below, draw structural formulae for four other structural isomers (at least one must be optically active). [4]

(c) Using your answer in (b), draw diagrams to illustrate how one of the above

molecules gives rise to optical isomerism. [1]

End of Paper

1 mark for any structure (can be condensed, skeletal, full, displayed)

CH

H

H

C

C

H

C

H

H

C

H

H

Br

H

H

H

CBr

H

H

C

H

H

C

H

H

C

H

H

C

H

H

H

CH

H

H

C*

Br

H

C

H

H

C

H

H

C

H

H

H

CH

H

H

C

H

H

C

Br

H

C

H

H

C

H

H

H

C

Br

HH

C CC

C HH

H

H

H

H

H

H

H

CH

H

H

C

C

H

C*

Br

H

C

H

H

H

H

H

H

CBr

H

H

C*

C

H

C

H

H

C

H

H

H

H

H

H

C

C2H5

BrH2C CH3H C

C2H5

CH2BrH3CH C

CH(CH3)2

H3C BrH C

CH(CH3)2

CH3BrH

C

CH2CH2CH3

H3C BrH C

CH2CH2CH3

CH3BrH

1-bromopentane

2-bromopentane

3-bromopentane

1-bromo-2,2-dimethylpropane

1-bromo-2-methylbutane

2-bromo-3-methylbutane

1-bromo-3-methylbutane

2-bromopentane

1-bromo-2-methylbutane 2-bromo-3-methylbutane