EdExcel a Level Chemistry Unit 7 Mark Scheme Jun 2000

8/17/2019 EdExcel a Level Chemistry Unit 7 Mark Scheme Jun 2000

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Summer 2000 CH2 Mark Scheme

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1. (a) (i) Reagent: potassium dichromate (VI)/potassium manganate (VII) (1) or

formulae sulphuric acid or hydrochloric acid (1) or formulae

If potassium manganate(VII) chosen not HCl or conc H2SO4 for second mark

‘Acidfied dichromate’ or H+ / Cr2O72–(1) 2

(ii) amount of propanol = 5.67/60 = 0.0945 mol (1)

amount of propanoic acid produced = 0.64 × 0.0945

= 0.06048 mol (1)yield of propanoic acid = 74 × 0.06048 = 4.5 / 4.48 / 4.476 g (1)

OR by mass ratio: ratio acid/alcohol = 74/60 = 1.23 (1)

100% yield = 1.23 × 5.67g = 6.99 g (1)

64% yield = 6.99 g × 0.64 = 4.5 / 4.48 / 4.476 g (1) 3

(b) (i) increase in temperature:

(position of ) equilibrium goes to the right (1)

as endothermic left to right (1)

on the addition of sodium propanaoatethe position of equilibrium goes to left (1)

higher concentration of / more propanoate ions

orsodium propanoate produces propanoate ions (1) 4

(ii) pH rises (consequential on above) (1) 1

(c) (i)

14

12

9

6

3

10 20 30 40 50

start pH 2 to 4 ( )

correct general shape ( )

vertical 6/7.6 to 10/12 ( )

at 25 cm ( )

1

1

1

13

4

(ii) indicator : thymol blue (1) consequential on vertical part of graph

reason: pH change sharp around pKin value / its colour changes around end point

pH / band pH8 to 10 shown on graph (1) 2

(d) (i) fully dissociated and reactions identical

OR

H+ + OH–➤

H2O (1) 1

(ii) HCN weak acid / partially dissociated (1)

❉

Hionisation of HCN endothermic (1) 2

[19]

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2. (a) (i) enthalpy change/ energy change/heat change/heat given out/ energy releasedper mole of substance burning in excess oxygen / burning completely (in oxygen )under standard conditions (2)

All three points in bold (2 marks) two points (1 mark) 2

(ii) C2H6(g) + 3½ O2(g) ➤ 2CO2(g) + 3H2O(l) ❉ H = –1560.0(kJ mol–1 )

correct stioichiometry (1 3½ = 2 3) (1)

correct state symbols (g g g l) water must be liquid (1)

Correct value of❉

H (if 2 C2H6 etc. in equation

❉

H must be –3120 kJ mol–1) (1) 3

(iii)

2C(s) + 3H (g) + (3½O ) C H (g) + (3½0 ) =heat of formation

H 2

2 2

2 2 26

2CO (g) + 3H O ( )1

Cycle (fully balanced) (1)

❉ H formation ethane = 2(–393.5) + 3(–285.8) – (– 1560)

= – 787 – 857.4 + 1560

= – 84.4 (kJ mol–1)

Working (1) answer (1) 3

(b) This mixture is thermodynamically unstable

Either as the reaction is very exothermic

Or energy level of products below energy level of reactants (1)

This reaction is kinetically stable because the activation energy is high (1) 2

(c) (i) atom/species with an unpaired electron (1) 1

(ii) where one atom or group replaces (not substitutes) another (1)

reference to ‘molecule’ rather than ‘group or atom’ scores (0)if the word ‘substitutes’ is used the mark can beawarded for a relevant example 1

(iii) one electron goes to the carbon atom (to form a radical) (1)the other goes to form a bond with the chlorine atom (1)

The second point could be shown on the diagram on the question paper 2

(iv)➲

Cl +➲

Cl➤

Cl2

or CH3CH2 ➲ + CH3CH2➲ ® C4H10

or Cl➲

+ CH3CH2➲ ®CH3CH2Cl (1) 1

(d) break C–H + 412make H–Br – 366 (1)

❉

H = + 46 (kJ mol–1) (1)

Correct answer with no working (2) 2

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(e) (from expt 1 & 2:/ [CN–] const ) [C2H5Br] doubles, rate doubles

1st order wrt C2H5Br (1)

(from expt 1 & 3:/ [C2H5Br] const) [CN–] doubles, rate doubles

1st order wrt CN– (1)

Rate = k[CN–][C2H5Br] (1)

The rate equation is consequential on the answers to the first part of the section 3

(f) (i) Route I as both reactants appear in rate equation rate determining stepOR Route I as route II would be zero order wrt CN– (1)

(ii) reactants

Uncatalysed

Catalysed

products

Correct profile for uncatalysed (1) ignore intermediate if shown

Correct profile for catalysed (1)

Labelled products lower energy than reactants

A correctly drawn unlabelled diagram can score maximum 2 marks 3[24]

3. (a) (i) KOH/NaOH (1)

ethanol/alcoholic/ specified alcohol (1)

heat (under reflux) (1) If use aqueous maximum 1 mark

If use aqueous alcohol max 2 marks 3

(ii) LiAlH4 (1) dry ether (1) 2

(iii) aqueous ( concentrated or dilute) NaOH /KOH or NaOH/KOH solution (1)

boil/heat (1) 2

(b) (i) 2,4–d n.p. ; carbonyl/ C=O group / aldehyde and ketone / aldehyde or ketone (1)

not just aldehyde or just ketone

Ag+ /NH3 ; aldehyde /CHO group (1)

PCl5 ; OH group /.acid and alcohol / acid or alcohol (1)

not just alcohol

Br2 (aq); C=C / alkene (1) 4

(c) (i)CH CH CH CHO CH CHCH

CHO

3 3 32 2

P

( )1 ( )1

2

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(ii)2

3

2

3Q

CH CH H C CH

C C

CH CH

H HOH HO

Correct identification of Q (1)Isomers correctly drawn (1)

The use of C2H3 in formula is acceptable 2

(iii)3 3

S

CH COOH CH

COOH

C C C C

H HH

H

Correct identification of S (1) Isomers correctly drawn (1)

Two correctly drawn isomers of R scores 1 mark 2[17]

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EdExcel a Level Chemistry Unit 7 Mark Scheme Jun 2000