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A-level Chemistry Paper 2 (7405/2): Organic and Physical Chemistry Mark scheme

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A-level Chemistry Paper 2 (7405/2): Organic and Physical Chemistry Mark scheme 7405 Specimen paper Version 0.5 Question Marking guidance Mark AO Comments 01.1 Consider experiments 1 and 2: [B constant] [A] increases × 3: rate increases by 32 therefore 2nd order with 1 AO3 1a respect to A Consider experiments 2 and 3: [A] increases × 2: rate should increase × 22 but only increases × 2 Therefore, halving [B] halves rate and so 1st order with respect to B 1 AO3 1a Rate equation: rate = k[A]2[B] 1 AO3 1b 01.2 rate = k [C]2[D] therefore k = rate / [C]2[D] 1 AO2h k = 7.2 ×10-4 = 57.0 (1.9 ×10- 2 )2 × (3.5 ×10- 2 ) 1 AO2h Allow consequential marking on incorrect transcription ................................................continued.............................................................................

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A-level
Chemistry
Paper 2 (7405/2): Organic and Physical Chemistry
Mark scheme

7405
Specimen paper


Version 0.5

, MARK SCHEME – A-level Chemistry – Specimen paper
2




Question Marking guidance Mark AO Comments

01.1 Consider experiments 1 and 2: [B constant]
[A] increases × 3: rate increases by 32 therefore 2nd order 1 AO3 1a
with
respect to A
Consider experiments 2 and 3:
[A] increases × 2: rate should increase × 22 but only increases
×2
Therefore, halving [B] halves rate and so 1st order with 1 AO3 1a
respect to B
Rate equation: rate = k[A]2[B] 1 AO3
1b

01.2 rate = k [C]2[D] therefore k = rate / [C]2[D] 1 AO2h

k= 7.2 ×10
-4
= 57.0 1 AO2h Allow consequential marking on
-2 2 -2 incorrect transcription
(1.9 ×10 ) × (3.5 ×10 )

–2 +6 –1 1 AO2h Any order
mol dm s

01.3 rate = 57.0 × (3.6 × 10–2)2 × 5.4 × 10–2 = 3.99 × 10–3 (mol 1 AO2h
dm–3 s–1)
OR
Their k × (3.6 × 10–2)2 × 5.4 × 10–2



2 of 20

, MARK SCHEME – A-level Chemistry – Specimen paper
2




01.4 Reaction occurs when molecules have E≥Ea 1 AO1a
Doubling T causes many more molecules to have this E 1 AO1a
Whereas doubling [E] only doubles the number with this E 1 AO1a


01.5 Ea = RT(lnA – lnk)/1000 1 AO1b Mark is for rearrangement of equation and
factor of 1000 used correctly to convert J into
kJ
E = 8.31 × 300 (23.97 – (–5.03))/1000 = 72.3 (kJ mol–1) 1 AO1b
a




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