OCR A-Level Chemistry Unified chemistry (03), June 2025: Question 6

Question Answer Marks Guidance

6 Please refer to the marking instructions on page 5 of this mark scheme 6 TWO APPROACHES: from n(Na2CO3) and from n(HNO3)

* for guidance on how to mark this question.

Level 3 (5–6 marks) SEE NEXT DOUBLE PAGE FOR GUIDANCE.

Calculates CORRECT enthalpy change ∆H2 using m = 33.5 g

in reaction 6.2. PAGE 16 NEEDS ‘SEEN’

AND

CORRECT enthalpy change ∆H1 in reaction 6.1

There is a well-developed line of reasoning which is clear and logically ALLOW omission of trailing zeroes

structured. ALLOW minor slips

The information presented is relevant and substantiated.

3 SF or more throughout

Level 2 (3–4 marks)

Using m = 30 – 33.5 g, c = 4.18/4.2 and ∆T = 4 calculates a value

for energy change per mole of Na2CO3 OR energy change per

mole of HNO3

AND

the CORRECT amount in mol of Na2CO3 OR HNO3

There is a line of reasoning presented with some structure.

The information presented is relevant and supported by some evidence.

Level 1 (1–2 marks)

Processes experimental data to obtain:

Calculates energy change correctly from mc∆T

OR

Attempts to calculate energy change from mc∆T

with two of m (30–33.5g), c (4.18/4.2) and ∆T (4) correct

AND amount in mol of Na2CO3 OR HNO3

There is an attempt at a logical structure with a line of reasoning.

The information is in the most part relevant.

0 marks – No response or no response worthy of credit.

Guidance Guidance

A: APPROACH based on n(Na2CO3) B: APPROACH using n(HNO3)

Indicative scientific points may include: Indicative scientific points may include:

1. Energy change from mcΔT 1. Energy change from mcΔT

33.50 × 4.18 × 4.0 = 560.12 (J) OR 0.56012 (kJ) 33.50 × 4.18 × 4.0 = 560.12 (J) OR 0.56012 (kJ)

OR 30.0 – 33.15, e.g. OR 30.0 – 33.15, e.g.

30 × 4.18 × 4.0 = 501.6 (J) OR 0.5016 (kJ) 30 × 4.18 × 4.0 = 501.6 (J) OR 0.5016 (kJ)

---------------------------------------------------- ----------------------------------------------------

Amount of Na2CO3 Amount of HNO3 and the Na2CO3

1.97 1.00 30.0

n(Na2CO3) = 106 = 0.01858…. (mol) n(HNO3) = = 0.0300 (mol)

1000

Amount of Na2CO3

0.0300

n(Na2CO3) = 2 = 0.0150 (mol)

---------------------------------------------- ----------------------------------------------

2. Energy change per mole of Na CO in kJ mol–1 –1

23 2. Energy change per mole of Na2CO3 in kJ mol

0.56012 0.56012

From m = 33.50 g = ± 0.01858…. = ±30.138… From m = 33.50 g = ± = ± 37.34

0.0150

ALLOW 30.11 → 30.15 0.5016

0.5016 From m = 30 g = ± = ±33.44

0.0150

From m = 30 g = ± = ±26.9896…

0.01858…. ** Some may calculate energy change per mole of HNO3

ALLOW ±27.0 OR ±26.9 giving:

0.56012 0.5016

--------------------------------------------- ± 0.0300 = ± 18.67 and ± 0.0300 = ± 16.67

3. ∆H2 and ∆H1 CORRECT with signs (From 33.50 ONLY) ----------------------------------------------

3. ∆H2 and ∆H1 CORRECT with signs (From 33.50 ONLY)

Reaction 6.2: ∆H = –30.138 (kJ mol–1)

ALLOW –30.11 → –30.15 Reaction 6.2: ∆H = –37.34 (kJ mol–1)

Reaction 6.1

∆H1 = 2 ∆H3 – ∆H2 Reaction 6.1

= 2 19.8 – (–30.138….) = ∆H = (+)69.7 (kJ mol–1)

1 ∆H1 = 2 ∆H3 – ∆H2

= 2 19.8 – (–37.34) ∆H = (+)76.94 (kJ mol–1)

NOTE: A clear and logically structured response might include an NOTE: A clear and logically structured response might include

energy cycle and accurate use of SF, 4.18 and signs throughout an energy cycle and accurate use of SF, 4.18 and signs

throughout