OCR A-Level Chemistry Unified chemistry (03), June 2023: Question 2
11 marks · Hard difficulty · Structured Questions
Calculate the enthalpy change of solution for rubidium chlorate, determine the initial rate for a diluted hydrochloric acid solution, and complete the reaction mechanism for the formation of phenol from a benzenediazonium ion.
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Question text
2 These questions are from different areas of chemistry.
(a) This question is about two salts of rubidium (atomic number 37): RbClO3 and RbClO4.
(i) The oxidation number of chlorine is different in the two rubidium salts, RbClO3 and
RbClO4.
What is the name of RbClO4?
… [1]
(ii) A student carries out an experiment to determine the enthalpy change of solution of
RbClO3 using the method below.
• A 2.00 g sample of solid RbClO3 is added to water in a well-insulated container.
The initial temperature is 23.0 °C.
• The mixture is stirred until all the RbClO3 has dissolved.
The final temperature is 21.5 °C.
The final solution has a mass of 102 g.
Determine the enthalpy change of solution, Δ H, of RbClO in kJ mol–1.
sol 3
Assume that the specific heat capacity of the solution is the same as that of pure water.
Δ H (RbClO ) = … kJ mol–1 [3]
sol 3
(b) A student investigates the rate of a reaction that is 1st order with respect to hydrochloric
acid, HCl(aq).
• The student carries out a reaction using 0.680 mol dm–3 HCl(aq). The initial rate is
9.52 × 10–4 mol dm–3 s–1.
• The student dilutes a different sample of 0.680 mol dm–3 HCl(aq) with water.
The pH of this diluted acid is 1.50.
• The student repeats the reaction using the same volume of this diluted acid.
Determine the initial rate using this diluted acid.
initial rate = … mol dm–3 s–1 [3]
(c) The benzenediazonium ion, shown below, is stable at temperatures below 10 °C.
+
N N
Benzenediazonium
Above 10 °C, the benzenediazonium ion reacts with water to form phenol.
The reaction proceeds in a three-step mechanism.
Step 1 Elimination of nitrogen gas to form a carbocation.
Step 2 Nucleophilic attack by water.
Step 3 Proton loss to form the organic product.
Complete the boxes below with intermediates and curly arrows to show the mechanism for
this reaction.
+ Step 1
N N + …
Step 2
Step 3
OH + H+
[4]
Mark scheme
Show the mark scheme
Question Answer Marks AO Guidance
element
2 (a) (i) Rubidium chlorate(VII) ✓ 1 AO1.1 ALLOW Rubidium(I) chlorate(VII)
Rubidium chloroate(VII)
IGNORE Rubidium (VII)chlorate
Rubidium chlorate(IIV)
Rubidium chlorate (7)
Rubidium perchlorate
(ii) FIRST CHECK THE ANSWER ON ANSWER LINE 3 AO2.8 ALLOW ECF throughout
If answer = 54.0 OR 54.1 OR 54.2 (kJ mol–1) award 3 marks ×3
---------------------------------------------------------------------------------
Energy change from mcΔT
Energy in J OR kJ IGNORE sign
= 102 4.18 1.5 OR 639.54 (J) OR 0.63954 (kJ) IGNORE RE and SF in 1st 2 marks
----------------------------------------------
Amount in mol of RbClO3
2.00
n(RbClO3) = 169 OR 0.0118 … (mol) 0.01183431953 unrounded
----------------------------------------------
∆solH(RbClO3) ALLOW 54 (from 54.0)
CARE 54.00 is a rounding error
0.63954 ---------------------------------------------------
= 0.0118….. = (+) 54.0 COMMON ERRORS
52.98 OR 53.14 2 marks
From unrounded values, ∆H = 54.04113 100 instead of 102:
Energy = 100 4.18 1.5 = 627 J
Examples of mixed acceptable intermediate rounding,
e.g. From unrounded n,
0.640 0.627 –1
∆H = 54.237 → 54.2 ∆H = 0.0118….. = 52.98 kJ mol
0.0118
OR 53.0 (3SF) OR 53
0.63954 From rounded 0.0118,
0.01183 ∆H = 54.06 → 54.1 0.627
∆H = 0.0118 = 53.14 OR 53.1
-----------------------------------
element
0.02078 OR 0.0208 1 mark
102 and 2 swapped:
Energy = 2 4.18 1.5 = 12.54 J
n = 169 = 0.60355……
0.01254 –1
ECF ∆H = 0.60355….. = 0.0208 kJ mol
-----------------------------------
1.06 2 marks
102 for n instead of 2.00:
n = 169 = 0.60355……
0.63954 –1
∆H = 0.60355….. = 1.06 kJ mol
OR
2 for energy instead of 102
Energy = 2 4.18 1.5 = 12.54 J
0.01254 –1
∆H = 0.0118….. = 1.06 kJ mol
-----------------------------------
107.4 – 107.7 2 marks
8.314 for c instead of 4.18:
Energy = 102 8.314 1.5 = 1272 J
Energy = 102 8.31 1.5 = 1271.4 J
∆H = 107.4 – 107.7 kJ mol–1
depends on intermediate rounding
CHECK
-----------------------------------
Apply ECF for any other comparable
responses. If in doubt contact TL
Question Answer 11 Marks AO Guidance
element
(b) FIRST CHECK THE ANSWER ON ANSWER LINE 3 AO3.1
If range = 4.4 10–5 – 4.5 10–5 (kJ mol–1) award 3 marks 3
---------------------------------------------------------------------------------
[H+] = 10–1.50 OR 0.0316 … OR 0.032 mol dm–3 1 mark Calculator: 0.0316227766
ALLOW 10–1.5
THEN 2 APPROACHES:
EITHER: ECF possible from incorrect [H+]
Factor that concentration changes by 1 mark
0.0316….. From unrounded [H+],
Factor = 0.680 = 0.0465… times Calculator: 0.04650408324
0.680
OR = 21.5….. times From [H+] = 0.032, Factor = 21.25
0.0316…..
Initial rate with diluted acid 1 mark
From unrounded [H+],
9.52 10–4
= 0.0465… 9.52 10–4 OR –5
21.5….. Calculator = 4.427188724 10
= 4.43 10–5 (mol dm–3 s–1)
From [H+] = 0.032, rate = 4.48 10–5
OR:
Rate concentration (1st order) 1 mark -----------------------------------------------
rate 9.52 10–4
k = = = 1.4(0) 10–3 ECF possible from incorrect [H+]
[HCl] 0.680
0.680
OR Constant = –4 = 714.2857… DO NOT ALLOW ECF unless derived
9.52 10 from concentration and rate
Initial rate with diluted acid
–3 0.0316…
= 1.4(0) 10 0.0316 … OR 714.2857…
= 4.43 10–5 (mol dm–3 s–1)
SUMMARY M1 [H+] 0.0316…. OR 0.032 1 mark
M2 Working 0.0465 OR 21.5 OR 1.4 10–3 OR 714 1 mark
M3 Initial rate Range: 4.4 10–5 – 4.5 10–5 2 SF or more depends on intermediate rounding CHECK 1 mark
element
(c) Mechanism: 4 AO3.2 ANNOTATE ANSWER TICKS AND
4 CROSSES
-----------------------------------------------------
N N Step 1 N2 NOTE: Curly arrows can be straight,
+ …
snake-like, etc. but NOT half arrows
H2O
1st curly arrow must start from, OR be
traced back to, any part of C–N+ bond
and go to N OR + of N+
Step 2
N N N N
H
Step 3 N N
OH + H+ O
H
2nd curly arrow must
• start from, OR be traced back to
M1: Curly arrow from C−N bond to N+ any point across width of lone
pair on O of H2O
• go to the C or + of C+ of C H +
M2: AND N 6 5
M3: Curly arrow from lone pair of O of H O to C+
H H O H O H O H2O
22 2
M4 O AND Curly arrow from O–H bond to O+
H
3rd curly arrow must
• start from ‘–‘ of O–H of –OH +
For all marks, treat additional curly arrows as CON • go to O or + of O+
H H H
ALLOW M3 shown in bottom box O
O O
IGNORE partial charges H H H
element
ALLOW M3 AND M4 combined
e.g. For + DO NOT ALLOW M2 for carbocation
BUT
ALLOW for M3 and/or M4 by ECF, e.g.
--------------------------------------------------------------
For DO NOT ALLOW M2 for carbocation
BUT
ALLOW for M3 and/or M4 by ECF, e.g.
How to answer it
OCR A-Level Chemistry Multi-Topic Study Guide
What this question tests
This synoptic-style paper tests your understanding across three distinct areas of chemistry: inorganic nomenclature and calorimetry calculations, reaction kinetics (pH, concentration, and initial rates), and organic reaction mechanisms (diazonium ion substitution and curly arrow conventions).
Part (a)(i) - Naming Rubidium Salts
Determine the systematic IUPAC name for RbClO₄
✅ Correct Answer
Rubidium chlorate(VII)
Also accepted: Rubidium chlorate(V) for the other salt if contextualized, but specifically for RbClO₄, Roman numerals indicate oxidation state +7.
💡 Key Knowledge
- Oxygen has an oxidation state of -2, and rubidium (Group 1) is +1.
- In RbClO₄ , let chlorine be x : +1 + x + 4(-2) = 0 , meaning x = +7 .
- Roman numerals in brackets denote the oxidation state of the central non-metal atom.
Part (a)(ii) - Enthalpy Change of Solution
Calculate enthalpy change of solution ( ΔsolH ) for RbClO₃
📐 Step-by-Step Calculation
- Calculate moles of RbClO₃ :
Molar mass = 85.47 + 35.45 + (3 × 16.00) = 169.92 g mol⁻¹ (or 169 g mol⁻¹)
Moles = mass / Mr = 2.00 / 169.2 = 0.0118 mol - Calculate heat energy change ( q ):
q = m × c × ΔT
Mass of solution ( m ) = 102 g, c = 4.18 J g⁻¹ K⁻¹, ΔT = 23.0 - 21.5 = 1.5 K
q = 102 × 4.18 × 1.5 = 639.54 J = 0.63954 kJ - Calculate enthalpy change per mole ( ΔsolH ):
ΔsolH = -q / moles = -0.63954 / 0.0118 = -54.0 kJ mol⁻¹
*(Note: temperature fell, meaning the process is endothermic, so ΔsolH is positive: +54.0 kJ mol⁻¹ )*
❌ Common Calculation Traps
- Mass error: Using 100 g instead of 102 g (adding the solute mass to water mass).
- Sign confusion: Forgetting that a temperature drop indicates an endothermic reaction requiring a positive +/+ sign.
- Rounding too early: Rounding intermediate values leads to answers like 52.98 or 53.14 instead of 54.0 .
Part (b) - Reaction Kinetics
Determine the initial rate using diluted HCl(aq)
📐 Step-by-Step Calculation
- Find [H⁺] in the diluted acid from pH:
[H⁺] = 10⁻ᵖᴴ = 10⁻¹.⁵⁰ = 0.0316 mol dm⁻³ - Determine the dilution factor or rate constant:
Ratio = 0.680 / 0.03162 = 21.5 (concentration decreases by a factor of 21.5)
Alternatively, calculate rate constant k = rate / [HCl] = (9.52 × 10⁻⁴) / 0.680 = 1.40 × 10⁻³ s⁻¹ - Calculate new initial rate:
Since the reaction is 1st order: Rate = k × [HCl]
Rate = (1.40 × 10⁻³) × 0.03162 = 4.43 × 10⁻⁵ mol dm⁻³ s⁻¹
🧠 Exam Technique & Correct Answer
Initial Rate = 4.40 × 10⁻⁵ to 4.50 × 10⁻⁵ mol dm⁻³ s⁻¹
- Always state units clearly: mol dm⁻³ s⁻¹ for initial rate.
- Look out for logarithmic calculations ( pH to [H⁺] ); ensure proper use of inverse log functions on your calculator.
Part (c) - Organic Mechanism
Complete the three-step mechanism for the reaction of benzenediazonium ion with water
💡 Mechanism Breakdown (Curly Arrows)
- Step 1 (Elimination): A curly arrow starts from the C-N bond and goes to the N⁺ nitrogen atom, showing the departure of neutral N₂ gas and formation of a phenyl carbocation ring intermediate C₆H₅⁺ .
- Step 2 (Nucleophilic Attack): A curly arrow starts from a lone pair on the oxygen atom of water ( H₂O ) and points directly to the positively charged carbon atom of the carbocation, forming a protonated phenol intermediate ( C₆H₅-OH₂⁺ ).
- Step 3 (Proton Loss): A curly arrow starts from the O-H bond of the protonated intermediate and goes to the oxygen atom (or a base like water), releasing H⁺ and yielding phenol ( C₆H₅OH ).
❌ Common Errors in Mechanisms
- Starting curly arrows in mid-air instead of precisely on bonds or lone pairs.
- Drawing arrows pointing *away* from electron-deficient centres rather than toward them.
- Omitting or misplacing formal charges (e.g., forgetting the positive charge on the carbocation or intermediate oxonium ion).
Topics
Module 5: Physical chemistry and transition elements · Module 6: Organic chemistry and analysis · Practical Activity Groups · 5.1 Rates, equilibrium and pH · 5.2 Energy · 6.1 Aromatic compounds, carbonyls and acids · PAG 3: Enthalpy determination · PAG 9: Rates of reaction – continuous monitoring method
Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.