OCR A-Level Chemistry Unified chemistry (03), June 2017: Question 1
11 marks · Medium difficulty · Structured Questions
Explain bond angles in water/ice, state concerns about methane release, determine formula of methane hydrate, calculate volume of released methane using ideal gas equation, and suggest an industrial interest in methane hydrate.
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Question text
1 Within the permafrost in Arctic regions of the Earth, large amounts of methane are trapped within
ice as ‘methane hydrate’, CH4•xH2O. Methane makes up about 13.4% of the mass of ‘methane
hydrate’.
Scientists are concerned that global warming will melt the permafrost, releasing large quantities of
methane into the atmosphere.
(a) The H–O–H bond angle in ice is about 109° but about 105° in gaseous H2O.
Explain why there is this difference.
… [3]
(b) Why are scientists concerned about the release of methane into the atmosphere?
… [1]
(c) Determine the formula of ‘methane hydrate’, CH4•xH2O.
In the formula, show the value of x to two decimal places.
formula = … [2]
(d) Calculate the volume of methane, in dm3, that would be released from the melting of each
1.00 kg of ‘methane hydrate’ at 101 kPa and 0 °C.
Give your answer to three significant figures.
volume = … dm3 [4]
(e) Suggest why some industries are interested in the presence of ‘methane hydrate’ in regions
of the Earth.
… [1]
Mark scheme
Show the mark scheme
Question Answer Marks Guidance
1 (a) Throughout
ALLOW bonding regions for bonded pairs
ALLOW diagrams for communicating two bonds, two lone pairs and hydrogen bonding in ice
IGNORE responses about open lattice/tetrahedral structure in ice
Ice 3
Ice has hydrogen bonds/bonding ALLOW more hydrogen bonding/H bonds
H2O(g)
2 bonded pairs AND 2 lone pairs For H2O(g),
ALLOW water
Repulsion IGNORE hydrogen bonding
Lone pairs repel more (than bonded pairs)
(b) It increases/causes/contributes to global warming 1 ALLOW it is a greenhouse gas/increases temp
OR C–H bonds vibrate OR absorb IR
IGNORE ozone, radicals OR acid rain
(c) FIRST CHECK THE ANSWER ON THE ANSWER LINE 2 Working to at least 3 SF but IGNORE ‘trailing
IF answer = CH4•5.74 H2O OR 5.74 award 2 marks zeroes’, e.g. ALLOW 16 for 16.0
---------------------------------------------------------------------------- ------------------------------------------------------------------
Mole ratio ALLOW algebraic approach, e.g.
13.4 86.6 n(CH4) = n(CH4•xH2O)
n(CH4) : n(H2O) = 16.0 : 18.0 13.4 100
16.0 = 16.0 + 18x
OR 0.8375 : 4.811
x = 5.74
Formula ALLOW ECF from incorrect mole ratio
CH4•5.74 H2O OR 5.74 -----------------------------------------------
For 1 mark, ALLOW x with < 2 DP:
x = 5.7
x = 6
x = 5.73 from 0.8375 and 4.8
from 0.84 and 4.811
x = 5.71 from 0.84 and 4.8
(d) FIRST CHECK THE ANSWER ON THE ANSWER LINE 4
IF answer = 188 (dm3) AND use of ideal gas equation ALLOW use of M(answer to (c) OR 119.32
Award 4 marks for calculation Examples
---------------------------------------------------------------------------- From n(CH4•5.74 H2O)
n(CH ) in 1 kg 1 103
1 103 13.4 119.32 = 8.38(1) 188 (dm )
n(CH4) = 16.0 100 = 8.375 OR 8.38 (mol)
From n(CH4•5.7 H2O)
Rearranging ideal gas equation 1 103
= 8.43(2) 189 (dm3)
nRT 118.6
V = p
nRT From n(CH4•6 H2O)
Substitution of values into V = p : 3
1 10 3
Calculated value of n(CH ) (Use ECF) 124.0 = 8.06 (mol) 181 (dm )
R = 8.314 OR 8.31 ----------------------------------------------------------------
T in K: 273 K nRT
35 IF V = is omitted, ALLOW when values are
p in Pa OR kPa 101 OR 101 10 OR 1.01 10 p
substituted into rearranged ideal gas equation.
8.375 8.314 273 8.375 8.314 273
e.g. 3 OR 101
(101 10 )
Final volume in dm3 to 3 SF
V = 188 (dm3)
COMMON ERRORS
Use of 298 K ALLOW ECF 3 marks max
8.375 8.314 298 3
Example n(CH4•5.74 H2O) = 8.375 V = 3 205 (dm )
101 10
Use of 24.0 dm3 OR 22.4 dm3 ALLOW ECF from n(CH ) 2 marks max for n(CH ) and V in dm3
24.0 dm3 n(CH •5.74 H O) = 8.375 V = 8.375 24.0 = 201 (dm3)
22.4 dm3 n(CH •5.74 H O) = 8.375 V = 8.375 22.4 = 188 (dm3)
13.4% (13.4/100) omitted 3 marks
1 103 62.5 8.314 273
n = = 62.5 (mol) V = 1400 (dm3)
16 101 103
(e) For fuel OR energy 1 ALLOW responses linked with energy. e.g.
to generate electricity
for burning/heat
ALLOW (chemical) feedstock
IGNORE cooking
Total 11
How to answer it
Methane Hydrate in Arctic Permafrost Study Guide
What this question tests
This multi-part exam question evaluates your understanding of shape and bond angles in molecules (VSEPR theory), environmental chemistry relating to greenhouse gases, empirical formula calculations involving water of crystallisation, applying the ideal gas equation (pV = nRT), and industrial applications of fossil resources.
Explaining Shape and Bond Angles (3 Marks)
✅ Correct Answer Requirements
- Ice has hydrogen bonding/bonds between molecules (1 mark).
- Gaseous H₂O has 2 bonded pairs and 2 lone pairs (1 mark).
- Lone pairs repel more than bonded pairs, compressing the angle (1 mark).
💡 Key Knowledge
In ice, molecules are held in a fixed, rigid tetrahedral lattice largely dictated by directional hydrogen bonding. In isolated gaseous water molecules, electron pair repulsion theory applies fully: lone pair-lone pair repulsion > lone pair-bond pair repulsion > bond pair-bond pair repulsion, giving a smaller 105° angle.
🧠 Exam Technique
Always structure your explanation of bond angles systematically: state the number of bonding pairs and lone pairs, state the relative repulsions explicitly, and explicitly link lone pair repulsion to the reduction of the bond angle.
❌ Common Errors
Students often lose marks by simply stating "lone pairs repel" without comparing them to bonding pairs, or by discussing open crystal structures in ice without mentioning hydrogen bonding directly.
Methane as a Greenhouse Gas (1 Mark)
✅ Correct Answer
Methane is a greenhouse gas / it increases/causes global warming / C-H bonds absorb IR radiation / vibrate.
🧠 Exam Technique & Guidance
Keep your answer direct. Examiners accept "it is a greenhouse gas" or "increases global temperature". Do not confuse methane release with ozone depletion or acid rain.
Determining Water of Crystallisation in Methane Hydrate (2 Marks)
📐 Step-by-Step Calculation
Step 1: Use percentage composition (13.4% CH₄ means 86.6% H₂O).
Step 2: Calculate moles for each component per 100g of sample:
n(CH₄) = 13.4 / 16.0 = 0.8375 mol
n(H₂O) = 86.6 / 18.0 = 4.811 mol
Step 3: Find the simplest whole-number ratio:
Ratio = 0.8375 : 4.811 → 1 : 5.74
✅ Final Answer Line
CH₄ • 5.74 H₂O
Calculating Gas Volume from Mass (4 Marks)
📐 Step-by-Step Calculation
Step 1: Find moles of CH₄ in 1.00 kg (1000 g) of methane hydrate:
M(CH₄ • 5.74 H₂O) = 16.0 + (5.74 × 18.0) = 119.32 g mol⁻¹
n(CH₄) = (1000 / 119.32) × (13.4 / 100) = 8.375 mol
Step 2: Rearrange the Ideal Gas Equation:
pV = nRT → V = nRT / p
Step 3: Substitute values with correct units:
n = 8.375 mol, R = 8.314 J K⁻¹ mol⁻¹, T = 273 K, p = 101 × 10³ Pa (or 101 kPa)
V = (8.375 × 8.314 × 273) / (101 × 10³) = 188.13 dm³
❌ Common Calculation Traps
- Temperature trap: Using 298 K instead of 273 K (0°C).
- Pressure unit trap: Forgetting to convert kPa to Pa when using R in J K⁻¹ mol⁻¹.
- Molar volume shortcut trap: Using 24.0 dm³ mol⁻¹ instead of the ideal gas equation (conditions are not standard laboratory conditions of 298K/101kPa).
✅ Final Answer
188 dm³ (to 3 significant figures)
Commercial Significance of Methane Hydrate (1 Mark)
✅ Correct Answer
Used as a fuel / for energy generation / burning / chemical feedstock.
🧠 Examiner Insight
Accept any concise answer linking the vast reserves of trapped methane to energy use or combustion, but ignore vague answers like "cooking".
Topics
Module 2: Foundations in chemistry · Module 3: Periodic table and energy · 2.2 Electrons, bonding and structure · 2.1 Atoms and reactions
Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.