DP Chemistry (last assessment 2024)

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Question 23M.3.SL.TZ2.11

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Date May 2023 Marks available [Maximum mark: 6] Reference code 23M.3.SL.TZ2.11
Level SL Paper 3 Time zone TZ2
Command term Determine, Explain, Identify, Suggest, Write Question number 11 Adapted from N/A
11.
[Maximum mark: 6]
23M.3.SL.TZ2.11
(a)

Photosynthesis enables green plants to store energy from sunlight as glucose.

(a.i)

Write the equation for photosynthesis.

[1]

Markscheme

6CO2 (g) + 6H2O (l) → C6H12O6 (aq) + 6O2 (g) ✓

(a.ii)

Identify the structural feature that allows chlorophyll to absorb light.
Use section 35 of the data booklet.

[1]

Markscheme

conjugated «electronic» structure/delocalized «pi» electrons/alternate «single and» double bonds ✓

 

Accept “many/delocalized double bonds”.

Do not accept “tetrapyrrole group” alone without reference to idea of conjugation.

(a.iii)

Explain how photosynthesis is being employed to control global warming.

[2]

Markscheme

reduces/sequesters CO2/carbon dioxide «concentration from atmosphere» ✓

«planting» more plants/trees ✓

 

Do not accept “carbon capture” alone for M1.

Do not accept just “plants/trees” alone for M2.

(b)

Glucose can be converted to ethanol through fermentation:

C6H12O6 (aq) → 2C2H5OH (aq) + 2CO2 (g)

(b.i)

Determine the energy efficiency of this conversion in terms of the enthalpies of combustion of the reactants and products. Use section 13 of the data booklet.

[1]

Markscheme

«(2x −1367 / −2803) × 100 =» 97.54 %
OR 2.46 % loss «in energy efficiency» ✓

(b.ii)

Suggest one reason, other than energy density and specific energy, why ethanol may be considered a more useful fuel than glucose.

[1]

Markscheme

liquid
OR
easier ignition
OR

more volatile ✓

 

Accept “complete combustion more likely”“
OR
“better octane rating”
OR
“engine must be converted in order to use glucose”.

Do not accept “less viscous”.