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Date May 2017 Marks available 1 Reference code 17M.3.sl.TZ1.15
Level SL Paper 3 Time zone TZ1
Command term Deduce Question number 15 Adapted from N/A

Question

Vegetable oils, such as that shown, require conversion to biodiesel for use in current internal combustion engines.

State two reagents required to convert vegetable oil to biodiesel.

[2]
a.

Deduce the formula of the biodiesel formed when the vegetable oil shown is reacted with the reagents in (a).

[1]
b.

Explain, in terms of the molecular structure, the critical difference in properties that makes biodiesel a more suitable liquid fuel than vegetable oil.

[2]
c.

Determine the specific energy, in kJ\(\,\)g−1, and energy density, in kJ\(\,\)cm−3, of a particular biodiesel using the following data and section 1 of the data booklet.

Density = 0.850 g\(\,\)cm−3; Molar mass = 299 g\(\,\)mol−1;

Enthalpy of combustion = 12.0 MJ\(\,\)mol−1.

[2]
d.

Markscheme

methanol
OR
ethanol

strong acid
OR
strong base

 

Accept “alcohol”.

Accept any specific strong acid or strong base other than HNO3/nitric acid.

[3 marks]

a.

CH3(CH2)16COOCH3 / CH3OCO(CH2)16CH3
OR
CH3(CH2)16COOC2H5 / C2H5OCO(CH2)16CH3

 

Product must correspond to alcohol chosen in (a), but award mark for either structure if neither given for (a).

[1 mark]

b.

lower viscosity

weaker intermolecular/dispersion/London/van der Waals’ forces
OR
smaller/shorter molecules

 

Accept “lower molecular mass/Mr” or “lower number of electrons”.

Accept converse arguments.

[2 marks]

c.

Specific energy: «\( = \frac{{12\,000{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}}}{{299{\text{ g}}\,{\text{mo}}{{\text{l}}^{ - 1}}}}\)» = 40.1 «kJ g−1»

Energy density: «= 40.1 kJ\(\,\)g−1 x 0.850 g\(\,\)cm−3» = 34.1 «kJ\(\,\)cm−3»

 

Award [1] if both are in terms of a unit other than kJ (such as J or MJ).

[2 marks]

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.

Syllabus sections

Core » Topic 10: Organic chemistry » 10.1 Fundamentals of organic chemistry
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