(i)     Calculate the heat required, in kJ, to raise the temperature of the water, using data in the table above and from Table 2 of the Data Booklet.

(ii)     Determine the enthalpy of combustion of the potato chip, in \({\text{kJ}}\,{{\text{g}}^{ - 1}}\).

This energy comes mainly from the combustion of triglycerides. State the name of one other type of lipid found in the body and one role, other than energy storage, of this type of lipid.

Name:

Role:

Explain why lipids have a higher energy content than carbohydrates.

(i)     \({\text{heat}} = \frac{{4.18 \times 20.0 \times (51.3 - 17.8)}}{{1000}}\);

\( = 2.80{\text{ (kJ)}}\);

(ii)     \({\text{enthalpy of combustion}} = \left( {\frac{{2.80}}{{0.421}} = } \right){\text{ }} - 6.65{\text{ (kJ}}\,{{\text{g}}^{ - 1}}{\text{)}}\);

Name:

steroids;

Role:

(sex) hormones;

OR

Name:

phospholipids;

Role:

membranes;

lipids less oxidized/contain less oxygen / carbohydrates partially/more oxidized/contain more oxygen / OWTTE;

This part was generally well answered but there were some cases where 33.5 °C was converted into Kelvin. Many candidates had serious problems with unit conversions and gave the answer as 2800 J or 2800 kJ. Some candidates had correct value for (ii) but lost the mark because of the omission of the negative sign.

Part (b) was well answered.

Very few candidates linked the fact that lipids have higher energy content due to being less oxidized.

Glucose, \({{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}\), is a monosaccharide. When 0.85 g of glucose was completely combusted in a calorimeter, the temperature of 200.10 g of water increased from 20.20 °C to 27.55 °C. Calculate the energy value of glucose in \({\text{J}}\,{{\text{g}}^{ - 1}}\).

(i)     Draw the straight chain structure of glucose.

(ii)     Draw the structural formula of \(\alpha \)-glucose.

(iii)     Distinguish between the structures of \(\alpha \)- and \(\beta \)-glucose.

(iv)     Two \(\alpha \)-glucose molecules condense to form the disaccharide maltose. Deduce the structure of maltose.

One of the major functions of carbohydrates in the human body is as an energy source. State one other function of a carbohydrate.

\(\Delta T = 7.35\) (K/°C);

\(q( = mc\Delta T = {\text{200.10 g}} \times {\text{4.18 J}}\,{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}} \times {\text{7.35 K}}) = 6.15 \times {10^3}{\text{ J}}\) (per 0.85 g of glucose heated);

energy value \( = 7.2 \times {10^3}{\text{ (J}}\,{{\text{g}}^{ - 1}}{\text{)}}\);

Award [3] for correct final answer.

(i)     

Accept CHO, CH₂OH and OH groups on either side of the carbon chain provided OH on C3 is on the opposite side to OHs on C2, C4 and C5.

(ii)      ;

Accept CH₂OH and OH groups on either face, as long as OH on C3 is on the opposite face to the OH’s on C1, C2 and C4.

No mark awarded if HOCH₂ is written, with H bonded to C or if HO is written for hydroxyl groups, with H bonded to C. Penalize this once only in (i), (ii) and (iv).

(iii)     the OH on carbon-1/C-1 is inverted / difference in position of OH on carbon-1/C-1;

(iv)      ;

energy reserve / can act as precursors in large number of metabolic reactions/for other biologically important molecules;

In (a) SD’s proved the major issue. Most candidates scored the mark for \(\Delta T\). The candidates who struggled made the following errors: incorrect mass (0.85 g) of water was used in \(q = mc\Delta T\), and failure to convert to J per g by dividing \(q\) by 0.85.

In (b)(i) many candidates struggled to draw the straight chain structure of glucose. In many instances the –OH groups were positioned incorrectly and some candidates were careless with bonding writing ‘-C–HO’ rather than ‘-C–OH’.

In (ii) a significant number of candidates mixed up the positions of the substituents on the two faces. In (iii) C1 was commonly omitted. It was surprising to see that quite a few candidates could not draw the structure of maltose in (iv). The most common mistake involved an incorrect linkage

Part (c) was answered well by the vast majority.

When 1.13 g of a granola bar was combusted in a bomb calorimeter, the temperature of \({\text{500 c}}{{\text{m}}^{\text{3}}}\) of water increased from 18.5 °C to 28.0 °C. Calculate the energy value, in kJ per 100 g, of the granola bar to the correct number of significant figures.

energy (released by granola) \( = (500 \times 4.18 \times 9.5 = ){\text{ }}19855{\text{ (J)}}/2.0 \times {10^4}{\text{ (J)}}\);

energy released (in kJ per 100 g) \(\left( {\frac{{19855}}{{1000}} \times \frac{{100}}{{1.13}} = } \right){\text{ }}1757.079\);

\( = 1.8 \times {10^3}{\text{ (kJ per 100 g)}}\);

Allow 1800 (kJ per 100 g).

Award [3] for correct final answer.

Award [2 max] for 1760 (kJ per 100 g).

Remember to allow ECF.

Candidates generally used the appropriate equations to solve for the energy value for part (a) however, many included 1.13g with the mass of water and on quite a few occasions, \({\text{22.4 d}}{{\text{m}}^{\text{3}}}\) was used in the question for to calculate energy per gram. Several missed the final mark for not expressing the final value in 2 significant figures. Candidates were not able to apply the rules for addition and subtraction within the mathematical process. Surprisingly defining dietary fibre in part (b) scored poorly. Many omitted reference to plant material or cellulose; there appears to be a general misunderstanding that any food that is not digested is dietary fibre. Most candidates were able to name two health problems for (b)(ii).

Deduce the equation for the cellular respiration of glucose.

Calculate the energy, in kJ, produced from 15.0g of glucose if its enthalpy of combustion is −2803kJmol⁻¹.

Glucose is the basic building block of starch which can be used to make bioplastics. Outline two advantages and two disadvantages of biodegradable plastics.

Two advantages:

Two disadvantages:
 
 

Bioplastics are broken down by enzyme catalysed reactions. Sketch a graph illustrating how the rate of this reaction varies with pH.

\(n\left( {{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}} \right)\left\langle { = \frac{{15.0}}{{180.18}}} \right\rangle  = 0.0833 \ll {\rm{mol}} \gg \)

«energy=0.0833×2803=»233«kJ»

Award [2] for correct final answer.
Accept -233«kJ».

State two reagents required to convert vegetable oil to biodiesel.

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

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

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.

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]

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]

lower viscosity

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

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

Accept converse arguments.

[2 marks]

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]

(i) Draw the structure of 2-methylpropene.

(ii) Deduce the repeating unit of poly(2-methylpropene).

Deduce the percentage atom economy for polymerization of 2-methylpropene.

(i) Suggest why incomplete combustion of plastic, such as polyvinyl chloride, is common in industrial and house fires.

(ii) Phthalate plasticizers such as DEHP, shown below, are frequently used in polyvinyl chloride.

With reference to bonding, suggest a reason why many adults have measurable levels of phthalates in their bodies.

i

OR
H₂C=C(CH₃)₂

ii

OR
−CH₂C(CH₃)₂−

Continuation bonds needed for mark.
No penalty if square brackets present or “n” appears after the bracket/formula.

«same mass of product as reactant, thus» 100«%»

Accept “less than 100%” only if a reason is given (eg, the catalyst is not converted into the product, or other reasonable answer).

i

due to stability of plastics/strong covalent bonds
OR
low volatility preventing good mixing with oxygen «gas»
OR
lack of/insufficient oxygen
OR
plastics are often parts of devices with non-combustible components «which mechanically prevent the combustion of plastic components»
OR
PVC already partly oxidised «because some C–H bonds are replaced with C–Cl bonds», so it cannot produce enough heat for complete combustion
OR
many industrial/household materials contain additives that reduce their flammability/act as flame retardants

ii

weakly bound to the PVC/no covalent bonds to PVC/only London/dispersion/instantaneous induced dipole-induced dipole forces between DEHP and PVC AND leach/evaporate «from PVC» to atmosphere/food chain
OR
has low polarity/contains non-polar hydrocarbon chains AND fat-soluble/deposits in the fatty tissues
OR
has unusual structural fragments/is a xenobiotic/difficult to metabolise AND stays in the body for a long time

Explain how the concentration may be calculated in this way.

Heat losses would make this method less accurate than the pH probe method. Outline why the thermometric method would always give a lower, not a higher, concentration.

Suggest how heat loss could be reduced.

State one other assumption that is usually made in the calculation of the heat produced.

Suggest why scientists often make assumptions that do not correspond to reality.

Outline why the thermochemical method would not be appropriate for 0.001 mol\(\,\)dm⁻³ hydrochloric acid and aqueous sodium hydroxide of a similar concentration.

heat change/evolved can be calculated from the «maximum» temperature increase and the mass of solution
OR
q = mcΔT

heat «evolved» gives the number of moles «of both acid and alkali present when neutralisation occurs»
OR
\(n = \frac{q}{{\Delta {H_{neut}}}}\)

volume «of acid and the volume of alkali required to just neutralise each other» can be used to calculate the concentration«s of both»
OR
\(\left[ {{\text{NaOH}}} \right] = \frac{n}{V}\)

[2 marks]

smaller temperature increase/ΔT
OR
heat released would «appear to» be less

amount of substance/n calculated is smaller

[2 marks]

using «expanded» polystyrene cup
OR
insulating beaker
OR
putting a lid on beaker

Do not accept calorimeter by itself.

Accept any other reasonable suggestion.

[1 mark]

«specific» heat capacity of the beaker/container/thermometer is ignored
OR
density of the solutions is assumed as 1.00 g\(\,\)cm^–3/same as water
OR
specific heat capacity of the solutions is assumed as 4.18 J g^–1\(\,\)K^–1/same as water

Accept “reaction goes to completion”.

Accept “reaction is conducted under standard conditions”.

Accept “no evaporation occurs”.

Accept any other relevant valid assumption.

Do not accept “heat is not released from other reactions”.

[1 mark]

allows simple theories to be applied to real life situations
OR
enables us to start to understand complex situations
OR
gives answers that are accurate to the required order of magnitude
OR
simplifies the calculations involved

Do not accept “to simplify the situation” without further detail.

Accept “errors do not have a major impact on the results”.

[1 mark]

temperature rise would be too small «to be accurately measured»

Accept “heat released would be too small «to be accurately measured»”.

[1 mark]