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
$n=\frac{q}{\mathrm{\Delta }{H}_{neut}}$

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]

(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

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: «$=\frac{12000\text{ kJ}\text{mo}{\text{l}}^{-1}}{299\text{ g}\text{mo}{\text{l}}^{-1}}$» = 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]

Estimate the time at which the powdered zinc was placed in the beaker.

State what point Y on the graph represents.

The maximum temperature used to calculate the enthalpy of reaction was chosen at a point on the extrapolated (dotted) line.

State the maximum temperature which should be used and outline one assumption made in choosing this temperature on the extrapolated line.

Maximum temperature:

Assumption:

To determine the enthalpy of reaction the experiment was carried out five times. The same volume and concentration of copper(II) sulfate was used but the mass of zinc was different each time. Suggest, with a reason, if zinc or copper(II) sulfate should be in excess for each trial.

The formula q = mcΔT was used to calculate the energy released. The values used in the calculation were m = 25.00 g, c = 4.18 J g⁻¹ K⁻¹.

State an assumption made when using these values for m and c.

Predict, giving a reason, how the final enthalpy of reaction calculated from this experiment would compare with the theoretical value.

100 «s»  [✔]

Note: Accept 90 to 100 s.

highest recorded temperature
OR
when rate of heat production equals rate of heat loss  [✔]

Note: Accept “maximum temperature”.

Accept “completion/end point of reaction”.

Maximum temperature:
73 «°C»  [✔]

Assumption:
«temperature reached if» reaction instantaneous
OR
«temperature reached if reaction occurred» without heat loss  [✔]

Note: Accept “rate of heat loss is constant” OR “rate of temperature decrease is constant”.

Any one of:
copper(II) sulfate AND mass/amount of zinc is independent variable/being changed.
OR
copper(II) sulfate AND with zinc in excess there is no independent variable «as amount of copper(II) sulfate is fixed»   [✔]

copper(II) sulfate AND having excess zinc will not yield different results in each trial  [✔]

zinc AND results can be used to see if amount of zinc affects temperature rise «so this can be allowed for» [✔]

zinc AND reduces variables/keeps the amount reacting constant  [✔]

Note: Accept “copper(II) sulfate/zinc sulfate” for “solution”.

lower/less exothermic/less negative AND heat loss/some heat not accounted for
OR
lower/less exothermic/less negative AND mass of reaction mixture greater than 25.00 g
OR
greater/more exothermic /more negative AND specific heat of solution less than water  [✔]

Note: Accept “temperature is lower” instead of “heat loss”.

Accept “similar to theoretical value AND heat losses have been compensated for”.

Accept “greater/more exothermic/more negative AND linear extrapolation overestimates heat loss”.

Almost all candidates identified 100 s as the time at which the reaction was initiated.

Many students gained this mark through stating this was the highest temperature recorded, though even more took advantage of the acceptance of the completion of the reaction, expressed in many different ways. Very few answered that it was when heat loss equalled heat production.

Even though almost all students recognised 100 seconds as the start time of the reaction less than 50% chose the extrapolated temperature at this time. Predictably the most common answer was the maximum of the graph, followed closely by the intercept with the y-axis. With regard to reasons, again relatively few gained the mark, though most who did wrote “no loss of heat”, even though it was rare to find this preceded by “the temperature that would have been attained if …”.

The correct answer depended on whether students considered the object of the additional trials was to investigate the effect of a new independent variable (excess copper(II) sulphate) or to obtain additional values of the same enthalpy change so they could be averaged (excess zinc). Answers that gave adequate reasons were rare.

Again relatively few gained these marks for stating that it was assumed the density and specific heat of the solution were the same as water.

Only about a third of the students correctly deduced that loss of heat to the environment means that the experimental value is lower than the theoretical one, though other answers, such as “higher because linear extrapolation over-compensates for the heat losses” were also accepted.

State the nuclear equation for the fusion reaction.

Explain why fusion is an exothermic process.

Beryllium-8 is a radioactive isotope with a half-life of 6.70 × 10⁻¹⁷ s.

Calculate the mass of beryllium-8 remaining after 2.01 × 10⁻¹⁶ s from a sample initially containing 4.00 g of beryllium-8.

${}_2^4\text{He + }{}_4^8\text{Be}\to {}_6^{12}\text{C}$✔

NOTE: Do not penalize missing atomic numbers.

ALTERNATIVE 1
binding energy per nucleon is larger in carbon-12/product «than beryllium-8 and helium-4/reactants» ✔

difference in «total» binding energy is released «during fusion» ✔

ALTERNATIVE 2
mass of carbon-12/product «nucleus» is less than «the sum of» the masses of helium-4 and beryllium-8 «nuclei»/reactants
OR
two smaller nuclei form a lager nucleus ✔

mass lost/difference is converted to energy «and released»
OR
E = mc² ✔

ALTERNATIVE 1
3 half-lives ✔
0.500 g «of beryllium-8 remain» ✔

ALTERNATIVE 2
$m=4.00{\left(\frac{1}{2}\right)}^{\frac{2.01\times {10}^{-16}}{6.70\times {10}^{-17}}}$ ✔
0.500 g «of beryllium-8 remain» ✔

ALTERNATIVE 3
λ = « $\frac{\ln 2}{6.70\times {10}^{-17}}$ »= 1.03 × 10¹⁶ «s⁻¹» ✔
m = « $4.00{\text{e}}^{-1.03\times {10}^{16}\times 2.01\times {10}^{-16}}\text{ = }$ » 0.500 «g» ✔

NOTE: Award [2] for correct final answer.