DP Chemistry (last assessment 2024)

Question 22N.2.hl.TZ0.1

Date	November 2022	Marks available	[Maximum mark: 35]	Reference code	22N.2.hl.TZ0.1
Level	hl	Paper	2	Time zone	TZ0
Command term	Calculate, Deduce, Determine, Justify, Predict, Sketch, State, Write	Question number	1	Adapted from	N/A

[Maximum mark: 35]

22N.2.hl.TZ0.1

Ammonium nitrate, NH₄NO₃, is used as a high nitrogen fertilizer.

(a)

Calculate the percentage by mass of nitrogen in ammonium nitrate. Use section 6 of the data booklet.

[1]

Markscheme

« $% N = \frac{2 \times 14.01 g m o l^{- 1}}{(2 \times 14.01 g m o l^{- 1} + 4 \times 1.01 g m o l^{- 1} + 3 \times 16.00 g m o l^{- 1})} \times 100 % =$ » $35.00$ « $%$ » ✔

Examiners report

Almost all students managed to calculate that ammonium nitrate contains 35.0% nitrogen.

(b)

State, with a reason, whether the ammonium ion is a Brønsted-Lowry acid or base.

[1]

Markscheme

«Brønsted-Lowry» acid AND can donate a proton/H⁺

«Brønsted-Lowry» acid AND cannot accept proton/H⁺ ✔

Examiners report

Not quite as well done as the previous and next questions, but it was unusual to find candidates who did not know the ammonium ion is a Brønsted-Lowry acid because it donates protons, though a handful seemed confused as to whether acids or bases donate protons.

A 0.20 mol dm⁻³ solution of ammonium nitrate is prepared.

(c.i)

Calculate the pH of an ammonium nitrate solution with [H₃O⁺] = 1.07× 10⁻⁵ mol dm⁻³. Use section 1 of the data booklet.

[1]

Markscheme

«pH = –log (1.07× 10⁻⁵) =» 4.97 ✔

Examiners report

Almost all students managed to use their calculators to work out that −log (1.07 × 10⁻⁶) is 4.97, though a few lost the mark by then rounding down to 4.9.

(c.ii)

Ammonium nitrate is neutralized with sodium hydroxide. Write the equation for the reaction.

[1]

Markscheme

NH₄⁺ (aq) + OH^– (aq) → NH₃ (aq) + H₂O (l)

NH₄NO₃ (aq) + NaOH (aq) → NH₃ (aq) + H₂O (l) + NaNO₃ (aq) ✔

Accept NH₄OH instead of NH₃ + H₂O.

Examiners report

A little more challenging as about a third of the candidates, even some who had correctly identified the ammonium ion as an acid, were unable to correctly work out the products of the reaction.

(c.iii)

A 20.00 cm³ sample of the 0.20 mol dm⁻³ solution of ammonium nitrate is titrated with a 0.20 mol dm⁻³ solution of sodium hydroxide. Determine the pH at the equivalence point, to two decimal places using section 1 and 21 of the data booklet.

[4]

Markscheme

«n(NH₄NO₃) = 0.20 mol dm⁻³ × 0.02000 dm³ =» 0.0040 «mol NH₄NO₃» ✔

«[NH₃] at equivalence point $= \frac{0.0040 m o l}{0.04000 d m^{3}} =$ 0.10 «mol dm⁻³» ✔

«K_b = 10^−pKb = 10^−4.75 = 1.8 × 10⁻⁵»

« $[O H^{-}] = \sqrt{K_{b} [N H_{3}]} = \sqrt{1.8 \times 10^{- 5} (0.10)} = 0.0013$ «mol dm⁻³» ✔

«pOH = –log (0.0013) = 2.89»

«pH = 14.00 – pOH =» 11.11✔

Award [4] for correct final answer.

Accept a range of 11.11 – 11.14.

Examiners report

Probably the most difficult question on the paper with an average mark of about 1½/4. Few candidates, were capable of working their way through this long calculation, though quite a number gained partial marks, showing the importance of setting out workings clearly. Many tried to, inappropriately, employ the Henderson-Hasselbach equation.

(c.iv)

Sketch the pH curve that would result from the titration of a 0.20 mol dm⁻³ solution of ammonium nitrate with sodium hydroxide.

[2]

Markscheme

non-symmetrical sigmoidal curve, starting pH 2–7 AND terminating pH>12 ✔

equivalence point pH approximately 11 AND at a volume 20 cm³ ✔

Examiners report

Most students managed to gain at least one mark, however it was a disappointing to see that many students did not use the information from the previous parts of the question to accurately inform their sketch. This question had a strong link between the mark and overall performance.

(c.v)

State, with a reason, if bromothymol blue is an appropriate indicator for this titration. Use section 22 of the data booklet.

[1]

Markscheme

no AND the end point is not in the sharp part of the curve

no AND the equivalence point does not fall within the end-point/pH range of the indicator

no AND there is a large difference in volume between end point and equivalence point

no AND no sharp rise in pH «near equivalence point» ✔

Examiners report

Just over half the candidates managed to correctly assess whether bromothymol blue would be act as an effective indicator for the titration curve they had drawn. From the ranges quoted, many students seem to have read bromophenol blue rather than bromothymol blue.

Cold packs contain ammonium nitrate and water separated by a membrane.

(d.i)

The mass of the contents of the cold pack is 25.32 g and its initial temperature is 25.2 °C. Once the contents are mixed, the temperature drops to 0.8 °C.

Calculate the energy, in J, absorbed by the dissolution of ammonium nitrate in water within the cold pack. Assume the specific heat capacity of the solution is 4.18 J g⁻¹ K⁻¹. Use section 1 of the data booklet.

[1]

Markscheme

«q = mcΔT = 25.32 g × 4.18 J g⁻¹ K⁻¹ × (25.2 °C – 0.8 °C) =» 2580 «J» ✔

Do not accept a negative value.

Examiners report

A quite simple calculation that about 80% of candidates managed to perform successfully, though a few read "drops by" rather than "drops to".

(d.ii)

Determine the mass of ammonium nitrate in the cold pack using your answer obtained in (d)(i) and and sections 6 and 19 of the data booklet.

If you did not obtain an answer in (d)(i), use 3.11 × 10³ J, although this is not the correct answer.

[2]

Markscheme

« $2.58 \times 10^{3} J \times \frac{1 k J}{1000 J} \times \frac{1 m o l}{25.69 k J} =$ » $0.100$ «mol» ✔

« $0.100$ mol $\times 80.06$ g mol⁻¹ =» 8.01 «g» ✔

Award [2] for the correct final answer.

Accept range of $8.0 - 8.1$ g.

If $3.11 \times 10^{3}$ J used then answer is $9.69$ g.

Examiners report

Almost ¾ of students managed to combine the previous answer with the enthalpy change of solution, from the data booklet, to calculate the amount, and hence the mass, of solute that had dissolved.

(d.iii)

The absolute uncertainty in mass of the contents of the cold pack is ±0.01 g and in each temperature reading is ±0.2 °C. Using your answer in (d)(ii), calculate the absolute uncertainty in the mass of ammonium nitrate in the cold pack.

If you did not obtain an answer in (d)(ii), use 6.55 g, although this is not the correct answer.

[3]

Markscheme

«fractional / % uncertainty in ΔT =» $\frac{0.4}{24.4}$ / 0.02 / 2«%» ✔

«fractional / % uncertainty in $m$ =» $\frac{0.01}{25.32}$ / 0.0004 / 0.04«%»

fractional / % uncertainty in $m$ is much smaller than uncertainty in ΔT ✔

«2% x 8.01 g =» 0.2 «g» ✔

Award [3] for correct final answer.

Accept range of 0.1 – 0.2 «g».

If 6.55 g used then answer is 0.1 «g».

Examiners report

Whilst this was better tackled than some uncertainty problems have been in recent years, students only gained about half marks on average, though there was a strong link with overall performance. The most common error was not to apply the temperature uncertainty to both recorded temperatures, to give an uncertainty of the change as ±0.4.

(d.iv)

The cold pack contains 9.50 g of ammonium nitrate. Calculate the percentage error in the experimentally determined mass of ammonium nitrate obtained in (d)(ii).

If you did not obtain an answer in (d)(ii), use 6.55 g, although this is not the correct answer.

[1]

Markscheme

«% error = $|\frac{9.50 g - 8.01 g}{9.50 g}| \times 100 % =$ » 15.7«%» ✔

Accept range 14.7 – 15.8«%».

If 6.55 g used then answer is 31.1«%».

Examiners report

About 70% of candidates correctly calculated the percentage error, though a significant minority divided the difference by the value obtained rather than the true value.

(d.v)

Calculate the standard entropy change, ΔS^⦵, for the dissolution of ammonium nitrate.

S^⦵NH₄NO₃ (s) = 151.1 J mol⁻¹ K⁻¹

S^⦵NH₄NO₃(aq) = 259.8 J mol⁻¹ K⁻¹

[1]

Markscheme

« $Δ S ° = 259.8 J m o l^{- 1} K^{- 1} - 151.1 J m o l^{- 1} K^{- 1} =$ » 108.7 «J mol⁻¹ K⁻¹» ✔

Examiners report

Almost all students managed to combine the standard entropies of the reactant and product to determine the increase in entropy, though a handful managed to end up with a negative sign.

(d.vi)

Calculate the standard Gibbs free energy change, ΔG^⦵, in kJ mol⁻¹, for the dissolution of ammonium nitrate at 298 K. Use sections 1 and 19 of the data booklet as well as your answer for question part (d)(v).

If you did not obtain an answer in (d)(v), use 102.3 J mol⁻¹ K⁻¹, although this is not the correct answer.

[1]

Markscheme

« $Δ G ° = Δ H ° - T Δ S ° = 25.69 k J m o l^{- 1} - 298 K (108.7 J m o l^{- 1} K^{- 1} \times \frac{1 k J}{1000 J}) =$

−6.70 «kJ mol⁻¹» ✔

If 102.3 J mol⁻¹ K⁻¹is used then answer is −4.80 KJ mol⁻¹.

Examiners report

About 80% of students applied the appropriate equation to calculate the free energy change for the reaction from the entropy and enthalpy change values previously found. As might have been predicted quite a few candidates failed to convert the J of the previous calculation into kJ, and a handful left the temperature as Celsius.

(d.vii)

Calculate the value of the equilibrium constant for the dissolution of ammonium nitrate at 298 K using the answer to question part (d)(vi) and section 1 of the data booklet.

NH₄NO₃(s) $⇌$ NH₄NO₃(aq)

If you did not obtain an answer in (d)(vi), use −7.84 kJ/mol, although this is not the correct answer.

[2]

Markscheme

« $Δ G ° = - R T \ln K$ »

« $- 6.70 k J \times \frac{1000 J}{k J} = - 8.31 J K^{- 1} (298 K) \ln K$ »

«ln K =» 2.71 ✔

«K = e^2.71 =» 15.0 ✔

Award [2] for correct final answer.

If –7.84 kJ is used then answer is 23.7.

Examiners report

About half the candidates managed to obtain the value of K corresponding to the free energy just calculated. Ensuring that the correct energy units (J) were being used was again the main problem, however some also appeared to have had difficulties with manipulations on the calculator.

(d.viii)

Deduce, with a reason, the position of the equilibrium.

[1]

Markscheme

product/right/solution/NH₄NO₃(aq) is favoured AND K>1 ✔

Accept K large.

Accept other valid ways of justifying equilibrium position such as ΔG<0/spontaneous/ΔH<0 AND ΔS>0.

Examiners report

About 70% of candidates correctly assessed the equilibrium position corresponding to constant calculated, however quite a few candidates seemed to think that K>0, not K>1, was the criterion.

(e)

Predict, using the given values, the reaction that would take place at the anode and cathode for the electrolysis of an aqueous solution of ammonium nitrate using graphite electrodes.

[2]

Markscheme

Anode:

H₂O (l) → 1/2O₂ (g) + 2H⁺ (aq) + 2e⁻ ✔

Cathode:

H⁺ (aq) + e⁻ → 1/2H₂ (g) ✔

Do not accept other equations.

Examiners report

A very poorly answered question with the average mark being ~0.6/2. Candidates seemed to be uncertain how to use standard electrode potential data to determine which species were most easily oxidized and reduced, as well as at which electrode electrons were gained and at which they were lost.

Solid ammonium nitrate can decompose to gaseous dinitrogen monoxide and liquid water.

(f.i)

Write the chemical equation for this decomposition.

[1]

Markscheme

NH₄NO₃ (s) → N₂O (g) + 2H₂O (l) ✔

Examiners report

Almost all students managed to write this simple equation, having been given the names of all reactants and products, though a handful managed to give the wrong formula for dinitrogen monoxide and some failed to balance the amount of water.

(f.ii)

Calculate the volume of dinitrogen monoxide produced at STP when a 5.00 g sample of ammonium nitrate decomposes. Use section 2 of the data booklet.

[2]

Markscheme

«5.00 g ÷ 80.06 g mol⁻¹ =» 0.0625 mol «NH₄NO₃» ✔

«1:1 mol ratio»

« $0.0625 m o l N_{2} O \times \frac{22.7 d m^{3}}{m o l} =$ » 1.42 «dm³» ✔

Award [2] for correct final answer.

Accept range 1.36 – 1.43 «dm³».

Accept calculations based on PV=nRT.

Examiners report

Over three quarters of the candidates managed this straightforward gas volume calculation, with an approximately equal split between those using the molar volume and the ideal gas law.

(f.iii)

Calculate the standard enthalpy change, $Δ H^{⦵}$ , of the reaction. Use section 12 of the data booklet.

$Δ H_{f}^{⦵}$ ammonium nitrate = −366 kJ mol⁻¹

$Δ H_{f}^{⦵}$ dinitrogen monoxide = 82 kJ mol⁻¹

[2]

Markscheme

2 x –285.8 «kJ mol⁻¹» ✔

«1 mol (82 kJ mol⁻¹) + 2 mol (– 285.8 kJ mol⁻¹) – 1 mol (–366 kJ mol⁻¹) =» –124 «kJ» ✔

Award [2] for correct final answer.

Examiners report

Slightly fewer students were able to negotiate this calculation, with a number forgetting to double the enthalpy of formation of water, whilst others ignored it all together. In addition some reversed values for the reactants and products.

(f.iv)

Predict, with a reason, the signs for the entropy change, ΔS^⦵, and Gibbs free energy change, ΔG^⦵, of the reaction.

[2]

Markscheme

Entropy change:

positive AND formation of gas «and liquid from solid» ✔

Gibbs free energy change:

negative AND increase in entropy/ΔS positive AND exothermic reaction/ΔH negative ✔

Examiners report

Candidates gained on average just half marks and the mark obtained seemed to correlate closely to their performance on the rest of the paper. Most candidates managed to predict that the formation of a gas resulted in an entropy increase, though a handful only mentioned the number of moles of product rather than noting their states. Slightly fewer were able deduce that the signs of the entropy and enthalpy changes meant that the reaction would be spontaneous at all temperatures.

(f.v)

Deduce the Lewis (electron dot) structure, including formal charges, and shape for dinitrogen monoxide showing nitrogen as the central atom.

[3]

Markscheme

Alternative 1

Lewis structure:

✔

+/1+/+1 on central N atom AND –/1–/–1 on O atom ✔

Alternative 2

Lewis structure:

✔

+/1+/+1 on central N atom AND –/1–/–1 on other N atom ✔

Shape: linear ✔

Accept .

Formal charges are not needed for M1.

Allow ECF for both formal charge and shape.

Only award M3 if the shape corresponds to that expected for the Lewis structure given.

Examiners report

Students on average gained about half marks. Many seemed to be in one of two camps; those that knew of the resonance structures of the molecule, and usually scored 3 marks, and those who gained zero because they did not, though occasionally the latter group would score one of the other marks as an ECF.