DP Chemistry (last assessment 2024)

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Question 22M.2.sl.TZ1.2

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Date May 2022 Marks available [Maximum mark: 15] Reference code 22M.2.sl.TZ1.2
Level sl Paper 2 Time zone TZ1
Command term Calculate, Deduce, Determine, Draw, Explain, State, Suggest Question number 2 Adapted from N/A
2.
[Maximum mark: 15]
22M.2.sl.TZ1.2

Ammonia, NH3, is industrially important for the manufacture of fertilizers, explosives and plastics.

(a)

Draw arrows in the boxes to represent the electron configuration of a nitrogen atom.

[1]

Markscheme

 

Accept all 2p electrons pointing downwards.

Accept half arrows instead of full arrows.

Examiners report

Most students realised that the three p-orbitals were all singly filled.

(b)

Draw the Lewis (electron dot) structure of the ammonia molecule.

[1]

Markscheme

 

Accept lines or dots or crosses for electrons, or a mixture of these

 

Examiners report

Even more candidates could draw the correct Lewis structure of ammonia, with omission of the lone pair being the most common error.

Ammonia is produced by the Haber–Bosch process which involves the equilibrium:

N2 (g) + 3 H2 (g)  2 NH3 (g)

(c(i))

Deduce the expression for the equilibrium constant, Kc, for this equation.

[1]

Markscheme

Kc=NH32N2H23 ✔

Examiners report

Most students could deduce the equilibrium constant expression from the equilibrium equation.

(c(ii))

Explain why an increase in pressure shifts the position of equilibrium towards the products and how this affects the value of the equilibrium constant, Kc.

[2]

Markscheme

shifts to the side with fewer moles «of gas»
OR
shifts to right as there is a reduction in volume✔

«value of » Kc unchanged ✔

 

Accept “Kc only affected by changes in temperature”.

Examiners report

Many students realised that increasing pressure shifts an equilibrium to the side with the most moles of gas (though the "of gas" was frequently omitted!) but probably less than half realised that, even though the equilibrium position changes, the value of the equilibrium constant remains constant.

(c(iii))

State how the use of a catalyst affects the position of the equilibrium.

[1]

Markscheme

same/unaffected/unchanged ✔

The effect of temperature on the position of equilibrium depends on the enthalpy change of the reaction.

(d(i))

Determine the enthalpy change, ΔH, for the Haber–Bosch process, in kJ. Use Section 11 of the data booklet.

[3]

Markscheme

bonds broken: N≡N + 3(H–H) / «1 mol×»945 «kJ mol–1» + 3«mol»×436 «kJ mol–1» / 945 «kJ» + 1308 «kJ» / 2253 «kJ» ✔

bonds formed: 6(N–H) / 6«mol»×391 «kJ mol–1» / 2346 «kJ» ✔

ΔH = «2253 kJ – 2346 kJ = » –93 «kJ» ✔

 

Award [2 max] for (+)93 «kJ»

Examiners report

It was pleasing to see that about a third of students gaining full marks and an equal number only lost a single mark because they failed to locate the correct bond enthalpy for molecular nitrogen.

(d(ii))

Calculate the enthalpy change, ΔH, for the Haber–Bosch process, in kJ, using the following data.

H fNH3=-46.2kJmol-1.

[1]

Markscheme

–92.4 «kJ» ✔

Examiners report

Very few students could determine the enthalpy change from enthalpy of formation data, with many being baffled by the absence of values for the elemental reactants and more than half who overcame this obstacle failed to note that 2 moles of ammonia are produced.

(d(iii))

Suggest why the values obtained in (d)(i) and (d)(ii) differ.

[1]

Markscheme

«N-H» bond enthalpy is an average «and may not be the precise value in NH3» ✔


Accept it relies on average values not specific to NH3

Ammonia is soluble in water and forms an alkaline solution:

NH3 (g) + H2O (l) NH4+ (aq) + HO– (aq)

(e(i))

State the relationship between NH4+ and NH3 in terms of the Brønsted–Lowry theory.

[1]

Markscheme

conjugate «acid and base» ✔

Examiners report

About half the candidates recognised the species as a conjugate acid-base pair, though some lost the mark by confusing the acid and base, even though this information was not asked for.

(e(ii))

Determine the concentration, in mol dm–3, of the solution formed when 900.0 dm3 of NH3 (g) at 300.0 K and 100.0 kPa, is dissolved in water to form 2.00 dm3 of solution. Use sections 1 and 2 of the data booklet.

[2]

Markscheme

amount of ammonia =P.VR.T=100.0kPa×900.0dm38.31JK-1mol-1×300.0K = 36.1 «mol» ✔

concentration =nV=36.12.00=18.1 «moldm-3» ✔

 

Award [2] for correct final answer.

Examiners report

About 40% of candidates gained full marks for the calculation and a significant number of others gained the second mark to calculate the concentration as an ECF.

(e(iii))

Calculate the concentration of hydroxide ions in an ammonia solution with pH = 9.3. Use sections 1 and 2 of the data booklet.

[1]

Markscheme

[OH=KWH+=10-1410-9.3=10-4.7=2.0 × 10-5 moldm-3 ✔

Examiners report

This question was very poorly answered with many candidates calculating the [H+] instead of [OH-].