DP Chemistry (first assessment 2025)

Question 23M.2.HL.TZ1.2

Date	May 2023	Marks available	[Maximum mark: 24]	Reference code	23M.2.HL.TZ1.2
Level	HL	Paper	2	Time zone	TZ1
Command term	Annotate, Calculate, Deduce, Determine, Explain, Outline, Sketch, State, Suggest	Question number	2	Adapted from	N/A

[Maximum mark: 24]

23M.2.HL.TZ1.2

The periodic table is a useful tool in explaining trends of chemical behaviour.

(a.i)

Annotate and label the ground state orbital diagram of boron, using arrows to represent electrons.

[1]

Markscheme

arrows AND identifies 2s AND 2p sub orbitals ✓

Accept “hooks” to represent the electrons.

(a.ii)

Sketch the shapes of the occupied orbitals identified in part (a)(i).

[2]

Markscheme

P_x,_y or _z can be used.
M2 cannot be awarded if labels of orbital types are missing or incorrect

Node of p orbital must be at the origin

(b)

Explain the decrease in first ionization energy from Li to Cs, group 1.

[2]

Markscheme

valence electron further from nucleus/«atomic» radius larger «down the group» ✓
«electron» more shielded/ less attractive force/easier to remove ✓

(c.i)

State the electron domain geometry of the ammonia molecule.

[1]

Markscheme

tetrahedral ✓

(c.ii)

Deduce the Lewis (electron dot) structure of ammonia and sketch its 3D molecular shape.

[2]

Markscheme

Accept a combination of dots /crosses /lines in the Lewis structure

Lone pair not required for shape

(c.iii)

Explain, with reference to the forces between molecules, why ammonia has a higher boiling point than phosphine (PH₃).

[3]

Markscheme

ammonia has intermolecular/IMF hydrogen bonds «phosphine does not» ✓
phosphine «and ammonia» dipole-dipole/London dispersion forces/instantaneous dipole attractions/Van der Waals forces ✓
hydrogen bonds stronger ✓

Accept converse argument.

Award 1 for stating that NH₃ is more polar than phosphine so the dipole-dipole forces are stronger

(d.i)

Ammonia is manufactured by the Haber process.

N₂ (g) + 3H₂(g) $\overset{}{⇌}$ 2NH₃(g) $Δ H_{f}^{⦵}$ = −92.0 kJ mol⁻¹

Outline what is meant by dynamic equilibrium.

[1]

Markscheme

«in a closed system» the rate of the forward reaction equals the rate of the reverse reaction. ✓

(d.ii)

Deduce the K_c expression for the reaction in part (d)(i).

[1]

Markscheme

[NH₃]²/([N₂][H₂]³) ✓

(d.iii)

Determine the entropy change, ΔS^⦵ for the forward reaction to four significant figures, using the data given.

Substance	Entropy (S^⦵) J K⁻¹ mol⁻¹
H₂	130.7
N₂	191.6
NH₃	192.8

[2]

Markscheme

ΔS^⦵ = ΔS^⦵_(products) −ΔS^⦵_(products)OR
(2 × 192.8«J mol⁻¹ K⁻¹») − (3 × 130.7 «J mol⁻¹ K⁻¹» + 191.6 «J mol⁻¹ K⁻¹» ✓
−198.1 «J K⁻¹ mol⁻¹» ✓

Award [2] for correct final answer with four significant figures.

(d.iv)

Calculate the temperature, in K, below which this reaction becomes spontaneous. Use section 1 of the data booklet. (If you were unable to obtain an answer for part (d)(iii) use −210.0 J K⁻¹ mol⁻¹, but this is not the correct value.)

[2]

Markscheme

«ΔG^_⦵ = ΔH^⦵ − TΔS^⦵»
ΔS^⦵ = −0.1981 kJ K⁻¹ mol⁻¹
AND
ΔH^⦵ = −92.0 kJ mol⁻¹✓
«0 kJ mol⁻¹ = (−92.0 kJ mol⁻¹) − (T K × −0.1981 kJ K⁻¹ mol⁻¹)»
464«K» ✓

Alternate:
ΔS^⦵ = −0.2100 kJ K⁻¹ mol⁻¹
AND
ΔH^⦵ = −92.0 kJ mol⁻¹✓
«0 kJ mol⁻¹ = (−92.0 kJ mol⁻¹) − (T K× −0.2100 kJ K⁻¹ mol⁻¹)»
438«K» ✓

M1 for conversion to common units for ΔH^⦵ and ΔS^⦵.

Award [2] for correct final answer.

(d.v)

The value of K_c for this reaction is 6.84 × 10⁻⁵ at 500 °C. Suggest, with a reason, how lowering the temperature affects the value of K_c.

[1]

Markscheme

«reaction» exothermic AND K_c increases «as equilibrium moves right» ✓

(d.vi)

Calculate the standard Gibbs free energy change, ΔG^⦵, in kJ mol⁻¹, for this reaction. Use sections 1 and 2 of the data booklet.

[2]

Markscheme

«ΔG^⦵ = −RT ln K_c»
«ΔG^⦵ = (−8.31 J K⁻¹ mol⁻¹ × 773 K × ln 6.84 × 10⁻⁵)/1000 =» «+» 61.6 «kJ mol⁻¹» ✓

OR
«ΔG^⦵ = ΔH^⦵ − TΔS^⦵»
«ΔG^⦵ = −92.0 kJ mol⁻¹ − 773 K × (−0.1981 kJ K⁻¹ mol⁻¹) =» + 61.1 «kJ mol⁻¹» ✓

Award [2] for the correct final answer.

(e.i)

The Haber process requires a catalyst. State how a catalyst functions.

[1]

Markscheme

alternate pathway AND lowers activation energy/E_a ✓

(e.ii)

Sketch a Maxwell–Boltzmann distribution curve showing the activation energies with and without a catalyst.

[2]

Markscheme

correct shape curve starting at the origin, without touching the x axis at high energy. ✓

(E_a) catalysed <(E_a) uncatalysed on x axis. ✓

Ignore any shading under the curve.

(e.iii)

Suggest how the progress of the reaction could be monitored.

[1]

Markscheme

change in AND
volume
OR
pressure
OR
temperature
OR
concentration of H₂/N₂/reactants/NH₃/product ✓

Do not accept pH.
Accept any valid method.

Question 23M.2.HL.TZ1.2

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