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Date May 2017 Marks available 2 Reference code 17M.2.hl.TZ1.5
Level HL Paper 2 Time zone TZ1
Command term Explain Question number 5 Adapted from N/A

Question

Two hydrides of nitrogen are ammonia and hydrazine, N 2 H 4 . One derivative of ammonia is methanamine whose molecular structure is shown below.

M17/4/CHEMI/HP2/ENG/TZ1/05

Hydrazine is used to remove oxygen from water used to generate steam or hot water.

N 2 H 4 (aq) + O 2 (aq) N 2 (g) + 2 H 2 O(l)

The concentration of dissolved oxygen in a sample of water is 8.0 × 10 3  g d m 3 .

Estimate the H−N−H bond angle in methanamine using VSEPR theory.

[1]
a.

State the electron domain geometry around the nitrogen atom and its hybridization in methanamine.

 

[2]
b.

Ammonia reacts reversibly with water.
N H 3 (g) + H 2 O(l) NH 4 + (aq) + O H (aq)
Explain the effect of adding H + (aq) ions on the position of the equilibrium.

[2]
c.

Hydrazine reacts with water in a similar way to ammonia. (The association of a molecule of hydrazine with a second H+ is so small it can be neglected.)

N 2 H 4 (aq) + H 2 O(l) N 2 H 5 + (aq) + O H (aq)

p K b  (hydrazine) = 5.77

Calculate the pH of a 0.0100  mol d m 3  solution of hydrazine.

[3]
d.i.

Suggest a suitable indicator for the titration of hydrazine solution with dilute sulfuric acid using section 22 of the data booklet.

[1]
d.ii.

Outline, using an ionic equation, what is observed when magnesium powder is added to a solution of ammonium chloride.

[2]
e.

Determine the enthalpy change of reaction, Δ H , in kJ, when 1.00 mol of gaseous hydrazine decomposes to its elements. Use bond enthalpy values in section 11 of the data booklet.

N 2 H 4 (g) N 2 (g) + 2 H 2 (g)

[3]
f.

The standard enthalpy of formation of N 2 H 4 (l) is + 50.6  kJ mo l 1 . Calculate the enthalpy of vaporization, Δ H vap , of hydrazine in kJ mo l 1 . N 2 H 4 (l) N 2 H 4 (g) (If you did not get an answer to (f), use 85  kJ but this is not the correct answer.)

[2]
g.

Calculate, showing your working, the mass of hydrazine needed to remove all the dissolved oxygen from 1000 d m 3 of the sample.

[3]
h.i.

Calculate the volume, in d m 3 , of nitrogen formed under SATP conditions. (The volume of 1 mol of gas = 24.8 d m 3 at SATP.)

[1]
h.ii.

Markscheme

107°

 

Accept 100° to < 109.5°.

Literature value = 105.8°

[1 mark]

a.

tetrahedral

sp3

 

 

No ECF allowed.

[2 marks]

b.

removes/reacts with O H

moves to the right/products «to replace O H ions»

 

Accept ionic equation for M1.

[2 marks]

c.

Kb = 10–5.77 / 1.698 x 10–6
OR
K b = [ N 2 H 5 + ] × [ O H ] [ N 2 H 4 ]

 [OH]2 «= 1.698 × 10–6 × 0.0100» = 1.698 × 10–8

OR

[OH] « = 1.698 × 10 8 » = 1.303 × 10–4 «mol dm–3»

pH « = lo g 10 1 × 10 14 1.3 × 10 4 » = 10.1

 

Award [3] for correct final answer.

Give appropriate credit for other methods containing errors that do not yield correct final answer.

[3 marks]

d.i.

methyl red

OR

bromocresol green

OR

bromophenol blue

OR

methyl orange

[1 mark]

d.ii.

bubbles

OR

gas

OR

magnesium disappears

2NH 4 + (aq) + Mg(s) M g 2 + (aq) + 2N H 3 (aq) + H 2 (g)

 

Do not accept “hydrogen” without reference to observed changes.

Accept "smell of ammonia".

Accept 2H+(aq) + Mg(s)  Mg2+(aq) + H2(g)

Equation must be ionic.

[2 marks]

e.

bonds broken:

E(N–N) + 4E(N–H)

OR

158   kJ mo l 1 + 4 × 391   kJ mo l 1 / 1722   kJ

bonds formed:

E(N N) + 2E(H–H)

OR

945   kJ mo l 1 + 2 × 436   kJ mo l 1 / 1817   kJ

Δ H = bonds broken bonds formed = 1722 1817 =≫ 95   kJ

 

Award [3] for correct final answer.

Award [2 max] for +95 «kJ».

[3 marks]

 

f.

M17/4/CHEMI/HP2/ENG/TZ1/05.g/M

OR

Δ H vap = 50.6  kJ mo l 1 ( 95  kJ mo l 1 )

Δ H v a p =≫ + 44   kJ mo l 1

 

Award [2] for correct final answer. Award [1 max] for 44 «kJ mol1».

Award [2] for:

ΔHvap = –50.6 kJ mol1 – (–85 J mol1) = 34 «kJ mol1».

Award [1 max] for –34 «kJ mol1».

[2 marks]

g.

total mass of oxygen  ≪= 8.0 × 10 3  g d m 3 × 1000  d m 3 ≫= 8.0   g

n( O 2 ≪= 8.0  g 32.00  g mo l 1 =≫   0.25   mol

OR

n( N 2 H 4 ) = n( O 2 )

mass of hydrazine = 0.25  mol × 32.06  g mo l 1 =≫   8.0   g

 

Award [3] for correct final answer.

[3 marks]

h.i.

n( N 2 H 4 ) = n( O 2 ) = 8.0  g 32.00  g mo l 1 =≫   0.25   mol

volume of nitrogen = 0.25  mol × 24.8  d m 3 mo l 1 ≫= 6.2   d m 3

 

Award [1] for correct final answer.

[1 mark]

h.ii.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.i.
[N/A]
d.ii.
[N/A]
e.
[N/A]
f.
[N/A]
g.
[N/A]
h.i.
[N/A]
h.ii.

Syllabus sections

Additional higher level (AHL) » Topic 14: Chemical bonding and structure » 14.2 Hybridization
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Core » Topic 4: Chemical bonding and structure » 4.3 Covalent structures
Additional higher level (AHL) » Topic 14: Chemical bonding and structure
Core » Topic 4: Chemical bonding and structure
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