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Date May 2012 Marks available 3 Reference code 12M.2.sl.TZ2.6
Level SL Paper 2 Time zone TZ2
Command term Determine Question number 6 Adapted from N/A

Question

Ammonia, \({\text{N}}{{\text{H}}_{\text{3}}}\), is a base according to both the Brønsted–Lowry and the Lewis theories of acids and bases.

The equation for the reaction between sodium hydroxide, NaOH, and nitric acid, \({\text{HN}}{{\text{O}}_{\text{3}}}\), is shown below.

\[\begin{array}{*{20}{l}} {{\text{NaOH(aq)}} + {\text{HN}}{{\text{O}}_3}{\text{(aq)}} \to {\text{NaN}}{{\text{O}}_3}{\text{(aq)}} + {{\text{H}}_2}{\text{O(l)}}}&{{\text{ }}\Delta H = - 57.6{\text{ kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}} \end{array}\]

Distinguish between the terms strong base and weak base, and state one example of each.

[3]
a.

State the equation for the reaction of ammonia with water.

[1]
b.i.

Explain why ammonia can act as a Brønsted–Lowry base.

[1]
b.ii.

Explain why ammonia can also act as a Lewis base.

[1]
b.iii.

(i)     When ammonium chloride, \({\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl(aq)}}\), is added to excess solid sodium carbonate, \({\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{(s)}}\), an acid–base reaction occurs. Bubbles of gas are produced and the solid sodium carbonate decreases in mass. State one difference which would be observed if nitric acid, \({\text{HN}}{{\text{O}}_{\text{3}}}{\text{(aq)}}\), was used instead of ammonium chloride.

(ii)     Deduce the Lewis structures of the ammonium ion, \({\text{NH}}_4^ + \), and the carbonate ion, \({\text{CO}}_3^{2 - }\).

Ammonium ion\(\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \)Carbonate ion

(iii)     Predict the shapes of \({\text{NH}}_4^ + \) and \({\text{CO}}_3^{2 - }\).

\({\text{NH}}_4^ + \):

\({\text{CO}}_3^{2 - }\):

[5]
c.

(i)     Sketch and label an enthalpy level diagram for this reaction.

(ii)     Deduce whether the reactants or the products are more energetically stable, stating your reasoning.

(iii)     Calculate the change in heat energy, in kJ, when \({\text{50.0 c}}{{\text{m}}^{\text{3}}}\) of \({\text{2.50 mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\) sodium hydroxide solution is added to excess nitric acid.

[6]
d.

When 5.35 g ammonium chloride, \({\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl(s)}}\), is added to \({\text{100.0 c}}{{\text{m}}^{\text{3}}}\) of water, the temperature of the water decreases from 19.30 °C to 15.80 °C. Determine the enthalpy change, in \({\text{kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}\), for the dissolving of ammonium chloride in water.

[3]
e.

Markscheme

a strong base: base/electrolyte (assumed to be almost) completely/100% dissociated/ionized (in solution/water) / OWTTE and a weak base: base/electrolyte partially dissociated/ionized (in solution/water) / OWTTE;

example of a strong base: any group I hydroxide / \({\text{Ba(OH}}{{\text{)}}_2}\);

example of a weak base: \({\text{N}}{{\text{H}}_3}\) / \({\text{C}}{{\text{H}}_3}{\text{N}}{{\text{H}}_2}\) / any reasonable answer;

a.

\({\text{N}}{{\text{H}}_3} + {{\text{H}}_2}{\text{O}} \rightleftharpoons {\text{NH}}_4^ +  + {\text{O}}{{\text{H}}^ - }\);

b.i.

accepts a proton/\({{\text{H}}^ + }\) / OWTTE;

b.ii.

donates an electron pair;

b.iii.

(i)     more vigorous reaction / more gas bubbles / OWTTE;

more heat released;

solid decreases more quickly;

(ii)     M12/4/CHEMI/SP2/ENG/TZ2/06.c/M

Accept any combination of lines, dots or crosses to represent electron pairs.

(iii)     NH4+:

tetrahedral;

CO32–:

trigonal/triangular planar;

c.

(i)     enthalpy on y-axis;

Do not accept energy.

reactants higher than products;

\(\Delta H\) labelled;

M12/4/CHEMI/SP2/ENG/TZ2/06.d/M

Accept appropriate formulas for reactants and products.

Arrow heads not needed.

57.6 is acceptable as an alternative to \(\Delta H\).

(ii)     products are more stable as they are at a lower enthalpy level / energy has been given off by the reactants / reaction is exothermic / OWTTE;

(iii)     \(n{\text{(NaOH)}} = 0.125{\text{ mol}}\);

change in heat energy \( = ( - 57.6 \times 0.125) =  - 7.20{\text{ (kJ)}}\) / heat released \( = (57.6 \times 0.125) = 7.20{\text{ (kJ)}}\);

d.

\(q = (mc\Delta T = ){\text{ }}100.0 \times 4.18 \times 3.50/1463{\text{ J}}/1460{\text{ J}}\);

\(n{\text{(N}}{{\text{H}}_{\text{4}}}{\text{Cl)}} = \frac{{5.35}}{{53.5}}/0.100{\text{ mol}}\);

\(\Delta H =  + 14.6/14.6{\text{ (kJ mo}}{{\text{l}}^{ - 1}}{\text{)}}\);

Accept q = 105.35 \( \times \) 4.18 \( \times \) 3.50 / 1541 J.

Accept \(\Delta H\) = +15.4 / 15.4 (kJ\(\,\)mol–1)

e.

Examiners report

Part (a) was answered well although some mentioned “dissolving” instead of “dissociating”.

a.

In (b), the equation was well done.

b.i.

In (b), the equation was well done as was (ii).

b.ii.

Inevitably, many omitted “pair” in (iii).

b.iii.

Part (c)(i) was generally correct. In (c)(ii) the carbonate ion was legitimately examined under AS 4.2.7; it was not well known – there were too many carbons with expanded octets and oxygens where the lone pairs had been missed. (In the HL specification, the carbonate ion‘s delocalization is considered.) In (iii), however, the shapes were well known.

c.

If there was to be an error made in (d)(i), it was to omit “enthalpy” from the y-axis and some unaccountably put the correct chemicals on the line and then reversed the names products and reactants. The calculations in (d)(iii) inevitably depended on an ability to calculate and think logically.

d.

The calculations in (e) inevitably depended on an ability to calculate and think logically.

e.

Syllabus sections

Core » Topic 5: Energetics/thermochemistry » 5.2 Hess’s Law
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