Methanol reacts with trinitramide to form nitrogen, carbon dioxide and water. Deduce the coefficients required to balance the equation for this reaction.

___ \({\text{N(N}}{{\text{O}}_2}{{\text{)}}_3}{\text{(g)}} + \) ___ \({\text{C}}{{\text{H}}_3}{\text{OH(l)}} \to \) ___ \({{\text{N}}_2}{\text{(g)}} + \) ___ \({\text{C}}{{\text{O}}_2}{\text{(g)}} + \) ___ \({{\text{H}}_2}{\text{O(l)}}\)

Calculate the enthalpy change, in \({\text{kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}\), when one mole of trinitramide decomposes to its elements, using bond enthalpy data from Table 10 of the Data Booklet. Assume that all the N–O bonds in this molecule have a bond enthalpy of \({\text{305 kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}\).

Outline how the length of the N–N bond in trinitramide compares with the N–N bond in nitrogen gas, \({{\text{N}}_{\text{2}}}\).

Deduce the N–N–N bond angle in trinitramide and explain your reasoning.

Predict, with an explanation, the polarity of the trinitramide molecule.

Methanol can also be burnt as a fuel. Describe an experiment that would allow the molar enthalpy change of combustion to be calculated from the results.

Explain how the results of this experiment could be used to calculate the molar enthalpy change of combustion of methanol.

Predict, with an explanation, how the result obtained would compare with the value in Table 12 of the Data Booklet.

\(\underline {{\text{ (1) }}} {\text{N(N}}{{\text{O}}_2}{{\text{)}}_3}{\text{(g)}} + \underline {{\text{  2  }}} {\text{C}}{{\text{H}}_3}{\text{OH(l)}} \to \underline {{\text{  2  }}} {{\text{N}}_2}{\text{(g)}} + \underline {{\text{  2  }}} {\text{C}}{{\text{O}}_2}{\text{(g)}} + \underline {{\text{  4  }}} {{\text{H}}_2}{\text{O(l)}}\);

bonds broken: \((6 \times 305) + (3 \times 158) = 1830 + 474 = 2304{\text{ }}({\text{kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}})\);

bonds made: \((2 \times 945) + (3 \times 498) = 1890 + 1494 = 3384{\text{ }}({\text{kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}})\);

enthalpy change: \(2304 - 3384 =  - 1080{\text{ }}({\text{kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}})\);

Award [3] for correct final answer.

Award [2 max] for +1080 (kJ mol^–1) .

Accept –234 kJ mol^–1 which arise from students assuming that 305 kJ mol^–1 refers to the strength of a single N–O bond. Students may then take N=O from the data book value (587 kJ mol^–1).

bonds broken: (3 \( \times \) 305) + (3 \( \times \) 587) + (3 \( \times \) 158) = 915 + 1761 + 474 = 3150 (kJ mol^–1)

bonds made: (2 \( \times \) 945) + (3 \( \times \) 498) = 1890 + 1494 = 3384 (kJ mol^–1)

enthalpy change: 3150 – 3384 = –234 (kJ mol^–1) .

Award [2 max] for correct calculation of the enthalpy change of reaction for the equation in part (a), which gives –2160 (kJ mol^–1).

Award [1] if the final answer is not –2160 but the candidate has correctly calculated the bonds broken in trinitramide as 2304 (kJ mol^–1).

(N–N bond in) trinitramide is longer/nitrogen (gas) is shorter / 0.145 nm in trinitramide versus 0.110 nm in nitrogen;

trinitramide has single (N–N) bond and nitrogen (gas) has triple bond;

106°–108°;

Accept \( < \)109°.

Any two for [2 max].

4 (negative) charge centres/electron pairs/electron domains around central nitrogen;

central nitrogen has a lone/non-bonding pair;

lone/non-bonding pairs repel more than bonding pairs;

molecule will be (trigonal/triangular) pyramidal;

(negative) charge centres/electron pairs/electron domains will be tetrahedrally arranged/orientated/ have tetrahedral geometry;

Do not apply ECF.

polar;

net dipole moment present in molecule / unsymmetrical distribution of charge / polar bonds do not cancel out / centre of negatively charged oxygen atoms does not coincide with positively charged nitrogen atom;

Marks may also be awarded for a suitably presented diagram showing net dipole moment.

Do not accept “unsymmetrical molecule”.

For polarity, apply ECF from part (e).

burn/combust a (known) mass/volume/quantity/amount of methanol (in a spirit burner) / weigh methanol/spirit burner before and after combustion;

use flame to heat a (known) mass/volume/quantity/amount of water;

measure the increase/rise/change in temperature (of the water);

calculate the heat gained by the water / calculate the heat evolved by the burning methanol / substitute in \(q = mc\Delta T\);

calculate the amount/moles of methanol / divide the mass of methanol by its molar mass;

divide the heat gained by the water by the amount/moles of methanol;

result would be less exothermic/less negative;

Accept “less/smaller/lower”.

heat loss / incomplete combustion;

Accept methanol is volatile/evaporates / beaker/material of calorimeter absorbs heat.

Most candidates got the correct stoichiometric coefficients for the equation in part (a).

In Part (c), the typical errors were using the incorrect bond enthalpies from the Data Booklet and using the sum of the bond enthalpies of bond forming (products) minus bond breaking (reactants) instead of the reverse. Some candidates surprisingly used the combustion equation from part (a) for their extensive calculations which was partially given credit.

Part (d) was well answered although a number of candidates thought that nitrogen has a single or double bond instead of a triple bond which was worrying. VSEPR theory however was exceptionally poor and most candidates demonstrated little or no understanding. Many incorrect geometries were cited, especially trigonal planar and even linear and v-shaped! Very few candidates related the geometry to four negative charge centres or electron domains around the central nitrogen atom.

In part (f), polarity typically involved just guess work and only few candidates could explain the reason for the polarity or gave a diagram showing the net dipole moment which suggested poor understanding of the topic.

Part (g) was generally well answered and of those that attempted the question they often scored full marks demonstrating good understanding of calorimetry.

Part (g) was generally well answered and of those that attempted the question they often scored full marks demonstrating good understanding of calorimetry.

Part (g) was generally well answered and of those that attempted the question they often scored full marks demonstrating good understanding of calorimetry.