IChO 1999 · Theory problem 1

Combustion calorimetry and molecular state of compound Q

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A compound Q (molar mass \(122.0\ \mathrm{g\,mol^{-1}}\)) consists of carbon, hydrogen and oxygen.

PART A

The standard enthalpy of formation of \(\ce{CO2(g)}\) and \(\ce{H2O(l)}\) at \(25.00\ ^\circ\mathrm C\) are −393.51 and −285.83 \(\mathrm{kJ\,mol^{-1}}\), respectively. The gas constant, \(R=8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}\).

(Relative atomic masses : H = 1.0; C = 12.0; O = 16.0)

A sample of solid Q that weighs 0.6000 g, is combusted in an excess of oxygen in a bomb calorimeter, which initially contains 710.0 g of water at \(25.000\ ^\circ\mathrm C\). After the reaction is completed, the temperature is observed to be \(27.250\ ^\circ\mathrm C\), and 1.5144 g of \(\ce{CO2(g)}\) and 0.2656 g of \(\ce{H2O(l)}\) are produced.

1.1 Determine the molecular formula and write a balanced equation with correct state of matters for the combustion of Q.

If the specific heat of water is \(4.184\ \mathrm{J\,g^{-1}\,K^{-1}}\) and the internal energy change of the reaction \((\Delta U^0)\) −3079 \(\mathrm{kJ\,mol^{-1}}\).

1.2 Calculate the heat capacity of the calorimeter (excluding the water).

1.3 Calculate the standard enthalpy of formation \((\Delta H_{\mathrm f}^{0})\) of Q.

PART B

The following data refer to the distribution of Q between benzene and water at \(6\ ^\circ\mathrm C\), \(C_{\mathrm B}\) and \(C_{\mathrm W}\) being equilibrium concentrations of the species of Q in the benzene and water layers, respectively:

Assume that there is only one species of Q in benzene independent of concentration and temperature.

Concentration \((\mathrm{mol\,dm^{-3}})\)

\(C_{\mathrm B}\) \(C_{\mathrm W}\)
0.0118 0.00281
0.0478 0.00566
0.0981 0.00812
0.156 0.0102

1.4 Show by calculation whether Q is monomer or dimer in benzene. Assume that Q is a monomer in water.

The freezing point depression, for an ideal dilute solution, is given by

\[ T_{\mathrm f}^{0}-T_{\mathrm f} =\frac{R(T_{\mathrm f}^{0})^{2}X_{\mathrm s}}{\Delta H_{\mathrm f}} \]

where \(T_{\mathrm f}\) is the freezing point of the solution, \(T_{\mathrm f}^{0}\) the freezing point of solvent, \(\Delta H_{\mathrm f}\) the heat of fusion of the solvent, and \(X_{\mathrm s}\) the mole fraction of solute. The molar mass of benzene is \(78.0\ \mathrm{g\,mol^{-1}}\). At 1 atm pure benzene freezes at \(5.40\ ^\circ\mathrm C\). The heat of fusion of benzene is \(9.89\ \mathrm{kJ\,mol^{-1}}\).

1.5 Calculate the freezing point \((T_{\mathrm f})\) of a solution containing 0.244 g of Q in 5.85 g of benzene at 1 atm.

Official English text of the problem.

Answers

  • 1.1 Molecular formula: \(\boxed{\ce{C7H6O2}}\). \[ \boxed{\ce{2 C7H6O2(s) + 15 O2(g) -> 14 CO2(g) + 6 H2O(l)}} \]
  • 1.2 Calorimeter heat capacity, excluding water: \(\boxed{3.759\ \mathrm{kJ\,K^{-1}}}\).
  • 1.3 Standard enthalpy of formation: \(\boxed{\Delta H_{\mathrm f}^{0}(\ce{Q(s)})\approx-532\ \mathrm{kJ\,mol^{-1}}}\).
  • 1.4 Q is a dimer, \(\ce{Q2}\), in benzene, since \(C_{\mathrm B}/C_{\mathrm W}^{2}\) is approximately constant.
  • 1.5 Freezing point: \(\boxed{T_{\mathrm f}=4.54\ ^\circ\mathrm C}\).

Solution

The specified molecular species are carbon dioxide O=C=O, water O, dioxygen O=O, and benzene c1ccccc1. The data determine Q’s molecular formula and association state. A unique chemical name or SMILES for Q or its dimer cannot be assigned without further structural information.

1.1 Molecular formula and combustion equation

The amount of Q burned is

\[ n_Q=\frac{0.6000\ \mathrm g}{122.0\ \mathrm{g\,mol^{-1}}} =4.91803\times10^{-3}\ \mathrm{mol}. \]

All carbon appears in carbon dioxide, and all hydrogen appears in water:

\[ n(\ce C)=n(\ce{CO2}) =\frac{1.5144\ \mathrm g}{44.0\ \mathrm{g\,mol^{-1}}} =0.0344182\ \mathrm{mol}, \]

\[ n(\ce H)=2n(\ce{H2O}) =2\frac{0.2656\ \mathrm g}{18.0\ \mathrm{g\,mol^{-1}}} =0.0295111\ \mathrm{mol}. \]

Their masses in the sample are therefore

\[ m(\ce C)=0.413018\ \mathrm g, \qquad m(\ce H)=0.0295111\ \mathrm g. \]

Since Q contains only carbon, hydrogen, and oxygen, its oxygen content follows by difference:

\[ m(\ce O) =0.6000-0.413018-0.0295111 =0.157471\ \mathrm g, \]

\[ n(\ce O) =\frac{0.157471\ \mathrm g}{16.0\ \mathrm{g\,mol^{-1}}} =0.00984192\ \mathrm{mol}. \]

Dividing each atomic amount by the amount of Q gives the number of atoms per molecule:

\[ \frac{n(\ce C)}{n_Q}=6.9984\approx7,\qquad \frac{n(\ce H)}{n_Q}=6.0006\approx6,\qquad \frac{n(\ce O)}{n_Q}=2.0012\approx2. \]

Thus,

\[ \boxed{Q:\ \ce{C7H6O2}}, \]

with molar mass

\[ 7(12.0)+6(1.0)+2(16.0)=122.0\ \mathrm{g\,mol^{-1}}, \]

in agreement with the given value.

Balancing carbon, hydrogen, and then oxygen gives

\[ \boxed{\ce{2 C7H6O2(s) + 15 O2(g) -> 14 CO2(g) + 6 H2O(l)}}. \]

1.2 Heat capacity of the calorimeter

A bomb calorimeter operates at constant volume, so the reaction heat is the internal-energy change:

\[ q_{\mathrm{rxn}} =n_Q\Delta U^0 =\frac{0.6000}{122.0}(-3079) =-15.1426\ \mathrm{kJ}. \]

The water and calorimeter therefore absorb \(15.1426\ \mathrm{kJ}\). Their temperature increase is

\[ \Delta T=27.250-25.000=2.250\ \mathrm K. \]

The water absorbs

\[ q_{\mathrm{water}} =m_{\mathrm{water}}c_{\mathrm{water}}\Delta T =(710.0)(4.184)(2.250) =6683.94\ \mathrm J =6.68394\ \mathrm{kJ}. \]

Energy conservation gives

\[ -q_{\mathrm{rxn}} =q_{\mathrm{water}}+C_{\mathrm{cal}}\Delta T. \]

Consequently,

\[ C_{\mathrm{cal}} =\frac{15.1426-6.68394}{2.250} =3.7594\ \mathrm{kJ\,K^{-1}}, \]

or

\[ \boxed{C_{\mathrm{cal}}=3.759\ \mathrm{kJ\,K^{-1}}}. \]

1.3 Standard enthalpy of formation of Q

For the combustion of one mole of Q, the change in gaseous stoichiometric amount is

\[ \Delta n_{\mathrm g}=7-\frac{15}{2}=-\frac12. \]

Water is liquid and therefore does not contribute to \(\Delta n_{\mathrm g}\). Using

\[ \Delta H_{\mathrm c}^{0} =\Delta U^{0}+\Delta n_{\mathrm g}RT \]

at \(T=298.15\ \mathrm K\),

\[ \begin{aligned} \Delta H_{\mathrm c}^{0} &=-3079 -\frac12(8.314\times10^{-3})(298.15)\\ &=-3080.239\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

By Hess’s law, with \(\Delta H_{\mathrm f}^{0}(\ce{O2(g)})=0\),

\[ \Delta H_{\mathrm c}^{0} = 7\Delta H_{\mathrm f}^{0}(\ce{CO2(g)}) +3\Delta H_{\mathrm f}^{0}(\ce{H2O(l)}) -\Delta H_{\mathrm f}^{0}(\ce{Q(s)}). \]

Therefore,

\[ \begin{aligned} \Delta H_{\mathrm f}^{0}(\ce{Q(s)}) &=7(-393.51)+3(-285.83)-(-3080.239)\\ &=-531.821\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

To a precision appropriate to the given combustion energy,

\[ \boxed{\Delta H_{\mathrm f}^{0}(\ce{Q(s)}) \approx-532\ \mathrm{kJ\,mol^{-1}}}. \]

1.4 Molecular state of Q in benzene

For monomeric Q in both solvents, the distribution equilibrium would require

\[ \frac{C_{\mathrm B}}{C_{\mathrm W}}=\text{constant}. \]

If Q forms a dimer in benzene, the overall equilibrium is

\[ \ce{2 Q(aq) <=> Q2(benzene)}, \]

and the corresponding concentration expression requires

\[ \frac{C_{\mathrm B}}{C_{\mathrm W}^{2}}=\text{constant}. \]

The two tests give:

\(C_{\mathrm B}\;(\mathrm{mol\,dm^{-3}})\) \(C_{\mathrm W}\;(\mathrm{mol\,dm^{-3}})\) \(C_{\mathrm B}/C_{\mathrm W}\) \(C_{\mathrm B}/C_{\mathrm W}^{2}\;(\mathrm{dm^3\,mol^{-1}})\)
0.0118 0.00281 4.20 \(1.494\times10^3\)
0.0478 0.00566 8.45 \(1.492\times10^3\)
0.0981 0.00812 12.1 \(1.488\times10^3\)
0.156 0.0102 15.3 \(1.499\times10^3\)

The monomer ratio varies substantially, whereas the dimer ratio is approximately constant:

\[ \frac{C_{\mathrm B}}{C_{\mathrm W}^{2}} \approx1.49\times10^3\ \mathrm{dm^3\,mol^{-1}}. \]

Hence,

\[ \boxed{\text{Q exists as a dimer in benzene.}} \]

1.5 Freezing point of the benzene solution

The stipulated independence of molecular state from concentration and temperature allows the dimer result to be used at the freezing point.

The solute particles are dimers, with molar mass

\[ M(\ce{Q2})=2(122.0)=244.0\ \mathrm{g\,mol^{-1}}. \]

Thus,

\[ n_{\mathrm s} =\frac{0.244\ \mathrm g}{244.0\ \mathrm{g\,mol^{-1}}} =0.00100\ \mathrm{mol}, \]

while the amount of benzene is

\[ n_{\mathrm{benzene}} =\frac{5.85\ \mathrm g}{78.0\ \mathrm{g\,mol^{-1}}} =0.0750\ \mathrm{mol}. \]

The mole fraction of solute particles is

\[ X_{\mathrm s} =\frac{n_{\mathrm s}}{n_{\mathrm s}+n_{\mathrm{benzene}}} =\frac{0.00100}{0.00100+0.0750} =0.0131579. \]

Convert the solvent freezing point and heat of fusion to the required units:

\[ T_{\mathrm f}^{0}=5.40+273.15=278.55\ \mathrm K, \qquad \Delta H_{\mathrm f}=9890\ \mathrm{J\,mol^{-1}}. \]

Using the supplied equation,

\[ \begin{aligned} \Delta T_{\mathrm f} &=\frac{R(T_{\mathrm f}^{0})^2X_{\mathrm s}} {\Delta H_{\mathrm f}}\\ &=\frac{(8.314)(278.55)^2(0.0131579)}{9890}\\ &=0.8582\ \mathrm K. \end{aligned} \]

Therefore,

\[ T_{\mathrm f}=5.40-0.8582=4.5418\ ^\circ\mathrm C, \]

giving

\[ \boxed{T_{\mathrm f}=4.54\ ^\circ\mathrm C}. \]

The solution was checked against the official solution.