IB DP Chemistry Reactivity 1.1 Measuring enthalpy changes HL Paper 2- Exam Style Questions - New Syllabus
Question
(i) Determine the standard enthalpy change for the reaction, \(\,\Delta H^\circ_{\mathrm{r}}\,\), in \(\,\mathrm{kJ\,mol^{-1}}\,\) for the oxidation of sulfur dioxide to sulfur trioxide, using the data. [\(\,2\,\)]
Enthalpy of formation \(\big(\Delta H^\circ_{\mathrm{f}}/\mathrm{kJ\,mol^{-1}}\big)\): \(\ \mathrm{SO_2(g)}:\,-296.8,\quad \mathrm{SO_3(g)}:\,-395.8\)
(ii) Formulate equations, including state symbols, showing how sulfur dioxide and sulfur trioxide lead to acid deposition. [\(\,2\,\)]
\(\mathrm{SO_3:}\ \underline{\hspace{120px}}\)
(iii) Explain the polarity of the \(\mathrm{S{-}O}\) bond in \(\mathrm{SO_2}\) and \(\mathrm{SO_3}\) using data booklet. [\(\,2\,\)]
The combustion of \(\,0.1\,\) moles of sulfur \(\,(\mathrm{S})\,\) was demonstrated in a school laboratory using the following apparatus in a fume cupboard.

(i) Calculate the enthalpy of combustion of sulfur, \(\,\Delta H_{\mathrm{c}}\,\), in \(\,\mathrm{kJ\,mol^{-1}}\,\) from this data. Use data booklet. [\(\,2\,\)]
Initial temperature of water \((^\circ\mathrm{C})\ \pm 0.5:\ \ \,20.0\)
Final temperature of water \((^\circ\mathrm{C})\ \pm 0.5:\ \ \,35.0\)
(v) Calculate the percentage difference between the accepted value, \(\,-297\,\mathrm{kJ\,mol^{-1}}\), and your value for \(\,\Delta H_{\mathrm{c}}\). [\(\,1\,\)]
▶️ Answer/Explanation
Balanced reaction: \(\ \mathrm{SO_2(g)+\tfrac{1}{2}O_2(g)\rightarrow SO_3(g)}\).
\(\displaystyle \Delta H^\circ_{\mathrm{r}}=\sum \Delta H^\circ_{\mathrm{f}}(\text{products})-\sum \Delta H^\circ_{\mathrm{f}}(\text{reactants})\) \(=\big(-395.8\big)-\big[-296.8+0\big]=\boxed{-99.0\ \mathrm{kJ\,mol^{-1}}}.\)
\(\mathrm{SO_2(aq)+H_2O(l)\rightleftharpoons H_2SO_3(aq)}\) (sulfurous acid).
\(\mathrm{SO_3(aq)+H_2O(l)\rightarrow H_2SO_4(aq)}\) (sulfuric acid).
Oxygen is more electronegative than sulfur \(\big(\chi_{\mathrm{O}}\approx 3.4,\ \chi_{\mathrm{S}}\approx 2.6\big)\), so the \(\mathrm{S{-}O}\) bond is polar with \(\delta^-\) on O and \(\delta^+\) on S. The bond dipole points from S to O (towards O).
Heat gained by water (assume all released heat warms the water): \(\ q=mc\Delta T\).
\(m=50.00\ \mathrm{g},\ \ c=4.18\ \mathrm{J\,K^{-1}\,g^{-1}},\ \ \Delta T=(35.0-20.0)^\circ\mathrm{C}=15.0\ \mathrm{K}.\)
\(\displaystyle q=50.00\times 4.18\times 15.0=3.14\times 10^3\ \mathrm{J}=3140\ \mathrm{J}.\)
For \(\,0.10\,\mathrm{mol}\) of S burned, the molar enthalpy (exothermic) is \(\displaystyle \Delta H_{\mathrm{c}}=-\frac{q}{n_{\mathrm{S}}}=-\frac{3140\ \mathrm{J}}{0.10\ \mathrm{mol}}=-3.14\times 10^4\ \mathrm{J\,mol^{-1}}=\boxed{-31.4\ \mathrm{kJ\,mol^{-1}}}.\)
One systematic error: significant heat loss to the surroundings (beaker, air, incomplete capture of hot gases).
Improvement: insulate and cover the calorimeter (lid), or use a calibrated/bomb calorimeter (or a windbreak around the flame/apparatus).
Uncertainty in \(\Delta T\): add the absolute thermometer uncertainties: \(\ \pm(0.5+0.5)=\pm 1.0^\circ\mathrm{C}.\)
\(\displaystyle \%\ \text{uncertainty}=\frac{1.0}{15.0}\times 100=\boxed{6.7\%}\ \) (accept \(\approx 5\%\) with root-sum-square method shown).
Reduce percentage uncertainty by increasing \(\Delta T\) (larger sample/volume), using a thermometer with finer divisions (greater precision), or improving insulation.
\(\displaystyle \%\ \text{difference}=\frac{\lvert -297-(-31.4)\rvert}{297}\times 100=\frac{265.6}{297}\times 100=\boxed{89.4\%}.\)