IChO 2011 · Theory problem 2

Combustion of ammonia

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Anhydrous ammonia is an ultra-clean, energy-dense alternative liquid fuel. It produces no greenhouse gases on combustion.

In an experiment, gaseous \(\ce{NH3}\) is burned with \(\ce{O2}\) in a container of fixed volume according to the equation given below.

Figure 1

The initial and final states are at 298 K. After combustion with 14.40 g of \(\ce{O2}\), some of \(\ce{NH3}\) remains unreacted.

2.1 Calculate the heat released during the process.

\[ \Delta_{\mathrm f}H^\circ(\ce{NH3(g)})=-46.11\ \mathrm{kJ\,mol^{-1}} \quad\text{and}\quad \Delta_{\mathrm f}H^\circ(\ce{H2O(l)})=-285.83\ \mathrm{kJ\,mol^{-1}} \]

To determine the amount of \(\ce{NH3}\) gas dissolved in water, produced during the combustion process, a 10.00 \(\mathrm{cm^3}\) sample of the aqueous solution was withdrawn from the reaction vessel and added to 15.0 \(\mathrm{cm^3}\) of a \(\ce{H2SO4}\) solution (\(c=0.0100\ \mathrm{mol\,dm^{-3}}\)). The resulting solution was titrated with a standard \(\ce{NaOH}\) solution (\(c=0.0200\ \mathrm{mol\,dm^{-3}}\)) and the equivalence point was reached at 10.64 \(\mathrm{cm^3}\).

\[ \left(K_b(\ce{NH3})=1.8\cdot10^{-5};\quad K_a(\ce{HSO4-})=1.1\cdot10^{-2}\right) \]

2.2 Calculate pH of the solution in the container after combustion.

2.3 At the end point of titration, \(\ce{NH4+}\) and \(\ce{SO4^2-}\) ions are present in the solution. Write the equations for the relevant equilibria to show how the presence of these two ions affects the pH and calculate their equilibrium constant(s).

2.4 Tick the correct statement for the pH of the solution at the equivalence point.

\[ \square\ \mathrm{pH}>7 \qquad \square\ \mathrm{pH}=7 \qquad \square\ \mathrm{pH}<7 \]

Official English text of the problem.

Answers

  • 2.1: Heat released: \(\boxed{227.7\ \mathrm{kJ}}\). Thus \(q_V=-227.7\ \mathrm{kJ}\) for the reacting system.
  • 2.2: \(\boxed{\mathrm{pH}=10.59\approx10.6}\). The usual weak-base approximation gives \(10.60\).
  • 2.3: Ammonium ion, [NH4+], undergoes acidic hydrolysis with \(\boxed{K_a=5.6\times10^{-10}}\). Sulfate ion, [O-]S(=O)(=O)[O-], undergoes basic hydrolysis with \(\boxed{K_b=9.1\times10^{-13}}\). The balanced equilibria are given below.
  • 2.4: Tick \(\boxed{\mathrm{pH}<7}\).

Solution

The chemical species used below have the following names and SMILES. Physical states are indicated in the equations; SMILES do not specify physical state.

Formula Name SMILES
\(\ce{NH3}\) Ammonia N
\(\ce{O2}\) Dioxygen O=O
\(\ce{N2}\) Dinitrogen N#N
\(\ce{H2O}\) Water O
\(\ce{H2SO4}\) Sulfuric acid OS(=O)(=O)O
\(\ce{NaOH}\) Sodium hydroxide [Na+].[OH-]
\(\ce{NH4+}\) Ammonium ion [NH4+]
\(\ce{SO4^2-}\) Sulfate ion [O-]S(=O)(=O)[O-]
\(\ce{HSO4-}\) Hydrogen sulfate ion OS(=O)(=O)[O-]
\(\ce{H3O+}\) Hydronium ion [OH3+]
\(\ce{OH-}\) Hydroxide ion [OH-]

2.1 — Heat released at constant volume

The balanced combustion equation is

\[ \ce{4NH3(g) + 3O2(g) -> 2N2(g) + 6H2O(l)}. \]

Since ammonia remains unreacted, oxygen is the limiting reagent and is completely consumed:

\[ n(\ce{O2})=\frac{14.40\ \mathrm g}{32.00\ \mathrm{g\,mol^{-1}}} =0.4500\ \mathrm{mol}. \]

The reaction extent is therefore

\[ \xi=\frac{0.4500}{3}=0.1500\ \mathrm{mol}. \]

The standard formation enthalpies of elemental \(\ce{N2(g)}\) and \(\ce{O2(g)}\) are zero. Thus, per mole of reaction as written,

\[ \begin{aligned} \Delta_{\mathrm r}H^\circ &=6\Delta_{\mathrm f}H^\circ(\ce{H2O(l)}) -4\Delta_{\mathrm f}H^\circ(\ce{NH3(g)})\\ &=6(-285.83)-4(-46.11)\\ &=-1530.54\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

At constant volume, with only pressure–volume work, the heat equals the internal-energy change. Using ideal-gas behavior and neglecting the liquid’s \(pV\) contribution,

\[ \Delta_{\mathrm r}U^\circ =\Delta_{\mathrm r}H^\circ-\Delta\nu_gRT. \]

Water is liquid, so the change in gaseous stoichiometric coefficients is

\[ \Delta\nu_g=2-(4+3)=-5. \]

Consequently,

\[ \begin{aligned} \Delta_{\mathrm r}U^\circ &=-1530.54-(-5)(8.314\times10^{-3})(298)\\ &=-1518.15\ \mathrm{kJ\,mol^{-1}}, \end{aligned} \]

and

\[ q_V=\xi\Delta_{\mathrm r}U^\circ =(0.1500)(-1518.15) =-227.7\ \mathrm{kJ}. \]

Therefore, the heat released is \(227.7\ \mathrm{kJ}\).

2.2 — pH of the solution after combustion

Sulfuric acid supplies two acid equivalents per mole. Ammonia consumes one acid equivalent per mole; the overall neutralization is

\[ \ce{H2SO4(aq) + 2NH3(aq) -> 2NH4+(aq) + SO4^2-(aq)}. \]

The excess acid is then neutralized by hydroxide during the back titration, through reactions including

\[ \ce{H3O+(aq) + OH-(aq) -> 2H2O(l)} \]

and

\[ \ce{HSO4-(aq) + OH-(aq) -> SO4^2-(aq) + H2O(l)}. \]

The initial amount of sulfuric acid is

\[ n(\ce{H2SO4}) =(0.0100)(15.0\times10^{-3}) =1.50\times10^{-4}\ \mathrm{mol}. \]

The amount of hydroxide used is

\[ n(\ce{OH-}) =(0.0200)(10.64\times10^{-3}) =2.128\times10^{-4}\ \mathrm{mol}. \]

Hence the total amount of dissolved ammonia in the aliquot is

\[ \begin{aligned} n_{\mathrm{ammonia}} &=2n(\ce{H2SO4})-n(\ce{OH-})\\ &=3.00\times10^{-4}-2.128\times10^{-4}\\ &=8.72\times10^{-5}\ \mathrm{mol}. \end{aligned} \]

Its analytical concentration in the original solution is

\[ C=\frac{8.72\times10^{-5}}{10.00\times10^{-3}} =8.72\times10^{-3}\ \mathrm{mol\,dm^{-3}}. \]

Ammonia establishes the equilibrium

\[ \ce{NH3(aq) + H2O(l) <=> NH4+(aq) + OH-(aq)}. \]

Neglecting water’s small contribution to the hydroxide concentration, let

\[ x=[\ce{OH-}]=[\ce{NH4+}],\qquad [\ce{NH3}]=C-x. \]

Then

\[ K_b=\frac{x^2}{C-x}=1.8\times10^{-5}, \]

so

\[ x=\frac{-K_b+\sqrt{K_b^2+4K_bC}}{2} =3.8728\times10^{-4}\ \mathrm{mol\,dm^{-3}}. \]

Using \(K_w=1.0\times10^{-14}\) at 298 K,

\[ \mathrm{pOH}=3.412, \qquad \boxed{\mathrm{pH}=14.00-3.412=10.59}. \]

The approximation \(x\approx\sqrt{K_bC}\) gives \(\mathrm{pH}=10.60\), consistent with reporting the result as approximately \(10.6\).

2.3 — Hydrolysis equilibria and their constants

Ammonium is the conjugate acid of ammonia. It produces hydronium ions and therefore lowers the pH:

\[ \boxed{\ce{NH4+(aq) + H2O(l) <=> NH3(aq) + H3O+(aq)}}. \]

Its acid-dissociation constant is

\[ \boxed{ K_a(\ce{NH4+}) =\frac{K_w}{K_b(\ce{NH3})} =\frac{1.0\times10^{-14}}{1.8\times10^{-5}} =5.6\times10^{-10} }. \]

Sulfate is the conjugate base of hydrogen sulfate. It produces hydroxide ions and therefore raises the pH:

\[ \boxed{\ce{SO4^2-(aq) + H2O(l) <=> HSO4-(aq) + OH-(aq)}}. \]

Its base-dissociation constant is

\[ \boxed{ K_b(\ce{SO4^2-}) =\frac{K_w}{K_a(\ce{HSO4-})} =\frac{1.0\times10^{-14}}{1.1\times10^{-2}} =9.1\times10^{-13} }. \]

2.4 — pH at the equivalence point

The acidic hydrolysis constant of ammonium is approximately

\[ \frac{K_a(\ce{NH4+})}{K_b(\ce{SO4^2-})} \approx6.1\times10^2 \]

times the basic hydrolysis constant of sulfate.

Their analytical amounts are comparable: the sulfate-to-ammonium ratio is only

\[ \frac{1.50\times10^{-4}}{8.72\times10^{-5}}=1.72. \]

Thus ammonium’s acidic effect dominates sulfate’s basic effect. Tick the third box:

\[ \boxed{\mathrm{pH}<7}. \]

The solution was checked against the official solution.