IChO 2025 · Theory problem 4

The life of tennis balls

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4.1 4.2 4.3 4.4 Total
3 16 5 16 40
Figure 1

This problem honours the incredible Rafael Nadal, a true tennis icon who concluded his remarkable career in 2024.

The Dubai Tennis Championships is a prestigious tournament held annually in UAE. Many of the world's top players, including Rafael Nadal, Roger Federer, Novak Djokovic, and Andy Murray have won titles there.

The composition of tennis balls has evolved over hundreds of years. To ensure a good bounce, the inside of modern tennis balls is pressurised above atmospheric pressure. However, old hollow tennis balls just contained air at atmospheric pressure.

In this problem, assume tennis balls have an inner radius of \(R = 3.0\ \mathrm{cm}\) that remains constant throughout the pressurisation process. Assume that the composition of air is 20% \(\ce{O2}\) and 80% \(\ce{N2}\) by volume. Consider that overpressure does not expand the ball.

4.1 Calculate the mass \(m\) (in g) of air inside an old tennis ball at \(T_0 = 25\ ^\circ\mathrm{C}\).

Chemists then discovered how to pressurise a ball above atmospheric pressure using a reaction between compounds A and B, which forms gas C (to create additional inner pressure) plus other non-gaseous products.

Figure 2

A capsule with a quantity \(m_{\mathrm{mix}} = 0.3389\ \mathrm{g}\) of a stoichiometric mixture of two white inorganic compounds A and B was placed inside a ball filled with air at atmospheric pressure. They reacted completely inside the ball upon heating to form gas C and other products (reaction 1) to produce an overall pressure inside the ball of \(p = 1.600\ \mathrm{atm}\) (at \(25\ ^\circ\mathrm{C}\)). Once this ball was opened, all residue inside was collected and dissolved in water. The dropping of silver nitrate in that solution gave a white precipitate (reaction 2). This collected precipitate could be partially dissolved in a concentrated aqueous solution of B. By increasing the pH of this mixture, the precipitate dissolved completely (reaction 3). If the same initial capsule of A+B is dissolved in water and concentrated solution of silver nitrate added, \(m_{\mathrm{precipitate}} = 0.8224\ \mathrm{g}\) of white precipitate formed which was approximately twice the mass of precipitate formed in reaction 2.

When the capsule of A+B dissolved in a solution of \(\ce{HI}\) the formation of gas D started, and the solution turned brown (reaction 4). Upon further addition of sodium thiosulfate this colour disappeared (reaction 5).

The densities of gases C and D differ by no more than 10%.

4.2 Write the chemical formulae of compounds A–D.

4.3 Write equations for reactions 1–5.

The pressure inside modern tennis balls is \(p_0 = 1.80\ \mathrm{atm}\). They are sold in a pressurised can, with the same pressure in the can as in the ball to stop gas coming out during storage.

Standard balls contain air and premium balls contain pure \(\ce{N2}\). Once the can is opened, depressurisation begins, and the gas diffuses out of the balls until the internal pressure reaches atmospheric pressure. Assume there is no diffusion of gas into the balls at any point. The activation energy for the depressurisation of a premium ball is \(E_A = 50.0\ \mathrm{kJ\,mol^{-1}}\).

The rate of ball depressurisation (pressure/time) could be described by similar equations of those used in chemical kinetics. At \(T_0 = 25.0\ ^\circ\mathrm{C}\), once the can is opened, the premium balls depressurise to \(p_1 = 1.40\ \mathrm{atm}\) after \(t_1 = 241\ \mathrm{h}\) and to \(p_2 = 1.19\ \mathrm{atm}\) after about \(t_2 = 21\) days. The initial depressurisation rate of regular balls is 10% faster than premium balls.

4.4a Demonstrate that the depressurisation process follows the first kinetic order by calculating the depressurisation rate constant \(k_{\ce{N2}}\) for \(\ce{N2}\) in \(\mathrm{h^{-1}}\).

Note: If you couldn't calculate the answer, use \(k_{\ce{N2}} = 10^{-3}\ \mathrm{h^{-1}}\) in further calculations.

4.4b Calculate the (depressurisation) rate constant \(k_{\ce{O2}}\) for \(\ce{O2}\) in \(\mathrm{h^{-1}}\), assuming the rates for both gases are additive.

As well as today being the IChO 2025 theory paper, it is also (perhaps less importantly) the Semi-Finals day at Wimbledon. Amanda Anisimova will start a match in \(t_{\mathrm{match}} = 12.0\ \mathrm{h}\) from now. The weather today in Wimbledon is predicted to be a constant \(T_W = 30.0\ ^\circ\mathrm{C}\). Assume a new set of premium balls, which were manufactured and placed in the can at \(T_0 = 25.0\ ^\circ\mathrm{C}\), are opened now and quickly change to \(T_W = 30.0\ ^\circ\mathrm{C}\).

4.4c Calculate the pressure \(p_{\mathrm{start}}\) (in atm) of the balls at the start of the match.

Periodic table and data sheet

Physical constants and formulae

Avogadro constant \(N_A\) \(6.022\times10^{23}\ \mathrm{mol^{-1}}\)
molar gas constant \(R\) \(8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}\)
Faraday constant \(F\) \(96485\ \mathrm{C\,mol^{-1}}\)
Planck constant \(h\) \(6.626\times10^{-34}\ \mathrm{J\,s}\)
speed of light in vacuum \(c\) \(2.998\times10^8\ \mathrm{m\,s^{-1}}\)
mass of electron \(m_e\) \(9.109\times10^{-31}\ \mathrm{kg}\)
charge of electron \(e\) \(1.602\times10^{-19}\ \mathrm{C}\)
atmospheric pressure \(p_{\mathrm{atm}}\) \(101325\ \mathrm{Pa}\)
ionic product of water at 298 K \(K_W\) \(10^{-14}\)
\(1\ \mathrm{nm}=1\times10^{-9}\ \mathrm{m}\)
\(1\ \text{Å}=1\times10^{-10}\ \mathrm{m}\)
\(1\ \mathrm{pm}=1\times10^{-12}\ \mathrm{m}\)
\(0\ ^\circ\mathrm{C}=273.15\ \mathrm{K}\)
\(1\ \mathrm{kWh}=3600\ \mathrm{kJ}\)
\(1\ \mathrm{W}=1\ \mathrm{J\,s^{-1}}\)
\(1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm{J}\)
\(1\ \mathrm{bar}=10^5\ \mathrm{Pa}\)
area of triangle \(A=\frac12 ab\cdot\sin C\)
volume of cube \(V=a^3\)
volume of sphere \(V=\frac43\pi r^3\)
surface area of sphere \(S=4\pi r^2\)
ideal gas equation \(pV=nRT\)
frequency of light \(\nu=\frac{c}{\lambda}\)
energy of a photon \(E=\frac{hc}{\lambda}\)
Beer-Lambert law \(A=\log\left(\frac{I_0}{I}\right)=\varepsilon cd\)
spin-only magnetic moment \(\mu=\sqrt{n(n+2)}\ \mathrm{BM}\)
Henderson–Hasselbalch equation \(\mathrm{pH}=\mathrm{p}K_a+\log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)\)
\(0^{\mathrm{th}}\) order integrated rate law \([A]=[A]_0-kt\)
\(1^{\mathrm{st}}\) order integrated rate law \(\ln[A]=\ln[A]_0-kt\)
Arrhenius equation \(k=A\cdot e^{-E_a/(RT)}\)
enthalpy \(H=U+pV\)
Gibbs energy \(G=H-TS\)
standard Gibbs energy change \(\Delta_rG^\circ=-RT\ln K=-nFE^\circ_{\mathrm{cell}}\)
Clausius-Clapeyron equation \(\ln\left(\frac{P_1}{P_2}\right)=\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\)
specific heat \(Q=mC_p\Delta T\)
boiling point elevation \(\Delta T=iK_bm_i\)
freezing point depression \(\Delta T=iK_fm_i\)

Consider all gases ideal unless told otherwise.

In equilibrium constant calculations, all concentrations are referenced to a concentration of \(1\ \mathrm{mol\,dm^{-3}}\).

Periodic table

Figure 3

Official English text of the problem.

Answers

  • 4.1: \(m=0.133\ \mathrm{g}\), or \(0.13\ \mathrm{g}\) to two significant figures.

  • 4.2: A: \(\ce{NaNO2}\), sodium nitrite, [Na+].[O-]N=O; B: \(\ce{NH4Cl}\), ammonium chloride, [NH4+].[Cl-]; C: \(\ce{N2}\), nitrogen, N#N; D: \(\ce{NO}\), nitrogen monoxide (nitric oxide), [N]=O.

  • 4.3: Balanced equations, using net ionic forms where appropriate: \[ \begin{aligned} (1)\quad&\ce{NaNO2 + NH4Cl ->[\Delta] N2 + NaCl + 2H2O}\\ (2)\quad&\ce{Ag+ + Cl- -> AgCl(s)}\\ (3)\quad&\ce{AgCl(s) + 2NH4+ + 2OH- -> [Ag(NH3)2]+ + Cl- + 2H2O}\\ (4)\quad&\ce{2NO2- + 4H+ + 2I- -> 2NO(g) + I2 + 2H2O}\\ (5)\quad&\ce{I2 + 2S2O3^2- -> 2I- + S4O6^2-} \end{aligned} \] Names and SMILES for the additional species are supplied in the solution.

  • 4.4a: First order in excess pressure \(p-p_{\mathrm{atm}}\), with \[ \boxed{k_{\ce{N2}}=2.88\times10^{-3}\ \mathrm{h^{-1}}}. \]

  • 4.4b: \[ \boxed{k_{\ce{O2}}=4.31\times10^{-3}\ \mathrm{h^{-1}}}. \]

  • 4.4c: \[ \boxed{p_{\mathrm{start}}=1.79\ \mathrm{atm}}. \]

Solution

4.1

To distinguish the ball radius from the gas constant, write the radius as \(r=0.030\ \mathrm{m}\). The internal volume is

\[ V=\frac43\pi r^3 =\frac43\pi(0.030\ \mathrm{m})^3 =1.13097\times10^{-4}\ \mathrm{m^3}. \]

At \(T_0=298.15\ \mathrm{K}\), the amount of air is

\[ n_{\mathrm{air}} =\frac{p_{\mathrm{atm}}V}{RT_0} =\frac{101325(1.13097\times10^{-4})} {8.314(298.15)} =4.62300\times10^{-3}\ \mathrm{mol}. \]

For ideal gases, volume fractions equal mole fractions. The components are oxygen, \(\ce{O2}\), O=O, and nitrogen, \(\ce{N2}\), N#N. Thus

\[ \overline M =0.20(2\times16.00)+0.80(2\times14.01) =28.816\ \mathrm{g\,mol^{-1}}. \]

Therefore,

\[ m=n_{\mathrm{air}}\overline M =(4.62300\times10^{-3})(28.816) =0.133216\ \mathrm{g}. \]

\[ \boxed{m\approx0.133\ \mathrm{g}\approx0.13\ \mathrm{g}\text{ to two significant figures}.} \]

4.2

The additional pressure comes from gas C. Because the volume and final temperature are unchanged,

\[ n_C =\frac{(p-p_{\mathrm{atm}})V}{RT_0} =0.600\,n_{\mathrm{air}} =2.77380\times10^{-3}\ \mathrm{mol}. \]

The qualitative observations identify the reactants as follows.

The white precipitate from the final residue is consistent with silver chloride, \(\ce{AgCl}\). Its dissolution upon increasing the pH in a solution of B indicates ammonia complex formation: raising the pH converts ammonium ions into ammonia, which dissolves silver chloride as the diamminesilver(I) ion. This points to ammonium chloride as B.

The other reactant must generate a gas with ammonium ions on heating and oxidize iodide to iodine in acid. Nitrite satisfies both requirements:

\[ \ce{NH4+ + NO2- ->[\Delta] N2 + 2H2O}, \]

while acidified nitrite oxidizes iodide and is reduced to nitrogen monoxide. Consequently, C is nitrogen and D is nitrogen monoxide.

Their molar masses are

\[ M_C=28.02\ \mathrm{g\,mol^{-1}}, \qquad M_D=30.01\ \mathrm{g\,mol^{-1}}. \]

At equal temperature and pressure, gas density is proportional to molar mass, so

\[ \frac{\rho_D-\rho_C}{\rho_C} =\frac{30.01-28.02}{28.02} =0.0710=7.10\%, \]

which satisfies the stated limit.

The silver-precipitation data identify the cation of A. Before heating, both chloride and nitrite precipitate:

\[ \ce{Ag+ + Cl- -> AgCl(s)}, \qquad \ce{Ag+ + NO2- -> AgNO2(s)}. \]

For a 1:1 mixture of a monovalent metal nitrite and ammonium chloride, each mole of reaction mixture gives one mole of each precipitate. Their molar masses are

\[ M(\ce{AgCl})=107.87+35.45=143.32\ \mathrm{g\,mol^{-1}}, \]

\[ M(\ce{AgNO2})=107.87+14.01+32.00 =153.88\ \mathrm{g\,mol^{-1}}. \]

Thus the initial precipitate corresponds to

\[ n=\frac{0.8224}{143.32+153.88} =2.76716\times10^{-3}\ \mathrm{mol}. \]

Since

\[ M(\ce{NH4Cl})=14.01+4(1.008)+35.45 =53.492\ \mathrm{g\,mol^{-1}}, \]

the inferred molar mass of A is

\[ M_A=\frac{0.3389}{2.76716\times10^{-3}}-53.492 =68.980\ \mathrm{g\,mol^{-1}}. \]

Subtracting the nitrite contribution gives the cation mass:

\[ 68.980-(14.01+32.00) =22.970\ \mathrm{g\,mol^{-1}}, \]

matching sodium, \(22.99\ \mathrm{g\,mol^{-1}}\). Hence A is sodium nitrite.

The complete identification is therefore:

Compound Formula Name SMILES
A \(\ce{NaNO2}\) Sodium nitrite [Na+].[O-]N=O
B \(\ce{NH4Cl}\) Ammonium chloride [NH4+].[Cl-]
C \(\ce{N2}\) Nitrogen N#N
D \(\ce{NO}\) Nitrogen monoxide (nitric oxide) [N]=O

A direct check using the proposed mixture gives

\[ n_{\mathrm{reaction}} =\frac{0.3389}{69.00+53.492} =2.76671\times10^{-3}\ \mathrm{mol}. \]

This predicts

\[ m_{\mathrm{initial\ precipitate}} =n_{\mathrm{reaction}}(143.32+153.88) =0.82227\ \mathrm{g}, \]

in agreement with \(0.8224\ \mathrm{g}\). After heating, only the chloride precipitate remains:

\[ m_{\mathrm{reaction\ 2}} =n_{\mathrm{reaction}}(143.32) =0.39653\ \mathrm{g}. \]

Their ratio is

\[ \frac{0.82227}{0.39653}=2.0737, \]

consistent with “approximately twice.” The predicted final pressure is \(1.5985\ \mathrm{atm}\), consistent with the stated \(1.600\ \mathrm{atm}\) given the two-significant-figure radius.

4.3

Reaction 1: thermal production of nitrogen

\[ \boxed{\ce{NaNO2 + NH4Cl ->[\Delta] N2(g) + NaCl + 2H2O}} \]

Nitrogen in ammonium has oxidation state \(-3\), while nitrogen in nitrite has oxidation state \(+3\). Both become nitrogen in oxidation state zero. At the final temperature, water is treated as a condensed product under the problem's stated model.

Reaction 2: precipitation of silver chloride

\[ \boxed{\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}} \]

The sodium chloride residue supplies the chloride ions.

Reaction 3: dissolution through ammonia complex formation

Increasing the pH releases ammonia:

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

Ammonia binds silver(I), shifting the dissolution equilibrium:

\[ \boxed{\ce{AgCl(s) + 2NH3(aq) <=> [Ag(NH3)2]+(aq) + Cl-(aq)}}. \]

Combining these equations gives an equivalent overall equation that explicitly includes B's ammonium ions and the added base:

\[ \ce{AgCl(s) + 2NH4+(aq) + 2OH-(aq) -> [Ag(NH3)2]+(aq) + Cl-(aq) + 2H2O(l)}. \]

Reaction 4: oxidation of iodide by acidified nitrite

Nitrite is reduced from nitrogen oxidation state \(+3\) to \(+2\), while iodide is oxidized to iodine:

\[ \ce{NO2- + 2H+ + e- -> NO + H2O}, \]

\[ \ce{2I- -> I2 + 2e-}. \]

Adding the half-reactions gives

\[ \boxed{\ce{2NO2-(aq) + 4H+(aq) + 2I-(aq) -> 2NO(g) + I2(aq) + 2H2O(l)}}. \]

The liberated iodine accounts for the brown colour.

Reaction 5: reduction of iodine by thiosulfate

\[ \boxed{\ce{I2(aq) + 2S2O3^2-(aq) -> 2I-(aq) + S4O6^2-(aq)}}. \]

Iodine is reduced to iodide, while thiosulfate is oxidized to tetrathionate, removing the brown colour.

The additional species used above have the following names and SMILES:

Formula Name SMILES
\(\ce{NaCl}\) Sodium chloride [Na+].[Cl-]
\(\ce{H2O}\) Water O
\(\ce{Ag+}\) Silver(I) ion [Ag+]
\(\ce{Cl-}\) Chloride ion [Cl-]
\(\ce{AgCl}\) Silver(I) chloride [Ag+].[Cl-]
\(\ce{AgNO2}\) Silver(I) nitrite [Ag+].[O-]N=O
\(\ce{NH4+}\) Ammonium ion [NH4+]
\(\ce{OH-}\) Hydroxide ion [OH-]
\(\ce{NH3}\) Ammonia N
\(\ce{[Ag(NH3)2]+}\) Diamminesilver(I) ion [NH3]->[Ag+]<-[NH3]
\(\ce{NO2-}\) Nitrite ion [O-]N=O
\(\ce{H+}\) Hydrogen ion, proton [H+]
\(\ce{I-}\) Iodide ion [I-]
\(\ce{I2}\) Iodine II
\(\ce{S2O3^2-}\) Thiosulfate ion [O-]S(=O)(=O)[S-]
\(\ce{S4O6^2-}\) Tetrathionate ion [O-]S(=O)(=O)SSS(=O)(=O)[O-]

Dot-separated salt SMILES specify constituent ions rather than a crystal lattice. The dative-bond SMILES for diamminesilver(I) does not specify its approximately linear \(\ce{N-Ag-N}\) coordination geometry.

4.4a

The pressure approaches atmospheric pressure. The quantity that decays to zero is therefore the excess pressure

\[ q=p-p_{\mathrm{atm}}. \]

For first-order depressurisation,

\[ -\frac{dq}{dt}=k_{\ce{N2}}q, \qquad q(t)=q_0e^{-k_{\ce{N2}}t}. \]

Hence

\[ k_{\ce{N2}} =\frac1t\ln\left(\frac{p_0-p_{\mathrm{atm}}} {p(t)-p_{\mathrm{atm}}}\right). \]

Using the first measurement,

\[ k_{\ce{N2}} =\frac1{241\ \mathrm{h}} \ln\left(\frac{1.80-1.00}{1.40-1.00}\right) =\frac{\ln2}{241\ \mathrm{h}} =2.87613\times10^{-3}\ \mathrm{h^{-1}}. \]

Using the second measurement, with \(21\ \mathrm{days}=504\ \mathrm{h}\),

\[ k_{\ce{N2}} =\frac1{504\ \mathrm{h}} \ln\left(\frac{1.80-1.00}{1.19-1.00}\right) =2.85236\times10^{-3}\ \mathrm{h^{-1}}. \]

These values agree within about \(0.8\%\), consistent with the approximate second time and rounded pressures. In particular, the first value predicts

\[ p(504\ \mathrm{h}) =1.00+0.80e^{-(2.87613\times10^{-3})(504)} =1.18774\ \mathrm{atm}, \]

which rounds to the stated \(1.19\ \mathrm{atm}\).

Thus the data support first-order decay of excess pressure, with

\[ \boxed{k_{\ce{N2}}=2.88\times10^{-3}\ \mathrm{h^{-1}}}. \]

4.4b

At the same initial excess pressure \(q_0\), the premium ball has initial rate

\[ v_{\mathrm{premium},0}=k_{\ce{N2}}q_0. \]

For the stated additive-rate model, the initial rate for air is the composition-weighted sum:

\[ v_{\mathrm{air},0} =\left(0.80k_{\ce{N2}}+0.20k_{\ce{O2}}\right)q_0. \]

The air-filled ball initially depressurises 10% faster, so

\[ 0.80k_{\ce{N2}}+0.20k_{\ce{O2}} =1.10k_{\ce{N2}}. \]

Therefore,

\[ k_{\ce{O2}} =\frac{1.10-0.80}{0.20}k_{\ce{N2}} =1.50k_{\ce{N2}}, \]

and

\[ \boxed{k_{\ce{O2}} =1.50(2.87613\times10^{-3}) =4.31\times10^{-3}\ \mathrm{h^{-1}}}. \]

4.4c

First, calculate the rate constant at \(30.0\ ^\circ\mathrm{C}\):

\[ T_0=298.15\ \mathrm{K}, \qquad T_W=303.15\ \mathrm{K}. \]

The Arrhenius equation gives

\[ \ln\left(\frac{k_W}{k_0}\right) =\frac{E_A}{R} \left(\frac1{T_0}-\frac1{T_W}\right). \]

Therefore,

\[ \frac{k_W}{k_0} =\exp\left[ \frac{50000}{8.314} \left(\frac1{298.15}-\frac1{303.15}\right) \right] =1.39471, \]

\[ k_W =(2.87613\times10^{-3})(1.39471) =4.01137\times10^{-3}\ \mathrm{h^{-1}}. \]

The rapid temperature change also raises the pressure immediately. At constant volume and before appreciable gas loss,

\[ p(0^+) =p_0\frac{T_W}{T_0} =1.80\frac{303.15}{298.15} =1.83019\ \mathrm{atm}. \]

The resulting excess pressure then decays for \(12.0\ \mathrm{h}\):

\[ p_{\mathrm{start}} =p_{\mathrm{atm}} +\left[p(0^+)-p_{\mathrm{atm}}\right] e^{-k_Wt_{\mathrm{match}}}. \]

Substitution gives

\[ p_{\mathrm{start}} =1.000 +(1.83019-1.000) e^{-(4.01137\times10^{-3})(12.0)} =1.79117\ \mathrm{atm}. \]

\[ \boxed{p_{\mathrm{start}}=1.79\ \mathrm{atm}}. \]

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.