IChO 2026 · Theory problem 4

The nuclear past of Uzbekistan

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T4. The nuclear past of Uzbekistan

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4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 \(\sum\)
2 3 3 2 2 2 2 2 4 22

Natural uranium consists primarily of two isotopes, \(\ce{^{235}U}\) and \(\ce{^{238}U}\). \(\ce{^{235}U}\) is responsible for sustaining nuclear fission reactions.

4.1 Calculate the atomic abundance of \(\ce{^{235}U}\) in natural uranium, assuming it consists only of isotopes \(\ce{^{235}U}\) (235.04 a.u.) and \(\ce{^{238}U}\) (238.05 a.u.). 2.0 pt

The main reaction occurring in nuclear power plants is the reaction where \(\ce{^{235}U}\) absorbs a neutron and undergoes fission, releasing energy and producing three additional neutrons, thus sustaining a chain reaction.

Figure 1

Neutrons that induce the chain reaction are produced by several ways:

a) Reaction of \(\ce{^{11}B}\) with \(\alpha\)-particles inside the reactor core

b) Interaction of \(\gamma\) rays with \(\ce{^{2}D}\) nuclei present in the reactor coolant water

c) Reaction of \(\ce{^{9}Be}\) with \(\alpha\)-particles produced in an externally installed neutron source.

4.2 Write the nuclear reaction equations for (a), (b), and (c). 3.0 pt

Figure 2

The graph above displays the percentage yield, \(P(A)\), of fission products by mass number, \(A\), resulting from \(\ce{^{235}U}\) fission.

4.3 Write the main nuclear fission equation for splitting of \(\ce{^{235}U}\) upon neutron absorption. The pair of elements formed in this reaction are in the same group of Periodic Table. 3.0 pt

4.4 Calculate the energy released in this reaction (\(\Delta E\), MeV), if the binding energy \(BE(\ce{^{235}U})=7.59\ \text{MeV/nucleon}\) and average binding energy for fission products \(BE(\text{fis.})=8.45\ \text{MeV/nucleon}\). Neglect the binding energy of free neutrons. 2.0 pt

Fission neutrons are produced with an average energy of 2 MeV and slow down through many scatterings with nuclei. After several collisions, their speed decreases until their average kinetic energy matches that of the surrounding atoms or molecules.

The logarithmic energy decrement, \(\xi\), shows the average decrease per collision in the logarithm of the neutron energy:

\[ \xi=\ln\left(\frac{E_{\mathit{initial}}}{E_{\mathit{final}}}\right) \]

\(E_{\mathit{initial}}\) and \(E_{\mathit{final}}\) are the initial and final neutron energies per collision.

\(\xi\) is a constant for each type of material and does not depend on the initial neutron energy.

4.5 Determine the average number of collisions required, \(n_c\), to slow a neutron from 2 MeV to 0.012 eV, using water as the moderator. \(\xi\) (water)=0.948 2.0 pt

In 1963, the Urtabulak gas field suffered an uncontrolled natural gas well blowout, releasing enormous volumes of burning methane.

4.6 Calculate \(\Delta_rH^\circ_{298}\) (\(\mathrm{kJ\,mol^{-1}}\)) for one mole of methane combustion reaction at 298 K. Assume all species are gaseous. Use the thermodynamic data below 2.0 pt

\(\Delta_fH^\circ_{298}(\ce{CH4})\) \(-74.8\ \mathrm{kJ\,mol^{-1}}\)
\(\Delta_fH^\circ_{298}(\ce{H2O},\text{ gas})\) \(-241.8\ \mathrm{kJ\,mol^{-1}}\)
\(\Delta_fH^\circ_{298}(\ce{CO2})\) \(-393.5\ \mathrm{kJ\,mol^{-1}}\)
\(C_P(\ce{CH4})\) \(35\ \mathrm{J\,mol^{-1}\,K^{-1}}\)
\(C_P(\ce{H2O},\text{ gas})\) \(34\ \mathrm{J\,mol^{-1}\,K^{-1}}\)
\(C_P(\ce{O2})\) \(29\ \mathrm{J\,mol^{-1}\,K^{-1}}\)
\(C_P(\ce{CO2})\) \(37\ \mathrm{J\,mol^{-1}\,K^{-1}}\)

If you did not get an answer for 4.6, use \(\Delta_rH^\circ_{298}=-750\ \mathrm{kJ\,mol^{-1}}\) for further calculations.

4.7 Calculate \(\Delta_rH_{2000}\) (\(\mathrm{kJ\,mol^{-1}}\)) per mole of methane combustion reaction at 2000 K. Assume all species are gaseous. 2.0 pt

If you did not get an answer for 4.7, use \(\Delta_rH_{2000}=-700\ \mathrm{kJ\,mol^{-1}}\) for further calculations.

Assume that methane is flowing out the well at 101.325 kPa and 298 K with volumetric flow \(Q=2.2\times10^5\ \mathrm{m^3}\) per day before it catches fire.

4.8 Calculate the total energy released per day, \(E\), (in \(\mathrm{J\,day^{-1}}\)) by complete isothermic combustion of methane at \(T=2000\ \mathrm K\) 2.0 pt

The Urtabulak gas leak was ended by detonating a 30-kiloton in TNT equivalent underground nuclear explosion that collapsed the reservoir and sealed the leak. 1 ton of TNT equivalent is 4.184 GJ of energy.

4.9 Using the value from 4.4, calculate the total number of fissions, \(TN\), during the nuclear explosion. Calculate the mass, \(m\), of enriched uranium used in the explosion (in kg), assuming it contained 90% by mass of the \(\ce{^{235}U}\) isotope, and 33% of the \(\ce{^{235}U}\) underwent fission.
If you did not get an answer for 4.4, use \(\Delta E=200\ \mathrm{MeV}\). 4.0 pt

Physical Constants and Equations

Constants

Avogadro constant \(N_A=6.022\times10^{23}\ \mathrm{mol^{-1}}\)
Universal gas constant \(R=8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}\)
Faraday constant \(F=96\,485\ \mathrm{C\,mol^{-1}}\)
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 \(e=1.602\times10^{-19}\ \mathrm C\)
Standard pressure \(P^\circ=1\ \mathrm{bar}=10^5\ \mathrm{Pa}\)
Ionic product of water at \(25\,^\circ\mathrm C\) \(K_w=1.00\times10^{-14}\)
Zero of the Celsius scale \(0\,^\circ\mathrm C=273.15\ \mathrm K\)
Electronvolt \(1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm J\)
Watt \(1\ \mathrm W=1\ \mathrm{J\,s^{-1}}\)
Coulomb \(1\ \mathrm C=1\ \mathrm{A\,s}\)
Nanometre \(1\ \mathrm{nm}=1\times10^{-9}\ \mathrm m\)
Ångstrom \(1\ \text{Å}=1\times10^{-10}\ \mathrm m\)
metric ton \(1\ \mathrm{ton}=1000\ \mathrm{kg}\)
litre \(1\ \mathrm L=1\ \mathrm{dm^3}\)

Equations

Ideal gas law \(PV=nRT\)
Dalton law \(P_i=x_iP\)
Beer–Lambert law \(A=\log_{10}\dfrac{I_0}{I}=\varepsilon cl\)
Energy of a photon \(E=h\nu=\dfrac{hc}{\lambda}\)
Heat and temperature change \(Q=nC_m\Delta T\)
Gibbs energy definition \(G=H-TS\)
Gibbs energy and equilibrium constant \(\Delta_rG^\circ=-RT\ln K\)
Enthalpy change dependence on temperature (at \(\Delta_rC_p=\mathit{const}\)) \(\Delta_rH_{T_2}=\Delta_rH_{T_1}+\Delta_rC_p(T_2-T_1)\)
Entropy change dependence on temperature (at \(\Delta_rC_p=\mathit{const}\)) \(\Delta_rS_{T_2}=\Delta_rS_{T_1}+\Delta_rC_p\ln\dfrac{T_2}{T_1}\)
Henderson–Hasselbalch equation \(\mathrm{pH}=\mathrm pK_a+\log_{10}\dfrac{[\ce{A-}]}{[\ce{HA}]}\)
Average reaction rate \(r=\dfrac{\Delta c}{\Delta t}\)
The rate law \(r=k[A]^x[B]^y\)
Arrhenius law \(k=A\exp\left(-\dfrac{E_a}{RT}\right)\)
Integrated rate law for a 1st order reaction \(c=c_0e^{-kt}\)
\(kt=\ln\dfrac{c_0}{c}\)
Lifetime of a reagent in a 1st order reaction \(\tau=\dfrac1k\)
Integrated rate law for n-th order reaction (\(n\ne1\)) \(kt=\dfrac1{n-1}\left(c^{1-n}-c_0^{1-n}\right)\) for \(n\ne1\)
Concentration of an intermediate in a consecutive reaction: \([B]=\dfrac{k_1[A]_0}{k_2-k_1}\left(e^{-k_1t}-e^{-k_2t}\right)\)
Figure 3
The ratio of products in the competitive reactions: \(\dfrac{[B]}{[C]}=\dfrac{k_1}{k_2}\)
Figure 4
Michaelis–Menten equation \(r_0=r_{\max}\dfrac{[S]_0}{K_M+[S]_0}\)
Specific surface area \(s=\dfrac Sm\)
Progressions: arithmetic geometric \(a_n=a_1+d(n-1)\)
\(a_n=a_1q^{n-1}\)
Sum of geometric progression \(\displaystyle\sum_{k=0}^{n}a_1q^k=a_1\dfrac{q^{n+1}-1}{q-1}\)
Volume of a sphere with radius \(r\) \(V=\dfrac43\pi r^3\)
Surface area of sphere with radius \(r\) \(S=4\pi r^2\)
Volume of a cone with base radius \(r\) and height \(h\) \(V=\dfrac13\pi r^2h\)
Volume of a cylinder with base radius \(r\) and height \(h\) \(V=\pi r^2h\)

Consider all gases ideal.

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

Periodic Table of Elements

Figure 5

Official English text of the problem.

Answers

  • 4.1: The atomic abundance of \(\ce{^{235}U}\) is \(\boxed{0.664\%\approx0.66\%}\).

  • 4.2: \[ \begin{aligned} \text{(a)}\quad&\ce{^{11}_{5}B + ^{4}_{2}He -> ^{14}_{7}N + ^{1}_{0}n}\\ \text{(b)}\quad&\ce{^{2}_{1}H + \gamma -> ^{1}_{1}H + ^{1}_{0}n}\\ \text{(c)}\quad&\ce{^{9}_{4}Be + ^{4}_{2}He -> ^{12}_{6}C + ^{1}_{0}n} \end{aligned} \]

  • 4.3: The fragments are rubidium-93 and cesium-140, both group 1 elements: \[ \boxed{\ce{^{235}_{92}U + ^{1}_{0}n -> ^{93}_{37}Rb + ^{140}_{55}Cs + 3 ^{1}_{0}n}} \]

  • 4.4: \(\boxed{\Delta E=185.2\ \mathrm{MeV}}\) per fission.

  • 4.5: \(\boxed{n_c=19.97\approx20\text{ collisions}}\).

  • 4.6: \(\boxed{\Delta_rH^\circ_{298}=-802.3\ \mathrm{kJ\,mol^{-1}}}\).

  • 4.7: \(\boxed{\Delta_rH_{2000}=-781.9\ \mathrm{kJ\,mol^{-1}}}\).

  • 4.8: \(\boxed{E\approx7.0\times10^{12}\ \mathrm{J\,day^{-1}}}\).

  • 4.9: \(\boxed{TN\approx4.23\times10^{24}\text{ fissions}}\); the enriched uranium mass is \(\boxed{m\approx5.6\ \mathrm{kg}}\).

Solution

4.1 — Atomic abundance

The supplied periodic table gives the average atomic mass of natural uranium as \(238.03\) a.u. Let \(x\) be the atomic fraction of \(\ce{^{235}U}\). The atomic fraction of \(\ce{^{238}U}\) is then \(1-x\), so

\[ 238.03=235.04x+238.05(1-x). \]

Therefore,

\[ x=\frac{238.05-238.03}{238.05-235.04} =\frac{0.02}{3.01} =0.0066445. \]

Thus, using the supplied rounded atomic masses,

\[ \boxed{100x=0.66445\%\approx0.66\%}. \]

4.2 — Neutron-producing reactions

In each nuclear reaction, both mass number \(A\) and atomic number \(Z\) are conserved. An \(\alpha\)-particle is a helium-4 nucleus, \(\ce{^{4}_{2}He}\).

(a) After emission of one neutron, the residual nucleus has

\[ A=11+4-1=14,\qquad Z=5+2=7. \]

Atomic number 7 identifies nitrogen, so the product is nitrogen-14:

\[ \boxed{\ce{^{11}_{5}B + ^{4}_{2}He -> ^{14}_{7}N + ^{1}_{0}n}}. \]

(b) A deuterium nucleus, \(\ce{^{2}_{1}H}\), contains one proton and one neutron. Absorption of a sufficiently energetic photon separates them:

\[ \boxed{\ce{^{2}_{1}H + \gamma -> ^{1}_{1}H + ^{1}_{0}n}}. \]

The hydrogen-1 nucleus is a proton.

(c) The residual nucleus has

\[ A=9+4-1=12,\qquad Z=4+2=6. \]

Atomic number 6 identifies carbon, giving carbon-12:

\[ \boxed{\ce{^{9}_{4}Be + ^{4}_{2}He -> ^{12}_{6}C + ^{1}_{0}n}}. \]

These nuclear equations specify nuclei; electron populations do not enter the balances.

4.3 — Identification of the fission fragments

The two maxima in the supplied graph occur at approximately

\[ A_{\text{light}}=93,\qquad A_{\text{heavy}}=140. \]

Their sum is \(233\), consistent with the stated emission of three neutrons:

\[ 235+1=93+140+3. \]

Conservation of atomic number requires

\[ Z_{\text{light}}+Z_{\text{heavy}}=92. \]

Rubidium and cesium satisfy the same-group condition: both are group 1 elements, and

\[ Z(\ce{Rb})+Z(\ce{Cs})=37+55=92. \]

Assigning rubidium to the lighter fragment and cesium to the heavier fragment gives

\[ \boxed{\ce{^{235}_{92}U + ^{1}_{0}n -> ^{93}_{37}Rb + ^{140}_{55}Cs + 3 ^{1}_{0}n}}. \]

Both mass number and atomic number are balanced.

4.4 — Energy released per fission

The energy released equals the increase in total nuclear binding energy.

Initially, the uranium nucleus contains \(235\) bound nucleons. The incoming neutron is free and contributes no binding energy under the stated approximation. Finally, the two fragments contain

\[ 93+140=233 \]

bound nucleons; the three emitted neutrons are free.

Consequently,

\[ \begin{aligned} \Delta E &=BE_{\text{products}}-BE_{\text{reactants}}\\ &=233(8.45)-235(7.59)\\ &=1968.85-1783.65\\ &=\boxed{185.2\ \mathrm{MeV}}. \end{aligned} \]

The positive value is the energy released per fission.

4.5 — Average number of collisions

The moderator is water, \(\ce{H2O}\), SMILES O.

Because the logarithmic energy decrements add over successive collisions,

\[ \ln\left(\frac{E_0}{E_f}\right)=n_c\xi. \]

Using the same energy unit in numerator and denominator,

\[ E_0=2\ \mathrm{MeV}=2\times10^6\ \mathrm{eV}. \]

Therefore,

\[ n_c= \frac{\ln\left(\dfrac{2\times10^6}{0.012}\right)}{0.948} =19.9699. \]

Thus,

\[ \boxed{n_c\approx20\text{ collisions}}. \]

4.6 — Methane combustion at 298 K

The molecular species are methane, \(\ce{CH4}\), SMILES C; dioxygen, \(\ce{O2}\), SMILES O=O; carbon dioxide, \(\ce{CO2}\), SMILES O=C=O; and water, \(\ce{H2O}\), SMILES O.

Complete combustion of one mole of methane is

\[ \ce{CH4(g) + 2 O2(g) -> CO2(g) + 2 H2O(g)}. \]

By Hess’s law,

\[ \Delta_rH^\circ_{298} =\sum_{\text{products}}\nu_i\Delta_fH^\circ_i -\sum_{\text{reactants}}\nu_i\Delta_fH^\circ_i. \]

The standard enthalpy of formation of \(\ce{O2(g)}\), an element in its reference state, is zero. Hence

\[ \begin{aligned} \Delta_rH^\circ_{298} &=\left[-393.5+2(-241.8)\right] -\left[-74.8+2(0)\right]\\ &=\boxed{-802.3\ \mathrm{kJ\,mol^{-1}}}. \end{aligned} \]

The negative sign indicates an exothermic reaction.

4.7 — Methane combustion at 2000 K

For the same balanced combustion reaction,

\[ \begin{aligned} \Delta_rC_P &=C_P(\ce{CO2})+2C_P(\ce{H2O}) -C_P(\ce{CH4})-2C_P(\ce{O2})\\ &=37+2(34)-35-2(29)\\ &=12\ \mathrm{J\,mol^{-1}\,K^{-1}}\\ &=0.012\ \mathrm{kJ\,mol^{-1}\,K^{-1}}. \end{aligned} \]

Treating the supplied heat capacities as constant, Kirchhoff’s law gives

\[ \begin{aligned} \Delta_rH_{2000} &=\Delta_rH^\circ_{298}+\Delta_rC_P(2000-298)\\ &=-802.3+0.012(1702)\\ &=-781.876\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

Thus,

\[ \boxed{\Delta_rH_{2000}\approx-781.9\ \mathrm{kJ\,mol^{-1}}}. \]

4.8 — Energy released per day

The volumetric flow is specified at \(298\ \mathrm K\), so that temperature must be used to convert the gas volume to an amount of methane.

From the ideal gas law, the molar flow is

\[ \begin{aligned} \dot n &=\frac{pQ}{RT}\\ &=\frac{ (101325\ \mathrm{Pa}) (2.2\times10^5\ \mathrm{m^3\,day^{-1}}) }{ (8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}) (298\ \mathrm K) }\\ &=8.9973\times10^6\ \mathrm{mol\,day^{-1}}. \end{aligned} \]

Here, \(\mathrm{Pa\,m^3}=\mathrm J\). The energy released is the positive magnitude of the combustion enthalpy:

\[ \begin{aligned} E &=-\dot n\,\Delta_rH_{2000}\\ &=(8.9973\times10^6\ \mathrm{mol\,day^{-1}}) (781876\ \mathrm{J\,mol^{-1}})\\ &=7.0348\times10^{12}\ \mathrm{J\,day^{-1}}. \end{aligned} \]

To two significant figures, consistent with the volumetric flow,

\[ \boxed{E=7.0\times10^{12}\ \mathrm{J\,day^{-1}}}. \]

4.9 — Number of fissions and uranium mass

The total energy represented by 30 kilotons of TNT is

\[ \begin{aligned} E_{\text{explosion}} &=(30\times10^3\ \mathrm{ton}) (4.184\times10^9\ \mathrm{J\,ton^{-1}})\\ &=1.2552\times10^{14}\ \mathrm J. \end{aligned} \]

Using the result from 4.4, the energy per fission in joules is

\[ \begin{aligned} \varepsilon_{\text{fission}} &=(185.2\ \mathrm{MeV}) (10^6\ \mathrm{eV\,MeV^{-1}}) (1.602\times10^{-19}\ \mathrm{J\,eV^{-1}})\\ &=2.966904\times10^{-11}\ \mathrm J. \end{aligned} \]

Therefore, the total number of fissions is

\[ TN=\frac{E_{\text{explosion}}}{\varepsilon_{\text{fission}}} =\frac{1.2552\times10^{14}}{2.966904\times10^{-11}} =\boxed{4.23\times10^{24}\text{ fissions}}. \]

Each fission consumes one \(\ce{^{235}U}\) nucleus. Thus, the mass of \(\ce{^{235}U}\) that actually underwent fission is

\[ \begin{aligned} m_{\text{fissioned}} &=\frac{TN}{N_A}M(\ce{^{235}U})\\ &=\frac{4.23067\times10^{24}} {6.022\times10^{23}\ \mathrm{mol^{-1}}} (235.04\ \mathrm{g\,mol^{-1}})\\ &=1651.24\ \mathrm g =1.65124\ \mathrm{kg}. \end{aligned} \]

If \(m\) is the initial mass of enriched uranium, its initial \(\ce{^{235}U}\) mass is \(0.90m\). Since \(33\%\) of that isotope underwent fission,

\[ m_{\text{fissioned}}=0.33(0.90m). \]

Hence

\[ m=\frac{1.65124}{0.33\times0.90} =5.5597\ \mathrm{kg}. \]

With sensible rounding,

\[ \boxed{m\approx5.6\ \mathrm{kg}}. \]

The solution was checked against the official solution.