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[
{
"name": "IChO-2025_4",
"background": {
"image": "images/4/4_bg.png",
"context": "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 cm that remains constant throughout the pressurisation process. Assume that the composition of air is 20% O_{2} and 80% N_{2} by volume. Consider that overpressure does not expand the ball."
},
"points": 40
},
{
"name": "IChO-2025_4.1",
"modality": "text",
"type": "Quantitative Calculation",
"evaluation": "Numeric Verification",
"points": 3,
"field": "",
"source": "IChO-2025",
"question": {
"context": "Calculate the mass m (in g) of air inside an old tennis ball at T_{0} = 25°C."
},
"answer": [
{
"step": 1,
"content": "The inner ball pressure is equal to the atmospheric pressure. The average molar mass of air, $M = 0.8 \\times 28.02 \\text{ g mol}^{-1} + 0.2 \\times 32.00 \\text{ g mol}^{-1} = 28.816 \\text{ g mol}^{-1}$",
"points": 1
},
{
"step": 2,
"content": "The ball volume is: $V = \\frac{4}{3}\\pi R^3 = \\frac{4}{3} \\cdot \\pi \\cdot (3.0)^3 = 113.1 \\text{ mL}$",
"points": 1
},
{
"step": 3,
"content": "Using the ideal gas law, we find the air mass as follows: $p = \\frac{nRT}{V} = \\frac{mRT}{MV} \\Rightarrow m = \\frac{pMV}{RT} = \\frac{101325 \\text{ Pa} \\times 28.816 \\text{ g mol}^{-1} \\times 1.131 \\times 10^{-4} \\text{ m}^3}{8.314 \\text{ J K}^{-1} \\text{ mol}^{-1} \\times 298.15 \\text{ K}} = 0.133 \\text{ g}$",
"points": 1
}
]
},
{
"name": "IChO-2025_4.2-3",
"modality": "image + text",
"type": "Mechanistic Reasoning",
"evaluation": "Reasoning Consistency",
"points": 21,
"field": "",
"source": "IChO-2025",
"question": {
"context": "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. A capsule with a quantity m_{mix} = 0.3389 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 atm (at 25°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_{precipitate} = 0.8224 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 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%. Write the chemical formulae of compounds A-D. Write equations for reactions 1-5.",
"image": "images/4/4.2_ques.png",
"requirement": "Compounds A-D must be identified through step-by-step reasoning with calculations and logical deductions. Simply stating the correct formulae without showing the reasoning process will not earn points. Reactions 1-5 must be written as complete, balanced chemical equations.",
"parsing_note": "The reaction diagram has been converted to reaction_scheme. 'unknown' indicates unspecified reactants or products in the reaction.",
"reaction_scheme": [
{
"reaction_number": 1,
"reactant_name": "A + B",
"conditions": "Δ",
"product_name": "C, unknown"
},
{
"reaction_number": 2,
"reactant_name": "unknown",
"conditions": "AgNO_{3(aq)}",
"products": "unknown"
},
{
"reaction_number": 3,
"reactant_name": "unknown",
"conditions": "B_{(aq)}, pH>7",
"products": "unknown"
},
{
"reaction_number": 4,
"reactant_name": "A + B",
"conditions": "HI_{(aq)}",
"product_name": "D, unknown"
},
{
"reaction_number": 5,
"reactant_name": "unknow",
"conditions": "Na_{2}S_{2}O_{3(aq)}",
"products": "unknown"
},
{
"reactant_name": "A + B",
"conditions": "AgNO_{3(aq)}",
"products": "unknown"
}
]
},
"answer": [
{
"step": 1,
"content": "We can find the quantity of gas C: $n = \\frac{pV}{RT} = \\frac{(161975-101325) \\text{ Pa} \\times \\frac{4}{3}\\pi(0.03 \\text{ m})^3}{8.314 \\text{ J K}^{-1} \\text{ mol}^{-1} \\times 298.15 \\text{ K}} = 2.767 \\times 10^{-3} \\text{ mol}$",
"points": 1
},
{
"step": 2,
"content": "The reaction with HI indicates a possible redox reaction with formation of $\\text{I}_2$ (which is confirmed by the positive result in the further reaction with $\\text{Na}_2\\text{S}_2\\text{O}_3$). Thus, A or B is an oxidant. If $\\text{Ag}^+$ is added to A+B there is the formation of precipitate(s) (likely containing Ag(I)) but we don't know if it is one or several compounds. We note that reaction 2 gives approximately half the amount of precipitate as the final reaction. This could mean that the precipitate from the final reaction is a mixture of two precipitates, one of which is not formed in reaction 2.",
"points": 0
},
{
"step": 3,
"content": "We might first guess an equimolar amount of each precipitate from reaction 2 and that the number of moles of each precipitate is the same the number of moles of gas C, $(2.767 \\times 10^{-3} \\text{ mol})$. If this is the case, we can calculate a molar mass for the precipitate: $M = \\frac{0.4112 \\text{ g}}{2.767 \\times 10^{-3} \\text{ mol}} = 148.61 \\text{ g mol}^{-1}$",
"points": 1
},
{
"step": 4,
"content": "Assuming the cation is Ag(I) (= 107.87 g mol$^{-1}$), this leaves an anion of 40.74 g mol$^{-1}$.",
"points": 1
},
{
"step": 5,
"content": "Remembering that this mass was only approximate, this is consistent with the anionic part being a chloride ion (= 35.45 g mol$^{-1}$) and the precipitate being AgCl. Upon addition of excess chloride, AgCl redissolves and thus this is consistent with B containing chloride ions.",
"points": 2
},
{
"step": 6,
"content": "Indeed, there is several qualitative indications that the precipitate could be AgCl. Now we know accurately the molar mass of the first precipitate, we can calculate the accurate molar mass of the second precipitate (= 153.90 g mol$^{-1}$). Assuming the cation is Ag(I), the anionic part has a molar mass of 46.03 g mol$^{-1}$.",
"points": 2
},
{
"step": 7,
"content": "There are no single atoms that fit this charge and mass combination. If we consider atoms of common elements then we can deduce the anion is nitrite (46.01 g mol$^{-1}$) and the precipitate as $\\text{AgNO}_2$.",
"points": 2
},
{
"step": 8,
"content": "Hence, A is a nitrite salt and B is a chloride salt. Again, taking equimolar ratios as a first guess, we can calculate the sum of the cation molar masses in A+B, $(122.48 - 35.45 - 46.01)$ g mol$^{-1}$ = 41.02 g mol$^{-1}$.",
"points": 1
},
{
"step": 9,
"content": "As we know A is an oxidant due to the nitrite, reaction 1 is likely a redox reaction. However, it is the cation in B that is likely oxidised (as oxidation of chloride is difficult). One possibility would be the ammonium cation, and this also has a low molar mass.",
"points": 2
},
{
"step": 10,
"content": "The remaining molar mass $(41.02 - (14.01+4 \\times 1.008))$ g mol$^{-1}$ = 22.98 g mol$^{-1}$ is consistent with the second cation being $\\text{Na}^+$, so this is a possible answer.",
"points": 1
},
{
"step": 11,
"content": "$\\text{NaNO}_2 + \\text{NH}_4\\text{Cl} \\rightarrow \\text{N}_2 + \\text{NaCl} + 2\\text{H}_2\\text{O}$. We then conclude that C would be $\\text{N}_2$.",
"points": 2
},
{
"step": 12,
"content": "If nitrite acts as an oxidant in reaction 4, then gas D has contains a nitrogen atom in an oxidation state of lower than +3. Examples include NO, $\\text{N}_2\\text{O}$ or $\\text{NH}_3$, but only NO has a density similar to $\\text{N}_2$.",
"points": 2
},
{
"step": 13,
"content": "The reaction 3 helps us understand that compound B is ammonium salt: $\\text{NH}_4\\text{Cl}$ thus A is $\\text{NaNO}_2$",
"points": 0
},
{
"reaction_number": 1,
"reactant_name": "NaNO_{2} + NH_{4}Cl",
"reactant_formula": "NaNO_{2} + NH_{4}Cl",
"conditions": "Δ",
"product_name": "N_{2} + NaCl + H_{2}O",
"product_formula": "N_{2} + NaCl + 2H_{2}O",
"balanced_equation": "$\\text{NaNO}_{2} + \\text{NH}_{4}\\text{Cl} \\rightarrow \\text{N}_{2} + \\text{NaCl} + 2\\text{H}_{2}\\text{O}$",
"points": 1,
"grading": {
"partial_credit": [
{
"condition": "If not balanced correctly",
"deduction": -0.5
}
]
}
},
{
"reaction_number": 2,
"reactant_name": "Ag^{+} + Cl^{-}",
"reactant_formula": "Ag^{+} + Cl^{-}",
"conditions": "AgNO_{3(aq)}",
"product_name": "AgCl",
"product_formula": "AgCl",
"balanced_equation": "$\\text{Ag}^{+} + \\text{Cl}^{-} \\rightarrow \\text{AgCl}$",
"points": 1,
"grading": {
"partial_credit": [
{
"condition": "If not balanced correctly",
"deduction": -0.5
}
]
}
},
{
"reaction_number": 3,
"reactant_name": "AgCl + NH_{4}^{+} + OH^{-}",
"reactant_formula": "AgCl + 2NH_{4}^{+} + 2OH^{-}",
"conditions": "pH>7",
"product_name": "Ag(NH_{3})_{2}^{+} + H_{2}O + Cl^{-}",
"product_formula": "Ag(NH_{3})_{2}^{+} + 2H_{2}O + Cl^{-}",
"balanced_equation": "$\\text{AgCl} + 2\\text{NH}_{4}^{+} + 2\\text{OH}^{-} \\rightarrow \\text{Ag}(\\text{NH}_{3})_{2}^{+} + 2\\text{H}_{2}\\text{O} + \\text{Cl}^{-}$",
"points": 1,
"grading": {
"partial_credit": [
{
"condition": "If not balanced correctly",
"deduction": -0.5
}
]
}
},
{
"reaction_number": 4,
"reactant_name": "NaNO_{2} + HI",
"reactant_formula": "2NaNO_{2} + 4HI",
"conditions": "HI_{(aq)}",
"product_name": "NO + I_{2} + NaI + H_{2}O",
"product_formula": "2NO + I_{2} + 2NaI + 2H_{2}O",
"balanced_equation": "$2\\text{NaNO}_{2} + 4\\text{HI} \\rightarrow 2\\text{NO} + \\text{I}_{2} + 2\\text{NaI} + 2\\text{H}_{2}\\text{O}$",
"points": 1,
"grading": {
"partial_credit": [
{
"condition": "If not balanced correctly",
"deduction": -0.5
}
]
}
},
{
"reaction_number": 5,
"reactant_name": "Na_{2}S_{2}O_{3} + I_{2}",
"reactant_formula": "2Na_{2}S_{2}O_{3} + I_{2}",
"conditions": "Na_{2}S_{2}O_{3(aq)}",
"product_name": "Na_{2}S_{4}O_{6} + NaI",
"product_formula": "Na_{2}S_{4}O_{6} + 2NaI",
"balanced_equation": "$2\\text{Na}_{2}\\text{S}_{2}\\text{O}_{3} + \\text{I}_{2} \\rightarrow \\text{Na}_{2}\\text{S}_{4}\\text{O}_{6} + 2\\text{NaI}$",
"points": 1,
"grading": {
"partial_credit": [
{
"condition": "If not balanced correctly",
"deduction": -0.5
}
]
}
}
]
},
{
"name": "IChO-2025_4.4",
"modality": "text",
"type": "Quantitative Calculation",
"evaluation": "Numeric Verification",
"points": 16,
"field": "",
"source": "IChO-2025",
"question": {
"context": "The pressure inside modern tennis balls is p_{0} = 1.80 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 N_{2}. 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 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°C, once the can is opened, the premium balls depressurise to p_{1} = 1.40 atm after t_{1} = 241 h and to p_{2} = 1.19 atm after about t_{2} = 21 days. The initial depressurisation rate of regular balls is 10% faster than premium balls. Demonstrate that the depressurisation process follows the first kinetic order by calculating the depressurisation rate constant k_{N_2} for N_{2} in h^{-1}. Note: If you couldn't calculate the answer, use $k_{N_2} = 10^{-3} h^{-1}$ in further calculations. Calculate the (depressurisation) rate constant k_{O_{2}} for O_{2} in 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_{match} = 12.0 h from now. The weather today in Wimbledon is predicted to be a constant T_{W} = 30.0 °C. Assume a new set of premium balls, which were manufactured and placed in the can at T_{0} = 25.0 °C, are opened now and quickly change to T_{W} = 30.0 °C. Calculate the pressure p_{start} (in atm) of the balls at the start of the match."
},
"answer": [
{
"step": 1,
"content": "Atmospheric pressure and therefore the final pressure is 1.00 atm. The pressure above atmospheric is therefore 0.80 atm initially, 0.40 atm after 241 h, and 0.19 atm after $t_2 = 21$ days (i.e. 504 hours). We remind that $t_2$ is approximate and using it we obtain an approximate data.",
"points": 0
},
{
"step": 2,
"content": "For the kinetics of the first order we have next equation: $\\ln \\frac{P_0}{P_t} = kt$. Thus we could write two equation to find dressurisation rate constant for $N_2$.",
"points": 0
},
{
"step": 3,
"content": "$\\ln \\frac{0.8}{0.4} = k_{N2} \\cdot 241 \\Rightarrow k_{N2} = \\frac{1}{241} \\cdot \\ln \\frac{0.8}{0.4} = 2.88 \\cdot 10^{-3} h^{-1}$",
"points": 1.5
},
{
"step": 4,
"content": "$\\ln \\frac{0.8}{0.19} = k_{N2} \\cdot 21 \\cdot 24 \\Rightarrow k_{N2} = \\frac{1}{504} \\cdot \\ln \\frac{0.8}{0.19} = 2.85 \\cdot 10^{-3} h^{-1}$",
"points": 1.5
},
{
"step": 5,
"content": "The rate constants are very close, thus the kinetics process is of the first order. Knowing that $t_2$ was approximate for the final version we use $k_{N2} = 2.88 \\cdot 10^{-3} h^{-1}$",
"points": 0
},
{
"step": 6,
"content": "Final rate constant: $k_{N2} = 2.88 \\cdot 10^{-3} h^{-1}$ or $k_{N2} = 2.85 \\cdot 10^{-3} h^{-1}$ or average of both constants",
"points": 1,
"grading": {
"partial_credit": [
{
"condition": "If only one equation is used",
"deduction": -0.5
}
]
}
},
{
"step": 7,
"content": "Next, we write equations for the initial processes for the $N_2$ ball ($v_{N2}$) and air ball ($v_{air}$) considering partial pressures and the initial difference of pressure $\\Delta P_0$: $v_{0\\,N_2} = k_{N_2}\\Delta P_0$",
"points": 1,
"grading": {
"partial_credit": [
{
"condition": "If the absolute pressure $P_0$ is considered",
"deduction": -0.25
}
]
}
},
{
"step": 8,
"content": "$v_{0\\,air} = k_{N_2} \\cdot 0.8\\Delta P_0 + k_{O_2} \\cdot 0.2\\Delta P_0$",
"points": 1,
"grading": {
"partial_credit": [
{
"condition": "If the absolute pressure considered",
"deduction": -0.25
}
]
}
},
{
"step": 9,
"content": "As we know that the second process is 10% faster than the first, we can find $k_{O2}$: $\\frac{v_{0\\,air}}{v_{0\\,N_2}} = \\frac{0.8 \\cdot k_{N_2}\\Delta P_0 + 0.2 \\cdot k_{O_2}\\Delta P_0}{k_{N_2}\\Delta P_0} = \\frac{1.10}{1.00}$",
"points": 1
},
{
"step": 10,
"content": "$0.3 \\cdot k_{N_2} = 0.2 \\cdot k_{O_2}$",
"points": 1
},
{
"step": 11,
"content": "$k_{O_2} = 1.5 \\cdot k_{N_2} = 4.31 \\times 10^{-3} \\text{ h}^{-1}$ or for $k_{O_2} = 1.5 \\times 10^{-3} \\text{ h}^{-1}$ the given value of $k_{N_2} = 1.0 \\times 10^{-3} \\text{ h}^{-1}$",
"points": 1
},
{
"step": 12,
"content": "We need to consider the following factors: (a) The initial inner ball pressure varies with the temperature (1 pt); (b) The final pressure is the same (1 atm) and taken into account (2 pts); (c) The rate constant varies with the temperature via Arrhenius equation (2 pts).",
"points": 0
},
{
"step": 13,
"content": "So next equations should be considered. (a) The initial pressure, $p_{0(T)}$, at another temperature, $T$, is $p_{0(T)} = p_{0(298.15)} \\cdot \\frac{T}{298.15 \\text{ K}}$",
"points": 1,
"grading": {
"partial_credit": [
{
"condition": "For correct calculation of factor (a)",
"deduction": -0.5
}
]
}
},
{
"step": 14,
"content": "(b) The pressure of the ball, $p$, over time, $t$, varies as $p = 1 + (p_0 - 1)e^{-kt}$",
"points": 2,
"grading": {
"partial_credit": [
{
"condition": "For correct factor (b)",
"deduction": -0.25
}
]
}
},
{
"step": 15,
"content": "(c) The variation of the rate constant with the temperature is: $k_{(T)} = k_{(298.15)}e^{\\left(\\frac{E_a}{R}\\left(\\frac{1}{298.15\\text{ K}} - \\frac{1}{T}\\right)\\right)}$",
"points": 2,
"grading": {
"partial_credit": [
{
"condition": "For correct factor (c)",
"deduction": -0.25
}
]
}
},
{
"step": 16,
"content": "to obtain the final equation: $p = 1 + \\left(p_0 \\cdot \\frac{T}{298.15\\text{ K}} - 1\\right)e^{-k \\cdot e^{\\left(\\frac{E_a}{R}\\left(\\frac{1}{298.15\\text{ K}} - \\frac{1}{T}\\right)\\right)} \\cdot t}$",
"points": 0
},
{
"step": 17,
"content": "$p = 1\\text{ atm} + \\left(1.80 \\cdot \\frac{302.15\\text{ K}}{298.15\\text{ K}} - 1\\right)\\text{ atm} \\cdot e^{-2.88 \\cdot 10^{-3}\\text{ h}^{-1} \\cdot e^{\\left(\\frac{50000\\text{ J mol}^{-1}}{8.314\\text{ J K}^{-1}\\text{ mol}^{-1}}\\left(\\frac{1}{298.15\\text{ K}} - \\frac{1}{302.15\\text{ K}}\\right)\\right)} \\cdot 12\\text{ h}} = 1.74\\text{ atm}$",
"points": 2,
"grading": {
"partial_credit": [
{
"condition": "For correct calculation",
"points": 0.5
}
],
"notes": "2 pt for the correct value (0.5 pts for correct calculation of factor (a), 0.75 pts for correct factor (b) and 0.75 pts for correct factor (c))."
}
}
]
}
]
|