IChO-2025_4.4
Type: Quantitative Calculation
Points: 16
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)).