What volume of gaseous (at STP) is needed to produce 23.739 kJ of useful work under standard conditi — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Relation between Gibbs Free Energy and Useful Work In thermodynamics, the maximum non-expansion work (or useful electrical work, $W_{\text{useful, max}}$) that can be obtained from a chemical reaction operating at constant temperature and pressure is directly equal to the negative of the Gibbs free energy change ($\Delta G^\circ$) of the process: $$W_{\text{useful, max}} = -\Delta G^\circ$$ For the overall fuel cell reaction described in the passage: $$\ce{2H2(g) + O2(g) -> 2H2O(l)}$$ The standard Gibbs free energy change ($\Delta G^\circ$) is given as: $$\Delta G^\circ = -237.39\text{ kJ}$$ This means that for the reaction of $2\text{ moles}$ of gaseous hydrogen ($\ce{H2}$) with $1\text{ mole}$ of oxygen to form $2\text{ moles}$ of liquid water, the maximum standard useful work obtained is: $$W_{\text{useful}} = 237.39\text{ kJ}$$ Step 2 - Calculate the Number of Moles of Hydrogen Gas Required To find the amount of hydrogen gas (in moles) required to produce exactly $23.739\text{ kJ}$ of useful work, we set up a stoichiometric ratio: $$\text{Moles of } \ce{H2} = \frac{\text{Required Work}}{\text{Total Work produced by } 2\text{ moles of } \ce{H2}} \times 2\text{ mol}$$ Substitute the given values into the formula: $$\text{Moles of } \ce{H2} = \frac{23.739\text{ kJ}}{237.39\text{ kJ}} \times 2\text{ mol}$$ $$\text{Moles of } \ce{H2} = 0.1 \times 2\text{ mol} = 0.2\text{ mol}$$ Step 3 - Determine the Volume of Gaseous Hydrogen at STP According to the modern IUPAC definition of Standard Temperature and Pressure (STP), where $T = 273.15\text{ K}$ and $P = 1\text{ bar}$, the molar volume ($V_m$) of an ideal gas is: $$V_m \approx 22.7\text{ L mol}^{-1}$$ The volume ($V$) of gaseous hydrogen required is calculated using the relation: $$V = n \times V_m$$ Substitute the moles of $\ce{H2}$ and the modern IUPAC molar volume: $$V = 0.2\text{ mol} \times 22.7\text{ L mol}^{-1} = \boxed{4.54\text{ L}}$$ *(Note: Under the classical definition of STP where standard pressure was taken as $1\text{ atm}$, the molar volume of an ideal gas is $22.4\text{ L mol}^{-1}$, which yields $0.2\text{ mol} \times 22.4\text{ L mol}^{-1} = 4.48\text{ L}$. Since $4.48\text{ L}$ is not among the choices and $4.54\text{ L}$ is explicitly provided, we use the IUPAC standard of $22.7\text{ L mol}^{-1}$.)* Step 4 - Evaluate and Explain the Options * **Option (A) is incorrect:** $2.27\text{ L}$ is obtained if we assume only $1\text{ mole}$ of hydrogen gas is consumed to produce $237.39\text{ kJ}$ of work, neglecting the stoichiometry of $2\text{ moles}$ of $\ce{H2}$ in the overall balanced equation. * **Option (B) is correct:** As mathematically shown, $0.2\text{ moles}$ of hydrogen gas corresponds to a volume of exactly $4.54\text{ L}$ under standard IUPAC STP conditions. * **Option (C) is incorrect:** $1.13\text{ L}$ is mathematically incorrect and represents a division error. * **Option (D) is incorrect:** $2.24\text{ L}$ represents $0.1\text{ moles}$ of gas under classical STP conditions, which would correspond to a stoichiometry error combined with old standard states. $$\text{Correct Option: } \boxed{\text{B}}$$