What is the approximate value of ΔS° for the fuel cell reaction at 25°C? — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Given Thermodynamic Parameters From the provided passage for the hydrogen-oxygen ($\ce{H2-O2}$) fuel cell reaction at $25^\circ\text{C}$ (which is $298.15\text{ K}$, commonly rounded to $298\text{ K}$): * Standard enthalpy change: $$\Delta H^\circ = -285.8\text{ kJ/mol} = -285.8 \times 10^3\text{ J/mol} = -285,800\text{ J/mol}$$ * Standard Gibbs free energy change: $$\Delta G^\circ = -237.39\text{ kJ/mol} = -237.39 \times 10^3\text{ J/mol} = -237,390\text{ J/mol}$$ * Absolute temperature: $$T = 25^\circ\text{C} = 25 + 273.15 = 298.15\text{ K} \approx 298\text{ K}$$ Step 2 - Apply the Gibbs-Helmholtz Equation The fundamental thermodynamic relation connecting Gibbs free energy ($\Delta G^\circ$), enthalpy ($\Delta H^\circ$), temperature ($T$), and entropy ($\Delta S^\circ$) is given by: $$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$ To solve for the standard entropy change ($\Delta S^\circ$), we rearrange this equation: $$T\Delta S^\circ = \Delta H^\circ - \Delta G^\circ$$ $$\Delta S^\circ = \frac{\Delta H^\circ - \Delta G^\circ}{T}$$ Step 3 - Calculate the Standard Entropy Change ($\Delta S^\circ$) Substitute the values with units into the rearranged equation: $$\Delta S^\circ = \frac{-285,800\text{ J/mol} - (-237,390\text{ J/mol})}{298\text{ K}}$$ $$\Delta S^\circ = \frac{-285,800\text{ J/mol} + 237,390\text{ J/mol}}{298\text{ K}}$$ $$\Delta S^\circ = \frac{-48,410\text{ J/mol}}{298\text{ K}}$$ $$\Delta S^\circ \approx -162.45\text{ J K}^{-1}\text{ mol}^{-1}$$ The negative sign indicates a decrease in entropy during the reaction: $$\ce{2H2(g) + O2(g) -> 2H2O(l)}$$ This is physically correct because three moles of gaseous reactants ($\ce{2H2}$ and $\ce{1O2}$) combine to form two moles of liquid water, resulting in a significantly more ordered state. Step 4 - Evaluate and Explain Each Option * **Option (A) is incorrect:** This value is $-0.1624\text{ J K}^{-1}$, which is off by a factor of $1000$ (representing the value in $\text{kJ K}^{-1}\text{ mol}^{-1}$ rather than $\text{J K}^{-1}\text{ mol}^{-1}$). * **Option (B) is correct:** This value is approximately $-162.4\text{ J K}^{-1}\text{ mol}^{-1}$ (or simply $\text{J K}^{-1}$ as printed in the options), which matches our calculation. * **Option (C) is incorrect:** This value has the correct magnitude but an incorrect positive sign. A positive sign would represent an increase in disorder, which is incorrect for a gas-to-liquid transformation. * **Option (D) is incorrect:** This value has an incorrect positive sign and is off by a factor of $1000$. $$\text{Correct Option: } \boxed{B}$$