A mixture of 1 mole of H2 and 1 mole of O2 is enclosed in a vessel at 300 K. The reaction H2 + O2 → H2O is initiated. Find the change in entropy of the system.
Correct: A
Using the formula for entropy change, ΔS = 2.303 nCv log(Tf/Ti) for an isothermal process. However, the given reaction is not isothermal, and the entropy change also involves the change in the number of moles of gas. Given the reaction 2H2 + O2 → 2H2O, where 3 moles of gas become 2 moles of liquid, there's a significant decrease in the number of moles of gas, leading to a decrease in entropy. Therefore, the entropy change is negative. Assuming standard conditions and using entropy values: S°(H2O) = 69.95 J/mol K, S°(H2) = 130.68 J/mol K, S°(O2) = 205.14 J/mol K. The total entropy of the reactants is 2*130.68 + 205.14 = 466.5 J/K. The total entropy of the products is 2*69.95 = 139.9 J/K. Therefore, ΔS = 139.9 - 466.5 = -326.6 J/K per mole of reaction. For 1 mole of H2 and 1 mole of O2 reacting, the change in entropy would be -326.6/2 = -163.3 J/K, which is closer to option A considering rounding.