The reaction 2 H2O(l) → 2 H2(g) + O2(g) is reactant favored under standard conditions at 282 K due to a positive standard free-energy change. To calculate the entropy change for 1.88 moles of H2O(l), we find the molar entropy change from G° = H° - T*S° and multiply by 1.88.
For the given reaction 2 H2O(l) → 2 H2(g) + O2(g) under standard conditions at 282 K, we want to determine if the reaction is reactant or product favored. The standard free-energy change (G°) is 479.5 kJ and the standard enthalpy change (H°) is 571.6 kJ. To find the standard entropy change (S°) for the reaction, we use the formula G° = H° - T*S°. Solving for S° gives us S° = (H° - G°)/T. Substituting the given values, we get S° = (571.6 kJ - 479.5 kJ) / 282 K which results in an entropy change for the reaction. Since the question asks for the entropy change for 1.88 moles of H2O(l), we need to use the molar entropy change calculated and multiply it by 1.88.
To determine if the reaction is reactant or product favored, we look at the sign of G°. A positive G° means that the reaction is reactant favored under standard conditions, while a negative G° means the reaction is product favored. In this case, the positive G° value of 479.5 kJ indicates the reaction is reactant favored under standard conditions at 282 K.
The complete question is- For the reaction 2 H2O(l) 2 H2(g) + O2(g) G° = 479.5 kJ and H° = 571.6 kJ at 282 K and 1 atm. This reaction is (reactant, product) favored under standard conditions at 282 K. The entropy change for the reaction of 1.88 moles of H2O(l) at this temperature would be J/K.