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Calculate the Gibbs free energy change at standard conditions for the reaction:2 H2(g) + O2(g) → 2 H2O(l)Given: ΔHºf[H2O(l)] = -285.8 kJ/mol, ΔHºf[H2(g)] = 0 kJ/mol, ΔHºf[O2(g)] = 0 kJ/mol, and ΔSº[H2O(l)] = -69.9 J/K·mol.

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To calculate the Gibbs free energy change  G  for the reaction, we first need to find the change in enthalpy  H  and the change in entropy  S  for the reaction. 1. Calculate H for the reaction:H =  Hf products  -  Hf reactants For the products, we have 2 moles of H2O l , so:H products  = 2 *  -285.8 kJ/mol  = -571.6 kJFor the reactants, we have 2 moles of H2 g  and 1 mole of O2 g , both with Hf = 0 kJ/mol, so:H reactants  = 2 * 0 kJ/mol + 1 * 0 kJ/mol = 0 kJNow, we can find H for the reaction:H = -571.6 kJ - 0 kJ = -571.6 kJ2. Calculate S for the reaction:We are given S[H2O l ] = -69.9 J/Kmol. However, we need the entropy change for the entire reaction. Since we don't have the entropy values for H2 g  and O2 g , we cannot directly calculate S for the reaction.3. Calculate G for the reaction:Since we don't have the entropy change for the reaction, we cannot use the equation G = H - TS directly. However, we know that the reaction of hydrogen and oxygen to form water is spontaneous under standard conditions, which means G must be negative. Therefore, the Gibbs free energy change for the reaction is negative, but we cannot determine the exact value without the entropy change for the reaction.

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