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What is the Gibbs free energy change for the reaction where 2 moles of hydrogen gas (H2) react with 1 mole of oxygen gas (O2) to form 2 moles of water vapor (H2O) at standard conditions (298 K and 1 atm pressure)?

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To calculate the Gibbs free energy change  G  for the reaction, we can use the following equation:G = H - TSwhere H is the change in enthalpy, T is the temperature in Kelvin, and S is the change in entropy.First, we need to find the standard enthalpy change  H  and standard entropy change  S  for the reaction. We can do this by using the standard enthalpies of formation and standard entropies of the substances involved in the reaction.Standard enthalpies of formation  Hf  at 298 K:H2 g  = 0 kJ/mol  since it is an element in its standard state O2 g  = 0 kJ/mol  since it is an element in its standard state H2O g  = -241.8 kJ/molStandard entropies  S  at 298 K:H2 g  = 130.7 J/molKO2 g  = 205.2 J/molKH2O g  = 188.8 J/molKNow, we can calculate the standard enthalpy change  H  and standard entropy change  S  for the reaction:H = [2 -241.8 kJ/mol ] - [2 0 kJ/mol  + 0 kJ/mol] = -483.6 kJ/molS = [2 188.8 J/molK ] - [2 130.7 J/molK  + 205.2 J/molK] = 101.6 J/molKNow, we can calculate the Gibbs free energy change  G  at 298 K:G = H - TS = -483.6 kJ/mol -  298 K  101.6 J/molK  1 kJ/1000 J   -457.3 kJ/molSo, the Gibbs free energy change for the reaction at standard conditions  298 K and 1 atm pressure  is approximately -457.3 kJ/mol.

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