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What is the change in entropy when 2 moles of hydrogen gas react with 1 mole of oxygen gas to form 2 moles of water vapor at 298 K and 1 atm pressure?

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To calculate the change in entropy for this reaction, we need to find the difference between the entropy of the products and the entropy of the reactants. The reaction is:2 H g  + O g   2 HO g First, we need to find the standard molar entropies  S  for each substance involved in the reaction. The standard molar entropies at 298 K are:S H  = 130.6 J/molKS O  = 205.1 J/molKS HO  = 188.8 J/molKNow, we can calculate the change in entropy  S  for the reaction:S = [2  S HO ] - [2  S H  + S O ]S = [2  188.8 J/molK] - [2  130.6 J/molK + 205.1 J/molK]S = [377.6 J/molK] - [466.3 J/molK]S = -88.7 J/molKThe change in entropy for this reaction is -88.7 J/molK.

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