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Calculate the standard entropy change for the reaction 2H2(g) + O2(g) → 2H2O(g) given the standard entropy values of H2(g) = 130.6 J/mol K, O2(g) = 205.0 J/mol K, and H2O(g) = 188.8 J/mol K. Also, determine whether the reaction is spontaneous or not at 298 K.

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To calculate the standard entropy change  S  for the reaction, we can use the formula:S = nS products  - nS reactants where n is the stoichiometric coefficient and S is the standard entropy of each species.For the given reaction, 2H2 g  + O2 g   2H2O g , we have:S = [2  S H2O g  ] - [2  S H2 g   + S O2 g  ]Now, we can plug in the given standard entropy values:S = [2  188.8 J/mol K] - [2  130.6 J/mol K + 205.0 J/mol K]S = [377.6 J/mol K] - [466.2 J/mol K]S = -88.6 J/mol KThe standard entropy change for the reaction is -88.6 J/mol K.To determine whether the reaction is spontaneous or not at 298 K, we need to calculate the Gibbs free energy change  G  using the formula:G = H - TSHowever, we don't have the enthalpy change  H  for the reaction. Instead, we can use the fact that a reaction is spontaneous if G < 0, and the sign of G depends on the signs of H and S.In this case, since S is negative, the reaction will be spontaneous if H is also negative and sufficiently large in magnitude to overcome the positive TS term. Without the value of H, we cannot definitively determine whether the reaction is spontaneous or not at 298 K.

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