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Calculate the entropy change (∆S) for the reaction between 2.0 moles of hydrogen gas (H2) and 1.0 mole of oxygen gas (O2) at constant pressure and a temperature of 298 K, given the standard molar entropy values for H2 and O2 to be 130.7 J/K•mol and 205.0 J/K•mol, respectively.

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To calculate the entropy change  ∆S  for the reaction, we first need to write the balanced chemical equation for the reaction between hydrogen gas  H2  and oxygen gas  O2 :

2H2 g  + O2 g  → 2H2O g

Next, we need to find the standard molar entropy value for water vapor  H2O . The standard molar entropy value for H2O g  is 188.8 J/K•mol.

Now we can calculate the entropy change  ∆S  for the reaction using the following formula:

∆S = ∑nS products  - ∑nS reactants

where n is the number of moles and S is the standard molar entropy value.

For the reactants, we have:

 2 moles H2 × 130.7 J/K•mol  +  1 mole O2 × 205.0 J/K•mol  = 261.4 J/K + 205.0 J/K = 466.4 J/K

For the products, we have:

 2 moles H2O × 188.8 J/K•mol  = 377.6 J/K

Now we can calculate the entropy change  ∆S :

∆S = 377.6 J/K - 466.4 J/K = -88.8 J/K

The entropy change for the reaction between 2.0 moles of hydrogen gas  H2  and 1.0 mole of oxygen gas  O2  at constant pressure and a temperature of 298 K is -88.8 J/K.

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