Question

What is the change in entropy when 2 moles of H2 (g) react with 1 mole of O2 (g) to form 2 moles of H2O (l) at a constant temperature of 298 K? (Assume standard molar entropy values for the reactants and products)

Answer 1

To calculate the change in entropy for this reaction, we can use the formula:S = nS products  - nS reactants where S is the change in entropy, n is the number of moles, and S is the standard molar entropy. We can find the standard molar entropy values for the reactants and products in a reference table.For this reaction:2 H2  g  + O2  g   2 H2O  l The standard molar entropy values at 298 K are as follows:H2  g  = 130.68 J/molKO2  g  = 205.15 J/molKH2O  l  = 69.95 J/molKNow, we can plug these values into the formula:S = [2 moles  69.95 J/molK  H2O ] - [2 moles  130.68 J/molK  H2  + 1 mole  205.15 J/molK  O2 ]S = [139.90 J/K] - [261.36 J/K + 205.15 J/K]S = 139.90 J/K - 466.51 J/KS = -326.61 J/KThe change in entropy for this reaction at 298 K is -326.61 J/K.