Question

Consider the reaction: 2H2O2(l)2H2O(l) + O2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.40 moles of H2O2(l) react at standard conditions. S°surroundings = J/K

Answer 1

2 H2O2(l) --------------> 2 H2O (l) + O2(g)

$\Delta$ Ho rxn = $\Delta$ Ho f products - $\Delta$ Hf reactants

               = (2 x –285.8 ) - 2 x (–187.8)

               = - 196 kJ/mol

$\Delta$ Sosurr = - $\Delta$ Ho rxn / T

      = 196 x 10^3 / 298

      = 657.7 J / K

S°surroundings = 657.7 J / K

2 mol -----------> 657.7 J /K

2.40 mol -----------> ??

S°surroundings = 789 J / K