Question

Consider the reaction: 2H2O(l)2H2(g) + O2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.88 moles of H2O(l) react at standard conditions. S°surroundings = J/K

Answer 1

2H2O(l) --->2H2(g) + O2(g)

first step is to get dHrxn = dHf(Products) - dHf(reactants)

take dHf of reactants and products from standard thermodynamic table

dHf(H2O) = -285.8 KJ/mol

dHf(H2) =0 kj/mol
dHf(O2)= 0 kj/mol

dHrxn = 2*dHf(H2) + dHf(O2) - 2*dHf(H2O)
= 2*0 + 0 - 2* [-285.8] kj = +571.6 KJ

this is for 2 mole of H2O
for 1.88 mols H2O, dHrxn = +571.6 KJ/2mol*1.88 mol = 537.304 kj
T =298 K
dSsurrounding = - dHrxn/T = -537304 j/298 K = -1803.03 J/K
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thanks!! .