Question

2H2O(l)2H2(g) + O2(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.92 moles of H2O(l) react at standard conditions.

Answer 1

Given:

Sof(H2O(l)) = 69.91 J/mol.K

Sof(H2(g)) = 130.684 J/mol.K

Sof(O2(g)) = 205.138 J/mol.K

Balanced chemical equation is:

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

ΔSo rxn = 2*Sof(H2(g)) + 1*Sof(O2(g)) - 2*Sof( H2O(l))

ΔSo rxn = 2*(130.684) + 1*(205.138) - 2*(69.91)

ΔSo rxn = 326.686 J/K

This is when 2 mol of H2O reacts as per equation

So,

for 1.92 mol, ΔSo rxn = 326.686 * 1.92 / 2 = 313.6 J/K

Answer: 313.6 J/K