Question

Consider the reaction: 2H,020) 2H 0(0) + O2(e) Using standard absolute entropies at 298K, calculate the entropy change for th
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Answer #1


Given:
Sof(H2O2(l)) = 109.6 J/mol.K
Sof(H2O(l)) = 69.91 J/mol.K
Sof(O2(g)) = 205.138 J/mol.K

Balanced chemical equation is:
2 H2O2(l) ---> 2 H2O(l) + O2(g)

ΔSo rxn = 2*Sof(H2O(l)) + 1*Sof(O2(g)) - 2*Sof( H2O2(l))
ΔSo rxn = 2*(69.91) + 1*(205.138) - 2*(109.6)
ΔSo rxn = 125.758 J/K

This is when 2 mol of H2O2 reacts as per given equation.
So,
When 2.15 mol of H2O2 reacts,
ΔSo rxn = 125.758 * 2.15 / 2 = 135.19 J/K


Answer: 135 J/K

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