I would rewrite the equation as
H2 + 1/2 O2 ==> H2O
dGof = dHof - TdSof
You know dGo and dHo and T, solve for dS.
That gives you dS for 1 mol H2O from 1 mol H2. You have 4.44g/2 = 2.22 mols
dS for 1 mol x 2.22 mol = ?
Calculate the entropy change when 4.44g of H2 reacts with O2 according to the reaction
2H2(g) + O2(g)→2H2O(ℓ)
at 298 K and 1 atm pressure. The stan-
dard molar enthalpy of formation of H
2O(ℓ) at 298 K is −285.8 kJ/mol. The corresponding free energy of formation is −237.2 kJ/mol.
Answer in units of J/K
1 answer