posted by Austin on .
Calculate the entropy change when 2.42 g of
H2 reacts with O2 according to the reaction
2 H2(g) + O2(g) → 2 H2O(ℓ)
at 298 K and 1 atm pressure. The standard molar enthalpy of formation of H2O(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
I would calculate dH rxn as
dHorxn = (n*dHproducts) - (n*dHreactants)
Also dGorxn = (n*dGproducts) - (n*dGreactants)
Then dG = dH - TdS.
You know dG, dH and T, solve for dS rxn. That will be dS for 4g H. Adjust with a ratio to obtain dS for 2.42g.
just forget about it dude. it's all pointless now