Calculate the entropy change when 4.88 g of H2 reacts with O2 according to the

Question

reaction 2H2(g) + O2(g)→2H2O(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 believe this is searching for deltaS but I don't think I'm rearranging the equation deltaG=deltaH-TdeltaS correctly if using that at all.

Answer 1

No, your equation looks ok to me.
I think what you need to do is to convert dG and dH from so many kJ/mol to so many kJ/4.88 g H2. To do that
dG = -237.2 kJ/mol x (4.88 g/4*atomic mass H) = ?
dH = -285.8 kJ/mol x (4.88 g H2/4*atomic mass H) = ?

Then dG = dH - TdS
You have dG for the 4.88 g H2.
You have dH for the 4.88 g H2.
Solve for dS for the 4.88 g H2 at 298 K.