Calculate the entropy change when 4.88 g of H2 reacts with O2 according to the
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.
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.
To calculate the entropy change (ΔS) for a reaction, you can use the equation:
ΔG = ΔH - TΔS
where ΔG is the change in Gibbs free energy, ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy.
Given that the standard molar enthalpy of formation (ΔH) of H2O(ℓ) at 298 K is -285.8 kJ/mol, and the corresponding free energy of formation (ΔG) is -237.2 kJ/mol, we can rearrange the equation to solve for ΔS.
Rearranging the equation, we have:
ΔS = (ΔH - ΔG) / T
To convert the units of the enthalpy and free energy from kJ/mol to J/mol, you need to multiply them by 1000.
ΔH = -285.8 kJ/mol * 1000 J/kJ = -285800 J/mol
ΔG = -237.2 kJ/mol * 1000 J/kJ = -237200 J/mol
Substituting the values into the equation:
ΔS = (-285800 J/mol - (-237200 J/mol)) / 298 K
ΔS = (-285800 J/mol + 237200 J/mol) / 298 K
Calculating the numerator:
ΔS = -48600 J/mol / 298 K
Finally, calculating ΔS:
ΔS ≈ -162.42 J/K
Therefore, the entropy change (ΔS) when 4.88 g of H2 reacts with O2 is approximately -162.42 J/K.