Question

Calculate the Gibbs Free Energy change for the following chemical reaction at 298 K:2H2(g) + O2(g) → 2H2O(l) Given the standard Gibbs free energy of formation values: ΔGf°[H2(g)] = 0 kJ/molΔGf°[O2(g)] = 0 kJ/molΔGf°[H2O(l)] = -237 kJ/mol

Answer 1

To calculate the Gibbs Free Energy change  G  for the reaction, we can use the following equation:G =  Gf[products]  -  Gf[reactants] For this reaction, the products are 2 moles of H2O l  and the reactants are 2 moles of H2 g  and 1 mole of O2 g .G = [2  Gf H2O ] - [2  Gf H2  + Gf O2 ]Given the standard Gibbs free energy of formation values:Gf[H2 g ] = 0 kJ/molGf[O2 g ] = 0 kJ/molGf[H2O l ] = -237 kJ/molSubstitute the given values into the equation:G = [2   -237 kJ/mol ] - [2   0 kJ/mol  +  0 kJ/mol ]G =  -474 kJ  -  0 kJ G = -474 kJThe Gibbs Free Energy change for the reaction at 298 K is -474 kJ.