Question

For each reaction, calculate Δ H ∘ rxn , Δ S ∘ rxn , and Δ G rxn at 29 ∘ C .

A) N2O4(g)→2NO2(g)

B) NH4Cl(s)→HCl(g)+NH3(g)

C) 3H2(g)+Fe2O3(s)→2Fe(s)+3H2O(g)

D) N2(g)+3H2(g)→2NH3(g)

Answer 1

For each reaction,

A) dHorxn = dHo(products) - dHo(reactants)

                 = 2 x 33.2 - 9.1 = 57.3 kJ/mol

dSorxn = dSo(products) - dSo(reactants)

             = 2 x 240 - 304 = 176 J/K.mol

T = 29 + 273 = 302 K

dGorxn = dHorxn - TdSorxn

             = 4.15 kJ/mol

B) dHorxn = dHo(products) - dHo(reactants)

                 = -92.3 - 45.9 - (-314.6) = 176.4 kJ/mol

dSorxn = dSo(products) - dSo(reactants)

             = 186.9 + 192.8 - 94.8 = 284.9 J/K.mol

T = 29 + 273 = 302 K

dGorxn = dHorxn - TdSorxn

             = 90.36 kJ/mol

C) dHorxn = dHo(products) - dHo(reactants)

                 = 3 x -241.8 - (-824.2) = 98.8 kJ/mol

dSorxn = dSo(products) - dSo(reactants)

             = (2 x 27.3 + 3 x 189) - (3 x 130.7 + 87.4) = 142.1 J/K.mol

T = 29 + 273 = 302 K

dGorxn = dHorxn - TdSorxn

             = 55.92 kJ/mol

D) dHorxn = dHo(products) - dHo(reactants)

                 = 2 x -45.9 = -91.8 kJ/mol

dSorxn = dSo(products) - dSo(reactants)

             = 2 x 192.8 - (191.6 + 3 x 130.7) = -198.1 J/K.mol

T = 29 + 273 = 302 K

dGorxn = dHorxn - TdSorxn

             = -32.00 kJ/mol