Calculate the δg°rxn using the following information. 2 h2s(g) + 3 o2(g) ⟶ 2 so2(g) + 2 h2o(g) δ g∘ = ? δ hf0 (kj/mol) -20.6 0.00 -296.8 -241.8 s0(j/mol k) 205.8 205.2 248.2 188.8

ΔG° =-990.17 kJ /mol

Explanation:

The equation is:

ΔG° = Δ H° -TΔS°  equation 1

Δ H°rxn = Δ H°f (products)- Δ H°f(reactants)

Δ H°rxn = [[2XΔ H°f(SO₂)]+[2Δ H°f(H₂O)]-[2Δ H°f(H₂S)]+[3Δ H°f(O₂)]

Δ H°rxn = [2X(-296.8)+2(-241.8)]-[2X(-20.6)] = -1036

Δ S°rxn = Δ S°f (products)- Δ S°f(reactants)

Δ S°rxn = [[2XΔ S°f(SO₂)]+[2Δ S°f(H₂O)]-[2Δ S°f(H₂S)]+[3Δ S°f(O₂)]

Δ S°rxn = [2(248.2)+2(188.8)]-[2(205.8)+3(205.2)] = -153.8 J/mol K

      = -0.1538 KJ /mol

Putting values in equation 1  ( T = 298 K)

ΔG° = Δ H° -TΔS° = (-1036)-[(298)(-0.1538)

ΔG° =-990.17 kJ /mol

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.