Consider the following reaction at 298 K. 4Al(s)+3O2(g)⟶2Al2O3(s)Δ𝐻∘=−3351.4 kJ

Question

Δ𝑆sys=

Δ𝑆surr=

Δ𝑆univ=

Answer 1

ΔG^o = -3164.6 kJ (from literature)

ΔG^o = ΔH^o - TΔS^o

-3164.6 kJ = -3154.4 kJ -( 298 K x ΔS^o)

therefore, ΔS^o = std entropy change of the system =0.03423kJ/K

= 34.23 J/K

It is an exothermic reaction and spontaneous reaction is taking place in the system, therefore, change in entropy of the surrounding, ΔS⁰_surr , is only due to the heat flow to the surroundings, therefore, ΔS^o_surr = +3154.4kJ/298K = 10.585 kJ/K = 10585 J/K

ΔS^o _univ = ΔS^o_sys + ΔS^o_surr

= 34.23 + 10585 = 10619 J/K