Final answer:
The question relates to the thermodynamics of an exothermic reaction involving aluminum and oxygen. The system entropy change (Δsys), surroundings entropy change (Δsurr), and universe entropy change (Δuniv) are discussed conceptually, though specific values cannot be provided without further data.
Step-by-step explanation:
The student has asked about the thermodynamics of the reaction 4Al(s) + 3O2(g) → 2Al2O3(s) with an enthalpy change (ΔH°) of −3351.4 kJ. To address this, we can discuss the concepts of system entropy change (Δsys), surroundings entropy change (Δsurr), and universe entropy change (Δuniv), although not enough information is provided in the question to calculate these values directly for this specific reaction. According to thermodynamics, Δsys is the change in entropy of the reaction, Δsurr is the change in entropy of the surroundings due to heat exchange with the system, and Δuniv is the sum of the entropy changes of both the system and the surroundings.
For an exothermic reaction like this one, ΔH° is negative because heat is released. This released heat increases the entropy of the surroundings, usually causing a positive Δsurr. The Δsys can be obtained by using standard molar entropy values from tables (not provided here). Finally, Δuniv is the sum of Δsys and Δsurr, and the Second Law of Thermodynamics states that for a spontaneous process, Δuniv must be positive.