G10 Final Exam Review Part 2

Which of the following represents the correct formula for calculating the standard entropy change (ΔS°) for a reaction?

• ΔS° = ΣS°reactants − ΣS°products

• ΔS° = ΣS°products − ΣS°reactants

• ΔS° = S°final − S°initial

• ΔS° = q/T

Given: S°(H₂O, g) = 188.8 J/mol·K S°(H₂, g) = 130.6 J/mol·K S°(O₂, g) = 205.0 J/mol·K Calculate ΔS° for the reaction: 2H₂(g) + O₂(g) → 2H₂O(g)

• −89.6 J/mol·K

• −326.4 J/mol·K

• +89.6 J/mol·K

• +326.4 J/mol·K

The units of standard molar entropy (S°) are:

• kJ/mol

• J/mol·K

• J/K

• kcal/mol

For the reaction: C(s) + O₂(g) → CO₂(g) Given: S°(C, s) = 5.7 J/mol·K S°(O₂, g) = 205.0 J/mol·K S°(CO₂, g) = 213.7 J/mol·K What is ΔS°?

• +4.3 J/mol·K

• −3.0 J/mol·K

• +213.7 J/mol·K

• −4.3 J/mol·K

Which of the following reactions would most likely have a positive ΔS°?

• 2SO₂(g) + O₂(g) → 2SO₃(g)

• H₂O(l) → H₂O(s)

• CaCO₃(s) → CaO(s) + CO₂(g)

• 2H₂(g) + O₂(g) → 2H₂O(g)

Why is the entropy of a gas generally higher than that of a solid or liquid?

• Gases have more mass.

• Gases are more disordered and have more accessible microstates.

• Gases have lower pressure.

• Gases conduct electricity better.

For the reaction: N₂(g) + 3H₂(g) → 2NH₃(g) Given: S°(N₂) = 191.5 J/mol·K S°(H₂) = 130.6 J/mol·K S°(NH₃) = 192.5 J/mol·K Calculate ΔS°.

• −198.3 J/mol·K

• +198.3 J/mol·K

• −383.0 J/mol·K

• +383.0 J/mol·K

A negative ΔS° for a reaction implies that:

• The disorder of the system increases.

• The system absorbs heat.

• The disorder of the system decreases.

• The reaction is spontaneous.

In which type of reaction is ΔS° most likely negative?

• A solid dissolves in water.

• Two gases combine to form a solid.

• A liquid boils.

• Ice melts to form water.

For a reaction with ΔS° = 0 J/mol·K, what does this imply about the products and reactants?

• They are at different temperatures.

• The entropy of the products equals the entropy of the reactants.

• The reaction is exothermic.

• The system releases energy.