To determine if the reactions are spontaneous or nonspontaneous and whether they are exothermic or endothermic, we can use the Gibbs Free Energy equation:

$\Delta G = \Delta H - T \Delta S$

• ΔG (Gibbs free energy): If ΔG < 0, the reaction is spontaneous; if ΔG > 0, the reaction is nonspontaneous.
• ΔH (Enthalpy): If ΔH < 0, the reaction is exothermic; if ΔH > 0, the reaction is endothermic.
• T (Temperature): Given in Kelvin.
• ΔS (Entropy): Convert to kJ/K by dividing J/K by 1000.

Let's do a sample calculation for case (a):

(a) $\Delta H = -128 \, \text{kJ}$, $\Delta S = +35 \, \text{J/K}$, $T = 500 \, \text{K}$

1. Convert ΔS to kJ/K: $\Delta S = \frac{35 \, \text{J/K}}{1000} = 0.035 \, \text{kJ/K}$

2. Apply the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$ $\Delta G = -128 \, \text{kJ} - (500 \, \text{K}) \times (0.035 \, \text{kJ/K})$ $\Delta G = -128 \, \text{kJ} - 17.5 \, \text{kJ}$ $\Delta G = -145.5 \, \text{kJ}$

Since $\Delta G = -145.5 \, \text{kJ}$, the reaction is spontaneous because ΔG < 0. Additionally, since ΔH is negative, it is an exothermic reaction.

(b) $\Delta H = +67 \, \text{kJ}$, $\Delta S = -140 \, \text{J/K}$, $T = 250 \, \text{K}$

1. Convert ΔS to kJ/K: $\Delta S = \frac{-140 \, \text{J/K}}{1000} = -0.14 \, \text{kJ/K}$

2. Apply the Gibbs free energy equation: $\Delta G = 67 \, \text{kJ} - (250 \, \text{K}) \times (-0.14 \, \text{kJ/K})$ $\Delta G = 67 \, \text{kJ} + 35 \, \text{kJ}$ $\Delta G = 102 \, \text{kJ}$

Since $\Delta G = 102 \, \text{kJ}$, the reaction is nonspontaneous because ΔG > 0. Additionally, since ΔH is positive, it is an endothermic reaction.

(c) $\Delta H = +75 \, \text{kJ}$, $\Delta S = +95 \, \text{J/K}$, $T = 800 \, \text{K}$

1. Convert ΔS to kJ/K: $\Delta S = \frac{95 \, \text{J/K}}{1000} = 0.095 \, \text{kJ/K}$

2. Apply the Gibbs free energy equation: $\Delta G = 75 \, \text{kJ} - (800 \, \text{K}) \times (0.095 \, \text{kJ/K})$ $\Delta G = 75 \, \text{kJ} - 76 \, \text{kJ}$ $\Delta G = -1 \, \text{kJ}$

Since $\Delta G = -1 \, \text{kJ}$, the reaction is spontaneous because ΔG < 0. Additionally, since ΔH is positive, it is an endothermic reaction.

Summary:

• (a) Spontaneous / Exothermic
• (b) Nonspontaneous / Endothermic
• (c) Spontaneous / Endothermic

Let me know if you would like more details or have any questions!

Here are some related questions:

1. How does temperature affect spontaneity in reactions?
2. Can a reaction be nonspontaneous at one temperature and spontaneous at another?
3. What is the physical meaning of ΔH and ΔS in a reaction?
4. How do exothermic reactions relate to heat transfer in systems?
5. What are real-world examples of spontaneous endothermic reactions?

Tip: Always remember to check the units when applying the Gibbs Free Energy equation!