Use each equation to predict whether the relative entropy change is S(products) > S(reactants) or S(products) < S(reactants) and whether ΔS > 0 or ΔS < 0 for the chemical reaction represented. Justify your answer in each case.

a. 2NH3(g)→N2(g) + 3H2(g)
b. 2Al(s) + 3F2(g)→2AlF3(s)

Thank you, but how do I solve for the number of moles? I believe the funny symbols are, respectively, are delta signs in front of the S and arrows in the chemical equations.

To determine the relative entropy change and the sign of ΔS (change in entropy) for each chemical reaction, you need to consider the number of moles of reactants and products.

1. Equation a: 2NH3(g) → N2(g) + 3H2(g)
In this equation, there are 2 moles of NH3 being converted into 1 mole of N2 and 3 moles of H2.

To analyze the entropy change:
- Consider the number of moles: Since the number of moles decreases from 2 to 1 (NH3 to N2), and increases from 0 to 3 (H2), we can conclude that the entropy of the reactants is greater than the entropy of the products (S(reactants) > S(products)).
- Determine the sign of ΔS: Since the reaction reduces the number of moles, it decreases the system's disorder. Therefore, ΔS is expected to be negative (ΔS < 0).

Thus, for equation a, S(products) < S(reactants) and ΔS < 0.

2. Equation b: 2Al(s) + 3F2(g) → 2AlF3(s)
In this equation, 2 moles of Al react with 3 moles of F2 to produce 2 moles of AlF3.

To analyze the entropy change:
- Consider the number of moles: Since the number of moles remains the same (2 moles of AlF3), we can conclude that the entropy of the reactants is equal to the entropy of the products (S(reactants) = S(products)).
- Determine the sign of ΔS: Since there is no net change in the number of moles, the entropy remains constant or nearly constant, resulting in ΔS ~ 0 or ΔS ≈ 0.

Thus, for equation b, S(products) = S(reactants) and ΔS ≈ 0.

To summarize:
a. S(products) < S(reactants) and ΔS < 0
b. S(products) = S(reactants) and ΔS ≈ 0.

These are predictions based on the number of moles of reactants and products and do not take into account other factors such as temperature, pressure, or the change in phase.

For the entropy, larger number of moles means more entropy. That will answer both a and b. I don't understand the rest of the question because of the funny symbols.

That answer I gave you are for gases only. You count the mols. For a you have 2 mols NH3 on the left and 4 mols on the right so an increase in mols means increase in entropy. b. You have 3 mols gas on left and zero on the right.