Question

Which of the following reactions have a positive ΔSrxn? Check all that apply.2A(g)+B(g)----->C(g)A(g)+B(g)----->3C(g)2A(g)+3B(g)----->4C(g)2A(g)+B(s)------->3C(g)

Answer 1

The reactions that have a positive ΔSrxn are the second and the fourth:

• A(g) + B(g) → 3C(g), and

• 2A(g) + B(s) → 3C(g)

Step-by-step explanation:

ΔSrxn stands for the change in entropy of a reaction.

Since ΔSrxn equals Final entropy - Initial entropy, a positive ΔSrxn (ΔSrxn > 0) means that the entropy increases.

Given the chemical equations for some reactions, you can identify the direction of the change of entropy (increase or decrease) by inspection, using basic definitions and principles.

These are:

• Entropy is a measure of the level of disorder of a system.

• The particles of the gases are in a state of greater disorder than the particles of the liquids, and the particles of liquids are in a state more disordered than those of the solids.

• In conclusion, under other conditions constant, the entropy of a system of gases is greater than the entropy of a system of liquids, and the entropy of a system of liquids is greaer than the entropy of a system of solids.

• A system with more particles has a greater entropy than a system of less particles.

With those principles, you get the following results for each of the given systems.

1) 2A(g) + B(g) → C(g)

Since in the product (right) side there are fewer molecules than in the reactant (left) side, you predict that the entropy decreases, and so the ΔSrxn is negative.

2) A(g) + B(g) → 3C(g)

Since in the product (right) side there are more molecules than in the reactant (left) side, you predict that the entropy increases, and so the ΔSrxn is positive.

3) 2A(g) + 3B(g) → 4C(g)

Since in the product (right) side there are fewer molecules than in the reactant (left) side, you predict that the entropy decreases, and so the ΔSrxn is negative.

4) 2A(g) + B(s) → 3C(g)

Since there are the same number of molecules in both sides, the determining factor of the change of entropy is the physical state of the components.

Since the reactant B (s) is in solid state and the only product is C(g), and it is in gas state, you predict that the entropy increases, and ΔSrxn is positive.

Conclusion: the reactions that have a positive ΔSrxn are the second and the fourth:

• A(g) + B(g) → 3C(g), and

• 2A(g) + B(s) → 3C(g)