Question

Which of the following reactions have a positive ΔSrxn? Check all that apply.

Answer 1

Answer:

The reactions that have a positive ΔS rxn are the first and the fourth choices:

• 2A(g) + B(s) → 3C(g)

• 2A(g) + 2B(g) → 5C(g)

Explanation:

ΔS rxn is the change of entropy of the chemical reaction.

ΔS rxn = S after reaction - S before reaction.

Therefore, a positive ΔS rxn  means that the entropy after the reaction is greater than the entropy before the reaction.

You may use some assumptions to predict whether a reaction will lead an increase or decrease of the entropy.

First, assume that all the non-shown conditions, such as temperature and pressure, are constant.

Under that assumption, and from the meaning of entropy as a measure of the disorder or randomness of a system you can predict the sign of the change of entropy.

• 2A(g) + B(s) → 3C(g)

        1)  The solid compounds, B(s) in this case, are very ordered and so they have low entropy.

        2) Gas molecules are highly disordered (scattered), and the greater the number of molecules of the gas the larger the entropy, S).

Hence, since the product side shows 3 gas molecules and the reactant side shows 2 gas molecules and 1 solid molecule, you predict that the products have a larger entropy than the reactants, meaning an increase in entropy: ΔS rxn is positive.

• 2A(g) + B(g) → C(g)

Using the same reasoning, 3 gas molecules in the  reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

• A(g) + B(g) → C(g)

Again, 2 gas molecules in the  reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

• 2A(g) + 2B(g) → 5C(g)

With the same reasoing, 5 molecules in the product side, lets you predict that will have more entropy than 4 molecules in the reactant side, and, the entropy will increase: ΔS rxn is positive.