Question

What is ΔS ∘

for B in the reaction
5A→4B if
rxn =−194.4J/mol⋅K? (
(A)=220.0J/mol⋅K)

A) -194.4 J/mol ･ K
B) -105.0 J/mol ･ K
C) 116.0 J/mol ･ K
D) 414.4 J/mol ･ K

Answer 1

To find the standard entropy change for B, we set up an equation using the reaction's entropy change and A's entropy, and solve for B's entropy. The resulting value for ΔS° for B in the reaction 5A → 4B is 226.4 J/mol·K.

Step-by-step explanation:

The problem is asking us to find the standard entropy change (ΔS°) for compound B in the reaction. Given that the total entropy change for the reaction (ΔS°rxn) is -194.4 J/mol·K and the standard entropy of A is 220.0 J/mol·K, we use the standard entropy relation for a reaction:

ΔS°rxn = Σ(standard entropy of products) - Σ(standard entropy of reactants)

This simplifies to:

-194.4 J/mol·K = (4 x ΔS°B) - (5 x 220.0 J/mol·K)

We solve for ΔS°B:

-194.4 J/mol·K = (4 x ΔS°B) - 1100 J/mol·K

4 x ΔS°B = -194.4 J/mol·K + 1100 J/mol·K

4 x ΔS°B = 905.6 J/mol·K

ΔS°B = 905.6 J/mol·K / 4

ΔS°B = 226.4 J/mol·K