Question

What is the value for ∆Soreaction for the following reaction, given the standard entropy values?

2H2S(g) + SO2(g) 3Srhombic(s) + 2H2O(g)

Answer 1

The value for ∆S°reaction for the given chemical reaction is -68.6 J/(mol·K).

Step-by-step explanation:

The value of the standard entropy change, ∆S°reaction, for the given reaction can be calculated by using the standard entropy values of the reactants and products. The formula is: ∆S°reaction = Σ(S°products) - Σ(S°reactants). For the given reaction: 2H2S(g) + SO2(g) → 3Srhombic(s) + 2H2O(g), the standard entropy values are:

• S°(H2S) = 206.7 J/(mol·K)
• S°(SO2) = 248.1 J/(mol·K)
• S°(Srhombic) = 42.9 J/(mol·K)
• S°(H2O) = 188.7 J/(mol·K)

Using the formula, ∆S°reaction = (3*S°(Srhombic)) + (2*S°(H2O)) - (2*S°(H2S)) - (S°(SO2)), we can calculate:

∆S°reaction = (3*42.9) + (2*188.7) - (2*206.7) - 248.1

∆S°reaction = -68.6 J/(mol·K)

Answer 2

To get the ∆S of the reaction, we simply have to add the ∆S of the reactants and the ∆S of the products. Then, we get the difference between the ∆S of the products and the ∆S of the products. If the ∆S is negative, then the reaction spontaneous. If the otherwise, the reaction is not spontaneous.