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Which relationship or statement best describes delta s for the following reaction:

2H2S(g) + 3O2(g) --> 2H2O(g)

a) Delta S = 0
b) Delta S < 0
c) Delta S> 0
d) Delta S = Delta H |T
e) More information is needed to make a reasonable prediction.

Respuesta :

dsdrajlin

Answer:

b) Delta S < 0

Explanation:

The change in the entropy (ΔS) is related to the change in the number of gaseous moles of the reaction: Δn(g) = n(g, products) - n(g, reactants).

  • If Δn(g) > 0, the entropy increases (ΔS > 0).
  • If Δn(g) < 0, the entropy decreases (ΔS < 0).
  • If Δn(g) = 0, there is little or no change in the entropy

Let's consider the following equation.

2 H₂S(g) + 3 O₂(g) → 2 H₂O(g)

Δn(g) = 2 - 5 = - 3. Since Δn(g) < 0, the entropy decreases and ΔS < 0.

Entropy in the reaction can be given as the degree of randomness and motion of the molecules. In the given chemical reaction, Delta S < O.

What is the relation between Delta S and the number of molecules?

In the chemical reaction, the increase in the number of product molecules results in a more degree of randomness, and the entropy of the reaction increases.

In the given chemical reaction, the number of reactant gaseous molecules is 6, while the product molecules are 2.

The reactant molecules are greater than the product molecules, thus the entropy of the reaction decreases. Hence, Delta S  < 0. Therefore, option B is correct.

Learn more about Delta S, here:

https://brainly.com/question/26433173

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