Respuesta :

Given reaction:

Mg(OH)2(s) + 2HCl(g) → MgCl2(s) + 2H2O(g)

Formula:

ΔS°rxn = ∑nS°(products) - ∑nS°(reactants)

           = [S°(MgCl2) + 2S°(H2O)] -[S°(Mg(OH)2 + 2S°(HCl)]

           = [89.3+2(188.7)] - [63.22+2(186.80] = 29.88 J/K

Thus, the standard entropy change for the reaction is around 30 J/K

Eduard22sly

The standard entropy change for the given reaction obtained at 25 °C is 29.88 J/K

Data obtained from the question

  • Mg(OH)₂(s) + 2HCl(g) → MgCl₂(s) + 2H₂O(g)
  • Change in entropy (ΔS°rxn) =?

Standard molar entropy

The standard molar entropy for each entity in the reaction is given below:

  • Entropy of Mg(OH)₂(s) = 63.22 J/K
  • Entropy of HCl(g) = 186.80 J/K
  • Entropy of MgCl₂(s) = 89.3 J/K
  • Entropy of H₂O(g) = 188.7 J/K

How to determine the change in entropy

Mg(OH)₂(s) + 2HCl(g) → MgCl₂(s) + 2H₂O(g)

ΔS°rxn = ∑nS°(products) - ∑nS°(reactants)

ΔS°rxn = [S°(MgCl₂) + 2S°(H₂O)] – [S°(Mg(OH)₂ + 2S°(HCl)]

ΔS°rxn = [89.3 + 2(188.7)] – [63.22 + 2(186.80]

ΔS°rxn = [89.3 + 377.4] – [63.22 + 373. 6]

ΔS°rxn = 466.7 – 436.82

ΔS°rxn = 29.88 J/K

Learn more about entropy:

https://brainly.com/question/13834300

Otras preguntas