Answer to Question #268066 in Chemistry for jena

Question

Calculate the standard molar entropy change for each of the following reactions at 25 ^oC.

1. Ca (s) + 2H₂O (l) → Ca(OH)₂ (aq) + 2H₂ (g)
2. Na₂CO₃ (aq) + 2HCl (aq) → 2NaCl (aq) + H₂O (l) + CO₂ (g)

Answer 1

Solution:

The change in the standard molar entropy of a reaction can be found by the difference between the sum of the molar entropies of the products and the sum of the molar entropies of the reactants.

ΔS°_reaction = Σn_pS°_products - Σn_rS°_reactants

(1):

Ca (s) + 2H₂O (l) → Ca(OH)₂ (aq) + H₂ (g)

S°(Ca, s) = 41.42 J mol^-1 K^-1

S°(H₂O, l) = 69.91 J mol^-1 K^-1

S°(Ca(OH)₂, aq) = -74.5 J mol^-1 K^-1

S°(H₂, g) = 130.684 J mol^-1 K^-1

Hence,

ΔS°_reaction = (S°_Ca(OH)2 + S°_H2) - (S°_Ca + 2×S°_H2O)

ΔS°_reaction = (-74.5 + 130.684) - (41.42 + 2×69.91) = -125.056

ΔS°_reaction = -125.06 J mol^-1 K^-1

Answer (1): ΔS°_reaction = -125.06 J mol^-1 K^-1

(2):

Na₂CO₃ (aq) + 2HCl(aq) → 2NaCl (aq) + H₂O (l) + CO₂ (g)

S°(Na₂CO₃, aq) = 61.1 J mol^-1 K^-1

S°(HCl, aq) = 56.5 J mol^-1 K^-1

S°(NaCl, aq) = 115.5 J mol^-1 K^-1

S°(H₂O, l) = 69.91 J mol^-1 K^-1

S°(CO₂, g) = 213.74 J mol^-1 K^-1

Hence,

ΔS°_reaction = (2×S°_NaCl + S°_H2O + S°_CO2) - (S°_Na2CO3 + 2×S°_HCl)

ΔS°_reaction = (2×115.5 + 69.91 + 213.74) - (61.1 + 2×56.5) = 340.55

ΔS°_reaction = 340.55 J mol^-1 K^-1

Answer (2): ΔS°_reaction = 340.55 J mol^-1 K^-1