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The combustion of ethene in the presence of excess oxygen yields carbon dioxide and water: c2h4 (g) + 3o2 (g) → 2co2 (g) + 2h2o (l) the value of δs° for this reaction is ________ j/k⋅mol.

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znk

Answer:

[tex]\boxed{-267.5}[/tex]

Explanation:

You can calculate the entropy change of a reaction by using the standard molar entropies of reactants and products.

The formula is

[tex]\Delta_{r} S^{\circ} = \sum_n {nS_{\text{products}}^{\circ} - \sum_{m} {mS_{\text{reactants}}^{\circ}}}[/tex]

The equation for the reaction is

                        C₂H₄(g) + 3O₂(g) ⟶ 2CO₂(g) + 2H₂O(ℓ)

ΔS°/J·K⁻¹mol⁻¹   219.5      205.0         213.6         69.9

[tex]\Delta_{r} S^{\circ} = (2\times213.6 + 2\times69.9) - (1\times219.5 + 3\times205.0)\\\\= 567.0 - 834.5 = \boxed{-267.5 \text{ J}\cdot\text{K}^{-1} \text{mol}^{-1}}[/tex]

Eduard22sly

The value of ΔS for the given reaction is –267.5 J/Kmol

The value of ΔS for the reaction given above can be calculated by using the standard entropies of the reactants and products.

The standard entropies for each compounds are given below:

  • ΔS for C₂H₄ = 219.5 J/Kmol
  • ΔS for O₂ = 205.0 J/Kmol
  • ΔS for CO₂ = 213.6 J/Kmol
  • ΔS for H₂O = 69.9 J/Kmol

Thus, we can obtain the change in the entropy, ΔS for the reaction as illustrated below:

C₂H₄ + 3O₂ —> 2CO₂ + 2H₂O

Change in entropy = entropy of product – entropy of reactant

ΔS = ΔSₚ – ΔSᵣ

ΔS = [(2×213.6) + (2×69.9)] – [(1×219.5) + (3×205)]

ΔS = [427.2 + 139.8] – [219.5 + 615]

ΔS = 567 – 834.5

ΔS = –267.5 J/Kmol

Learn more about entropy:

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