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Consider the reaction: C2H5OH(ℓ) + 3O2(g) → 2CO2(g) + 3H2O(ℓ); ∆H = –1.37 x 103 kJ Consider the following statements: I. The reaction is endothermic II. The reaction is exothermic. III. The enthalpy term would be different if the water formed was gaseous. Which of these statement(s) is (are) true?

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Answer:

Two statements are true:

  • II. The reaction is exothermic.
  • III. The enthalpy term would be different if the water formed was gaseous.

Explanation:

  • C₂H₅OH(ℓ) + 3O₂(g) → 2CO₂(g) + 3H₂O(ℓ); ∆H = –1.37 × 10³ kJ

1. ∆H = –1.37 × 10³ kJ it telling that the enthalpy of the reaction is negative.

That means that reaction releases heat, which, by definition, means that the reaction is exothermic.

Then statement I is false and statement II is true.

2. The enthalpy of a reaction is equal to the enthalpy of the products less the enthalpy of the reactants:

       [tex]\Delta H_{rxn}=\sum (H_{products})-\sum (H_{reactants})[/tex]

The enthalpy of each substance depends on its state, thus the enthalpy term for water will be different if it is formed as a gas than if it is formed a liquid.

Hence, the enthalpy of the reaction will be different in case the water formed was gaseous, and the third statement is also true.

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