C₂H₅OH(l)+ 3O₂(g)$\to$2CO₂(g)+ 3H₂O(l),$\Delta$H = -1.37 $\times$ 10³kJ For the combustion of ethyl alcohol as described in the above equation,which of the following is true? I.The reaction is exothermic. II.The enthalpy change would be different if gaseous water was produced. III.The reaction is not an oxidation-reduction one. IV.The products of the reaction occupy a larger volume than the reactants.

Question

Quizplus · Accepted Answer

I. The negative sign of the enthalpy change (-1.37 kJ) indicates that the reaction releases heat to the surroundings, making it exothermic.

II. The enthalpy change would indeed be different if gaseous water was produced instead of liquid water. The enthalpy of vaporization of water is not negligible, and would affect the total enthalpy change of the reaction.

III. The reaction involves the combustion of ethyl alcohol, which is an oxidation-reduction reaction. Hence, this statement is incorrect.

IV. There is no information in the equation that suggests that the products occupy a larger volume than the reactants, so this statement cannot be justified.

C2H5OH(l)+ 3O2(g) →\to→ 2CO2(g)+ 3H2O(l) Δ\DeltaΔ H = -137 ×\times× 103 kJ for the Combustion of Ethyl Alcohol as Described

C₂H₅OH(l)+ 3O₂(g) $\to$ 2CO₂(g)+ 3H₂O(l) $\Delta$ H = -137 $\times$ 10³kJ for the Combustion of Ethyl Alcohol as Described