The enthalpy of formation of liquid ethanol (c2h5oh) is −277.6 kj/mol. what is the equation that represents

B. 2 C(s) + 3 H₂(g) + ½ O₂(g) → C₂H₅OH(l)

Explanation:

The enthalpy of formation refers to the formation of 1 mole of a substance from the simple substances in their most stable states. The thermochemical equation that represents this reaction is:

2 C(s) + 3 H₂(g) + ½ O₂(g) → C₂H₅OH(l)     ΔH°f = -277.6kJ/mol

b. [tex]2 C(s) + 3 H_2(g) + \frac{1}{2} O_2(g) \rightarrow C_2H_5OH(l)[/tex]

Explanation:

Hello,

In this case, formation reactions are said to be undergone when only the elements constituting the involved compounds, react in order to yield it in a combination chemical reaction. In such a way, for ethanol, carbon, hydrogen and oxygen must react to form it. In addition, it is mandatory to remember that hydrogen and oxygen only exist diatomic as pure gaseous elements. Moreover, the reaction must be properly balanced therefore, ethanol's formation reaction turns out:

[tex]2 C(s) + 3 H_2(g) + \frac{1}{2} O_2(g) \rightarrow C_2H_5OH(l)[/tex]

Which have an enthalpy of formation of -277.6 kJ/mol (exothermic reaction).

Best regards.

The enthalpy of formation has a negative sign, which means that the reaction is exothermic. In other words, when ethanol is formed, heat is released to the surroundings as a by-product. The reaction for the formation of ethanol is written below:

C₂H₄ + H₂O → C₂H₅OH