Question

The enthalpy of formation of liquid ethanol (C2H5OH) is −277.6 kJ/mol. What is the equation that represents the formation of liquid ethanol? A. 2 C(s) + 6 H(g) + O(g) → C2H5OH(l) B. 2 C(s) + 3 H2(g) + ½ O2(g) → C2H5OH(l) C. 2CO2(g) + 3H2O(g) → C2H5OH(l) + 3 O2(g) D. 4 C(s) + 6 H2(g) + O2(g) → 2 C2H5OH(l)

Answer 1

b.
`2 C(s) + 3 H_2(g) + (1)/(2) O_2(g) \rightarrow C_2H_5OH(l)`

Step-by-step 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:

`2 C(s) + 3 H_2(g) + (1)/(2) O_2(g) \rightarrow C_2H_5OH(l)`

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

Best regards.

Answer 2

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

Step-by-step 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