Question

the thermochemical equation that represents the molar enthalpy of formation for ethanol is a c₂h₅oh(l) + 277.1 kj → 2c(s) + 3h₂(g) + 1/2 o₂(g) b c₂h₅oh(l) + 3o₂(g) → 2co₂(g) + 3h₂o(g) + 1235.3 kj c 2c(s) + 3h₂(g) + 1/2 o₂(g) → c₂h₅oh(l) + 277.1 kj d c₂h₅oh(l) + h₂o(l) ⇌ c₂h₅o⁻(aq) + h₃o⁺(aq) k = 1.3×10⁻¹⁶ my answer, including all work:

Explanation:

Step1: Recall definition of enthalpy of formation

The enthalpy of formation ($\Delta H_f$) is the change in enthalpy when one - mole of a substance is formed from its constituent elements in their standard states.

Step2: Analyze the reaction for ethanol formation

The reaction for the formation of ethanol ($C_2H_5OH$) from its elements in their standard states is $2C(s)+3H_2(g)+\frac{1}{2}O_2(g)
ightarrow C_2H_5OH(l)$. The enthalpy change for this reaction is the enthalpy of formation of ethanol.

Step3: Identify the correct equation

Among the given options, the equation that represents the formation of one - mole of ethanol from its elements in their standard states is $2C(s)+3H_2(g)+\frac{1}{2}O_2(g)
ightarrow C_2H_5OH(l)$ with an enthalpy change of $- 277.1\ kJ$. This corresponds to option A.

Answer:

A. $C_2H_5OH(l)+277.1\ kJ
ightarrow2C(s)+3H_2(g)+\frac{1}{2}O_2(g)$ (re - written in the form showing the formation of ethanol from its elements with the correct enthalpy value)