Q10.76CP - Chemistry: Molecular Nature Of Matter And Change

Question

Ethanol ( ${\mathbf{CH}}_3{\mathbf{CH}}_2\mathbf{OH}$ ) is being used as a gasoline additive or alternative in many parts of the world.

1. Use bond energies to find  $\mathbf{\Delta H}{\mathbf{°}}_{\mathrm{rxn}}$ for the combustion of gaseous ethanol. (Assume  ${\mathbf{H}}_2\mathbf{O}$ forms as a gas.)
2. In its standard state at  $\mathbf{25}\mathbf{°}\mathbf{C}$ , ethanol is a liquid. Its vaporization requires 40.5 kJ/mol. Correct the value from part (a) to find the heat of reaction for the combustion of liquid ethanol.
3. How does the value from part (b) compare with the value you calculate from standard heats of formation (Appendix B)?
4. “Greener” methods produce ethanol from corn and other plant material, but the main industrial method involves hydrating ethylene from petroleum. Use Lewis structures and bond energies to calculate $\mathbf{\Delta H}{\mathbf{°}}_{\mathrm{rxn}}$  for the formation of gaseous ethanol from ethylene gas with water vapor. 

Step-by-Step Solution

Answer

1. ${\text{ΔH°}}_{\text{rxn}}$  for the combustion of gaseous ethanol is   $\text{-1267kJ/mol}$.
2.  ${\text{ΔH°}}_{\text{rxn}}$  for the combustion of liquid ethanol is   $\text{-1226kJ/mol}$
3. The difference in heat of combustion of gaseous and liquid ethanol is around  $\text{8.35 kJ}$ .
4.  ${\text{ΔH°}}_{\text{rxn}}$ for the formation of gaseous ethanol is  $\text{-37KJ}$.

Step 1: Definition

The heat of combustion is the change that occurs in enthalpy when one mole of a substance is entirely burnt in the air.

 

The amount of heat absorbed or evolved when one molecule of a substance is formed is called heat of formation.

Step 2: Subpart (a) The ΔH ° rxn for the combustion of gaseous ethanol.

Ethanol reacts with three molecules of oxygen and forms two molecules of carbon dioxide and three molecules of water. The combustion goes as,

The heat of combustion or heat of formation is calculated by adding bond energies of atoms present in molecule.

 

The energy required in bond breaking 

                                                  $\begin{array}{l}\text{= 1}\left(\text{C-C}\right)\text{+5}\left(\text{C-H}\right)\text{ + 1}\left(\text{C-O}\right)\text{ + 1}\left(\text{O-H}\right)\text{ + 3}\left(\text{O=O}\right)\text{ bond}\\ \text{= 347 + 5}\left(\text{413}\right)\text{ +358 + 467+ 3}\left(\text{498}\right)\text{ kJ/mol}\\ \text{= 4731 kJ/mol}\end{array}$

The energy released when bonds formed 

                                                             $\begin{array}{l}\text{= 4}\left(\text{C=O}\right)\text{ + 6}\left(\text{O-H}\right)\text{ Bonds}\\ \text{= 4}\left(\text{799}\right)\text{ + 6}\left(\text{467}\right)\text{ kJ}\\ \text{= 5998kJ}\end{array}$

 $\begin{array}{c}\text{The heat of combustion = bonds broken energy-bonds formed energy}\\ \text{ = }\left(\text{4731-5998}\right)\text{ kJ}\\ \text{ = -1267kJ/mol}\end{array}$

 

Therefore, the heat of combustion of ethanol is  $\text{-1267kJ/mol}$.

Step 3: Subpart (b) The heat of reaction for the combustion of liquid ethanol.

$\begin{array}{c}\text{ Heat of combustion of liquid ethanol =}\left(\begin{array}{c}\text{ heat of vaporization }\\ \text{+ heat of combustion of gaseous ethanol}\end{array}\right)\\ \text{ = 40.5}-\text{1267 }\\ \text{ = -1226.5 kJ/mol}\end{array}$

Therefore, the heat of reaction for the combustion of liquid ethanol is  $\text{-1226kJ/mol}$.

Step 4: Subpart (c) The value from part (b) compare with the value you calculate from standard heats of formation.

The calculation from standard values.

 

 $\begin{array}{c}{\text{Heat of combustion = 2(ΔH}}_{\text{f}}{\text{(CO}}_{\text{2}}{\text{)) + 3(ΔH}}_{\text{f}}{\text{(H}}_{\text{2}}{\text{O)) - (ΔH}}_{\text{f}}{\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH)) - (ΔH}}_{\text{f}}{\text{(O}}_{\text{2}}\text{))}\\ \text{ = 2(-393.5) + 3(-241.826) - (-277.63)- 0}\\ \text{ = -1234.848kJ}\end{array}$

                                

                                

The difference in the two answers is around  $\text{8.35 kJ}$ 

Thus, there is a good agreement between the two values and the values of energies used to calculate the answer.

Step 5: Subpart (d) The ΔH ° rxn for the formation of gaseous ethanol from ethylene gas with water vapor.

Energy required when bonds broken

                                                   $\begin{array}{l}\text{= 1}\left(\text{C=C}\right)\text{ + 4}\left(\text{C-H}\right)\text{ +2}\left(\text{O-H}\right)\text{ bonds}\\ \text{= 614+4}\left(\text{413}\right)\text{+2}\left(\text{467}\right)\text{\hspace{0.17em}kJ}\\ \text{= 3200 kJ}\end{array}$

 

Energy released when bonds formed 

                                               $\begin{array}{l}\text{=1 }\left(\text{C-C}\right)\text{+5}\left(\text{C-H}\right)\text{ +1}\left(\text{C-O}\right)\text{ +1}\left(\text{O-H}\right)\text{ bonds}\\ \begin{array}{l}\text{=}\left(\text{347}\right)\text{+5}\left(\text{413}\right)\text{ +}\left(\text{358}\right)\text{+}\left(\text{467}\right)\text{ kJ}\\ \text{= 3237kJ}\end{array}\end{array}$                  

 

Therefore, the heat of reaction 

 $\begin{array}{l}\text{=}\left(\text{3200 - 3237}\right)\text{kJ}\\ \text{=}\text{ -37kJ}\end{array}$