Solveeit Logo
Login

Question

Question: Calculate the enthalpy of combustion of methyl alcohol at \[298{\text{ K}}\] from the following data...

Calculate the enthalpy of combustion of methyl alcohol at 298 K298{\text{ K}} from the following data

BondC−H{\text{C}} - {\text{H}}C−O{\text{C}} - {\text{O}}O−H{\text{O}} - {\text{H}}O=O{\text{O}} = {\text{O}}C=O{\text{C}} = {\text{O}}
Bond Enthalpy (kJ/mol)\left( {{\text{kJ/mol}}} \right)414351.5464.5494711

Resonance energy of CO2 = −143 kJ/mol{\text{C}}{{\text{O}}_{\text{2}}}{\text{ = }} - {\text{143 kJ/mol}}
Latent heat of evaporation of methyl alcohol = 35.5 kJ/mol = {\text{ }}35.5{\text{ kJ/mol}}
Latent heat of evaporation of water = 40.6 kJ/mol = {\text{ }}40.6{\text{ kJ/mol}}
A −166.7 kJ/mol - 166.7{\text{ kJ/mol}}
B −659.7 kJ/mol - 659.7{\text{ kJ/mol}}
C −136.7 kJ/mol - 136.7{\text{ kJ/mol}}
D −696.9 kJ/mol - 696.9{\text{ kJ/mol}}

Explanation

Solution

To solve this problem, you need to apply the Hess Law of constant heat summation. Divide the entire process into individual steps and calculate the energy change associated with each individual step by using the given data. Then add the energy changes for these individual steps to calculate the overall energy change.

Complete step by step answer: Latent heat of evaporation of methyl alcohol = 35.5 kJ/mol = {\text{ }}35.5{\text{ kJ/mol}}
Write an equation that shows evaporation of ethanol
CH3OH(l) → CH3OH(g) ΔH1= 35.5 kJ/mol{\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}\left( l \right){\text{ }} \to {\text{ C}}{{\text{H}}_{\text{3}}}{\text{OH}}\left( g \right){\text{ }}\Delta {H_1} = {\text{ }}35.5{\text{ kJ/mol}}
Write a balanced chemical equation for the combustion reaction of methyl alcohol.
CH3OH(g) + 2 O2→ CO2 + 2 H2O{\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}\left( g \right){\text{ + 2 }}{{\text{O}}_2} \to {\text{ C}}{{\text{O}}_{\text{2}}}{\text{ + 2 }}{{\text{H}}_2}{\text{O}}
One mole of methanol reacts with two moles of oxygen to produce one mole of carbon dioxide and two moles of water
Calculate the bond energy of reactants
3HC - H + HC - O+HO - H+2HO = O=3(414)+351.5+464.5+2(494)=3046 kJ/mol{\text{3}}{{\text{H}}_{{\text{C - H}}}}{\text{ + }}{{\text{H}}_{{\text{C - O}}}} + {{\text{H}}_{{\text{O - H}}}} + 2{{\text{H}}_{{\text{O = O}}}} = 3\left( {414} \right) + 351.5 + 464.5 + 2\left( {494} \right) = 3046{\text{ kJ/mol}}
Calculate the bond energy of products
2HC = O− resonance energy of CO2+4HO - H=2(711)−143+4(464.5)=3137 kJ/mol{\text{2}}{{\text{H}}_{{\text{C = O}}}} - {\text{ resonance energy of C}}{{\text{O}}_2} + 4{{\text{H}}_{{\text{O - H}}}} = 2\left( {711} \right) - 143 + 4\left( {464.5} \right) = 3137{\text{ kJ/mol}}
From the bond energy of reactants, subtract the bond energy of products to obtain the energy change associated with combustion of methanol vapours.
ΔH2=3046 kJ/mol−3137 kJ/mol=−91 kJ/mol\Delta {H_2} = 3046{\text{ kJ/mol}} - 3137{\text{ kJ/mol}} = - 91{\text{ kJ/mol}}
Latent heat of evaporation of water = 40.6 kJ/mol = {\text{ }}40.6{\text{ kJ/mol}}
During the combustion of one mole of methyl alcohol, two moles of water are formed.
Calculate the amount of energy needed for the vaporization of two moles of water
ΔH3=2× 40.6=81.2 kJ/mol\Delta {H_3} = 2 \times {\text{ }}40.6 = 81.2{\text{ kJ/mol}}

Calculate the enthalpy of combustion of methyl alcohol at 298 K298{\text{ K}}

⇒ΔH=ΔH1+ΔH2−ΔH3 ⇒ΔH=35.5−91−81.2 ⇒ΔH=−136.7 kJ/mol  \Rightarrow \Delta H = \Delta {H_1} + \Delta {H_2} - \Delta {H_3} \\\ \Rightarrow \Delta H = 35.5 - 91 - 81.2 \\\ \Rightarrow \Delta H = - 136.7{\text{ kJ/mol}} \\\

Hence, the correct option is the option C −136.7 kJ/mol - 136.7{\text{ kJ/mol}}.

Note: To break a bond, you need to supply the energy. During bond formation, energy is released. In a chemical reaction, you can calculate the enthalpy change by subtracting the total bond energy of products from the total bond energy of reactants.