Respuesta :
Answer:
a. ΔH rxn (fermentation) = -68.9 kJ
ΔH rxn (respiration) = -2538.5 kJ
b. CH₃CH₂OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)
Explanation:
The enthalpy of reaction ([tex]\Delta H_{rxn}[/tex] can be calculated using the standard enthalpy of formation (ΔHf) of products and reactants as follows:
[tex] \Delta H_{rxn} = \Sigma n_{p} \Delta H_{p} - \Sigma n_{r} \Delta H_{r} [/tex] (1)
where ΔHp and ΔHr are the enthalpy of formation of the products and reactants, respectively, and np and nr are the stoichiometric coefficient of the product and reactants, respectively, from the balanced reaction.
a. The ΔH rxn of fermentation of sugar is:
C₆H₁₂O₆ → 2CO₂ + 2CH₃CH₂OH
ΔHf (C₆H₁₂O₆) = -1273.3 kJ/mole
ΔHf (CO₂) = -393.5 kJ/mole
ΔHf (CH₃CH₂OH) = -277.6 kJ/mole
Using equation (1):
[tex] \Delta H_{rxn} = 2 \Delta H_{f}(CO_{2}) + 2 \Delta H_{f}(CH_{3}CH_{2}OH) - \Delta H_{f}(C_{6}H_{12}O_{6}) [/tex]
[tex] \Delta H_{rxn} = 2(-393.5) + 2(-277.6) - (-1273.3) [/tex]
[tex] \Delta H_{rxn} = -68.9 kJ [/tex]
The ΔH rxn of respiration (combustion) of sugar is:
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O
ΔHf (H₂O) = -241.8 kJ/mole
[tex] \Delta H_{rxn} = 6 \Delta H_{f}(CO_{2}) + 6 \Delta H_{f}(H_{2}O) - \Delta H_{f}(C_{6}H_{12}O_{6}) [/tex]
[tex] \Delta H_{rxn} = 6 (-393.5) + 6 (-241.8) - (-1273.3) [/tex]
[tex] \Delta H_{rxn} = -2538.5 kJ [/tex]
b. The ΔH rxn of combustion of ethanol is:
CH₃CH₂OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)
[tex] \Delta H_{rxn} = 2 \Delta H_{f}(CO_{2}) + 3 \Delta H_{f}(H_{2}O) - \Delta H_{f}(CH_{3}CH_{2}OH) [/tex]
[tex] \Delta H_{rxn} = 2 (-393.5) + 3 (-241.8) - (-277.6) [/tex]
[tex] \Delta H_{rxn} = -1234.8 kJ [/tex]
We can see that the combustion of sugar releases more heat than the combustion of ethanol.
I hope it helps you!