Question

The combustion of sucrose is used by aerobic organisms to provide energy to life sustaining processes. If all the energy capture is done by electrical processes, (non P-V work) then calculate the maximum available energy which can be captured by the combustion of 34.2 gm of sucrose.
Given: $\Delta H_{combustion}\left(sucrose\right)=6000{\text{kJmol}}^{-1}$ $\Delta S_{combustion}=180\text{J/K mol}$ and the body temperature is 300 K.

• A. 600 kJ
• B. 594.6 kJ
• C. 5.4 kJ
• D. 605.4 kJ

Answer 1

The correct option is D 605.4 kJ

Moles of sucrose$\left(C_{1}2H_{2}2O_{1}1\right)=34.2/342=0.1$
Enthalpy of combution of sucrose is $6000kJmo{l}^{-1}$
$\Delta H=-6000\times 0.1=-600kJl$
$\Delta S=180\times 0.1=18J$
we know that non-mechanical or non p-v work:
$\Delta G=\Delta H-T\Delta S$
$\Delta G=-600-\frac{300\times 18}{1000}=-605.4kJ$