Question

Combustion of sucrose is used by aerobic organisms for providing energy for the life sustaining processes. If all the capturing of energy from the reaction is done through electrical process (non P–V work) then calculate maximum available energy which can be captured by combustion of 34.2 g of sucrose

Given :$\Delta$Hcombustion (sucrose) = – 6000 kJ mol–1

$\Delta$Scombustion = 180 J/K - mol and body temperature is 300 K

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

Answer 1

Answer: (D)

605.4 kJ

Answer 2

No. of moles of sucrose = $\frac{34.2}{342}$= 0.1

– ($\Delta$G)T.P = useful work done by the system

–$\Delta$G = – $\Delta$H + T. $\Delta$S = + (6000 × 0.1) +

$\frac{180 \times 0.1 \times 300}{1000}$ = 605.4 kJ.