Combustion of sucrose is used by aerobic organisms for providing energy for the life substaining process. If all the capturing of energy from the reaction is done through electrical process (non $P - V$ work) then calculate maximum available energy (in kJ) which can be captured by combustion of 34.2 gm of sucrose
Given: $Δ H_{c o m b u s t i o n} (s u c r o s e) = - 6000 k J m o l^{- 1}$
$Δ S_{c o m b u s t i o n} = 180 J K^{- 1} m o l^{- 1}$ and body temperature is $300 K$ .

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Solution

$Molar mass of sucrose (C_{12} H_{22} O_{11}) = 342 g / m o l$
Number of moles of sucrose $= \frac{34.2}{342} = 0.1$
$- (Δ G) =$ useful work done by the system
We know,
$- Δ G = - Δ H + T \times Δ S$
$= (6000 \times 0.1) + \frac{180 \times 0.1 \times 300}{1000}$
$= 605.4 k J$

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Similar questions

Δ H_{c o m b u s t i o n} (s u c r o s e) = 6000 {kJmol}^{- 1}

Δ S_{c o m b u s t i o n} = 180 J/K mol

34.2 g m

Δ H_{c o m b u s t i o n} (s u c r o s e) = - 6000 k J . {m o l}^{- 1}

Δ S_{c o m b u s t i o n} = 180 J / K . m o l

300 K

Δ H_{c o m b u s t i o n} (s u c r o s e) = 6000 {kJmol}^{- 1}

Δ S_{c o m b u s t i o n} = 180 J/K mol

m o l^{- 1}

m o l^{- 1}

△ H_{\int}^{^{\circ}}

H_{2}

△ H_{c o m}

- 6000 k j m o l^{- 1}

V_{i n h a l e d a i r} = V_{e x h a l e d a i r}, P_{a t m} = 1 a t m]

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