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Combustion

The standard ...

Question

The standard enthalpy of combustion of sucrose is-5645 kJ $m o l^{- 1}$ .What is the advantage (in kJ $m o l^{- 1}$ of energy released as heat) of complete aerobic oxidation compared to anaerobic hydrolysis of sucrose to lactic acid? $△ H_{\int}^{^{\circ}}$ for lactic acid,CO_{2} and $H_{2}$ O is -694,-395 and -286.0 respectively

A

advantage=4356 kJ.

m o l^{- 1}

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B

advantage=5396 kJ.

m o l^{- 1}

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C

advantage=4756 kJ.

m o l^{- 1}

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D

advantage=5657 kJ.

m o l^{- 1}

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Solution

The correct option is B advantage=5396 kJ. $m o l^{- 1}$
Sucrose $= C_{12} H_{22} O_{11}$
Lactic acid $= C_{3} H_{6} O_{3} = L A$
Given, $C + O_{2} (g) \to {C O}_{2} (g)$ $Δ_{f} H_{{C O}_{2}} = - 395 k J / m o l$ ...(a)
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)$ $Δ_{f} H_{H_{2} O} = - 286 k J / m o l$ ...(b)
$3 C + 3 H_{2} + \frac{3}{2} O_{2} \to C_{3} H_{6} O_{3}$ $Δ_{f} H_{L A} = - 694 k J / m o l$ ...(c)
Now, multiply (a) by $12$ and (b) by $11$ and add both to get
$12 C + 11 H_{2} + \frac{35}{2} O_{2} \to 12 C O_{2} + 11 H_{2} O$ $Δ_{f} H = - 7886 k J / m o l$
or $12 {C O}_{2} + 11 H_{2} O \to 12 C + 11 H_{2} + \frac{35}{2} O_{2}$ $Δ_{f} H = + 7886 k J / m o l$ ...(d)
Now combustion or aerobic oxidation reaction is
$C_{12} H_{22} O_{11} + 12 O_{2} \to 12 {C O}_{2} + 11 H_{2} O$ $Δ_{C} H = - 5645 k J / m o l$ ...(e)
(d) $+$ (e) gives
$C_{12} H_{22} O_{11} \to 12 C + 11 H_{2} + \frac{11}{2} O_{2}$ $Δ H = 2241 k J / m o l$
or $12 C + 11 H_{2} + \frac{11}{2} O_{2} \to C_{12} H_{22} O_{11}$ $Δ_{f} H_{s u c r o s e} = - 2241 k J / m o l$ ...(f)
This equation gives heat of formation of sucrose.
Now consider anaerobic hydrolysis of sucrose
$C_{12} H_{22} O_{11} + H_{2} O (g) \to {4 C}_{3} H_{6} O_{3}$ $Δ_{h} H = ?$
$Δ_{h} H = Δ_{f} H_{p r o d u c t s} - Δ_{f} H_{r e a c t a n t s}$
$Δ_{f} H_{L A} = - 694 k J / m o l$ (given)
so $Δ_{h} H = 4 (- 694) - [Δ_{f} H_{s u c r o s e} + Δ_{f} H_{H_{2} O}]$
$= 4 (- 694) - [- 2241 - 286] = - 249 k J / m o l$
So enthalpy change in anaerobic hydrolysis $= - 249 k J / m o l$
enthalpy change in aerobic oxidation $= - 5646 k J / m o l$
So advantage of aerobic oxidation to anaerobic hydrolysis $= - 249 - (- 5645) = 5396 k J / m o l$

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