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Gibbs Free Energy & Spontaneity

The standard ...

Question

The standard enthalpies of formation of $C O_{2} (g), H_{2} O (l)$ and glucose(s) at $25^{\circ} C$ are – 400 kJ/mol, - 300 kJ/mol and – 1300 kJ/mol, respectively. The standard enthalpy of combusion per gram of glucose at $25^{\circ} C$ is

+ 2900kJ

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– 2900 kJ

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– 16.11 kJ

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+ 16.11 kJ

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Solution

The correct option is C – 16.11 kJ
$Δ_{c} H^{\circ}$ (Standard heat of combustion) is the standard enthalpy change when one mole of the substance is completely oxidised.
Also standard heat of formation $(Δ_{f} H^{\circ})$ can be taken as the standard of that substance.
$H_{C O_{2}}^{\circ} = Δ_{F} H^{\circ} (C O_{2}) = - 400 k J m o l^{- 1}$
$H_{H_{2} O}^{\circ} = Δ_{f} H^{\circ} (H_{2} O) = 300 k J m o l^{- 1}$
$H_{g l u c o s e}^{\circ} = Δ_{f} H^{\circ} (g l u c o s e) = - 1300 k J m o l^{- 1}$
$H_{O_{2}}^{\circ} = Δ_{f} H^{\circ} (O_{2}) = 0.00$
$C_{6} H_{12} O_{6} (s) + 6 O_{2} (g) \to 6 C O_{2} (g) + 6 H_{2} O (l)$
$Δ_{C} H^{\circ} (g l u c o s e) = 6 [Δ_{f} H^{\circ} (C O_{2}) + Δ_{f} H^{\circ} (H_{2} O)]$
$- [Δ_{f} H^{\circ} (C_{6} H_{12} O_{6}) + 6 Δ_{f} H^{\circ} (O_{2})]$
$= - 2900 k J m o l^{- 1}$
Molar mass of $C_{6} H_{12} O_{6} = 180 g m o l^{- 1}$
Thus,standard heat of combustion of glucose per gram
$= \frac{- 2900}{180} = - 16.11 k J g^{- 1}$
To solve such heat of combustion of glucose per gram
$= \frac{- 2900}{180} = - 16.11 k J g^{- 1}$
To solve such problem , students are advised to keep much importance in unit conversion. As here, value of R
$(8.314 J K^{- 1} m o l^{- 1}) i n J K^{- 1} m o l^{- 1}$ must be converted into kJ by dividing the unit by 1000.

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