The standard enthalpies of formation of ${C O}_{2} (g), H_{2} O (l)$ and glucose(s) at $25^{o} C$ are $- 400 k J / m o l, - 300 k J / m o l$ and $- 1300 k J / m o l$ respectively. The standard enthalpy of combustion per gram of glucose at $25^{o} C$ is:

+ 2900 k J

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- 2900 k J

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- 16.11 k J

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+ 16.11 k J

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Solution

The correct option is B $- 16.11 k J$
$C_{6} H_{12} O_{6} (s) + 6 O_{2} (g) \to 6 {C O}_{2} + 6 H_{2} O (l)$
$Δ H = [6 Δ_{f} H_{{C O}_{2}} + Δ_{f} H_{H_{2} O}] - [Δ_{f} H_{C_{6} H_{12} O_{6}} + 6 Δ_{f} H_{O_{2}}]$
$= [6 (- 400) + 6 (- 300)] - [- 1300 + 0]$
$= - 4200 + 1300 = - 2900 k J / m o l$
Enthalpy oc combustion per gram $= \frac{- 2900}{180} = - 16.11 k J / g$

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