The daily energy requirement of a teenager is $7800 k J$ . As calculated from the data given in the table below, the amount of glucose he has to consume (g) per day assuming that the entire energy he requires comes from the combustion of glucose is:
Molecules $Δ H_{f} (k J m o l^{- 1})$
$C_{6} H_{12} O_{6}$ $- 1273$
$C O_{2} (g)$ $- 394$
$H_{2} O$ $- 286$

262

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500

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131

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250

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Solution

The correct option is A $500$
Combustion of glucose follows the reaction,
$C_{6} H_{12} O_{6} + 6 O_{2} \to 6 C O_{2} + 6 H_{2} O$
$H e a t o f c o m b u s t i o n = Σ (H e a t o f f o r m a t i o n o f p r o d u c t s) - Σ (H e a t o f f o r m a t i o n o f r e a c t a n t s)$
$Δ {_{c} H}_{C_{6} H_{12} O_{6}} = [6 \times (- 394) + 6 \times (- 286)] - (- 1273) k J m o l^{- 1}$
$Δ {_{c} H}_{C_{6} H_{12} O_{6}} = - 4080 + 1273 = - 2807 k J m o l^{- 1}$
So, 1 mole of glucose (180g) produces energy $= 2807 k J$
Amount of glucose which produces 7800 kJ energy $= \frac{180}{2807} \times 7800 = 500 g$

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Q. Combustion of glucose takes place according to the equation.

C_{6} H_{12} O_{6} + 6 O_{2} \to 6 C O_{2} + 6 H_{2} O △ H = 0 = - 72 k - c a l

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Molecules	$Δ H_{f} (k J m o l^{- 1})$
$C_{6} H_{12} O_{6}$	$- 1273$
$C O_{2} (g)$	$- 394$
$H_{2} O$	$- 286$