The values of heat of combustion of graphite and $H_{2}$ are -395 and -269 kJ/mol respectively. If heat of formation of glucose is -1169 kJ/mol, then the heat of combustion of glucose is:

-2815 kJ/mol

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-1169 kJ/mol

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1169 kJ/mol

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2815 kJ/mol

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Solution

The correct option is B -2815 kJ/mol
(i) $C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g)$
(ii) $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$
(iii) $6 C (s) + 6 H_{2} (g) + 3 O_{2} (g) \to C_{6} H_{12} O_{6}$
Now,
6 X (i) + 6 X (ii) - (iii)
$C_{6} H_{12} O_{6} + 6 O_{2} \to 6 C O_{2} + 6 H_{2} O$
$Δ H = 6 \times (- 395) + 6 \times (- 269)$ $+ 1169$ $= - 2815$ kJ/mol

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