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Four grams of graphite is burnt in bomb calorimeter of heat capacity 30 KJ K^-1 in excess of Oxygen at 1 atm pressure. The temperature rises from 300 to 304 K. The heat of reaction at constant volume in $KJ / mole$ is

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– 720

– 400

– 1440

– 360

answer is B.

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Detailed Solution

$Δ U = \frac{- Z \times θ \times M}{m}$
$\frac{- 30 \times 4 \times 12}{4} = - 360 KJ / mole$

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