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1 gm of graphite is burnt in a bomb calorimeter in excess of oxygen at 298K and 1 atm pressure. During the reaction, temperature rises from 298 K to 299 k. If heat capacity of the bomb calorimeter is 20 kJ/K/gm
$∆ H$ for burning of 1 gm of graphite at 298K and 1 atm is

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240kJ

-20 kJ

-1.66kJ

-240kJ

answer is D.

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Detailed Solution

$q = ∆ E = - C_{v} ∆ T$ for combustion of 1 gm of carbon

$q = - 20 (299 - 298)$ = -20kJ

since $∆ n_{g}$ =0 for 1gm of graphite combustion both $∆ H; ∆ E$ are equal (and also for 1 mole also both are equal)

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