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A sample of ${CH}_{4} of 0.08 g$ was subjected to combustion at 27 $°$ C in a bomb calorimeter. The temperature of the calorimeter system was found to be raised by 0.25 C . If heat capacity of calorimeter is 18kJ, $∆ H$ for combustion of ${CH}_{4}$ at 27 $°$ C is

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−905kJ/mole

-900kJ/mole

−895kJ/mole

−890kJ/mole

answer is B.

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Detailed Solution

${CH}_{4} + 2 O_{2} \to {CO}_{2} + 2 H_{2} O + x . cal$

T=300 g = 18KJ

Q=0.25 m=0.08

$∆ H = 0.25 \times 18 \times \frac{16}{0.08}$

$∆ H = - 900 KJ$

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