Question

The heat of combustion of ethanol determined in a bomb calorimeter is $-670.48kcal$ ${mol}^{-1}$ at $298K$. What is the $\Delta U$ at $298K$ for the reaction?

• A. $-760kcal$ ${mol}^{-1}$
• B. $-670.48kcal$ ${mol}^{-1}$
• C. $+760kcal$ ${mol}^{-1}$
• D. $+670.48kcal$ ${mol}^{-1}$

Answer 1

The correct option is B $-670.48kcal$ ${mol}^{-1}$

Bomb calorimeter works on a basic principle that involves combustion of the organic compound at constant volume constraint.
Since, at constant volume the heat of combustion of compound = change in internal energy of that compound (as pressure-volume work will be zero).
Hence, $\Delta U$ $=\text{heat of combustion in a bomb calorimeter}=-670.48{\text{Kcal mol}}^{-1}.$
Hence, answer is option B.