Question

Heat of a reaction is defined as the amount of heat absorbed or evolved at a given temperature when the reactants have combined to form the products is represented by balanced chemical equation. If the heat is denoted by $q$ then the numerical value of $q$ depends on the manner in which
the reaction is performed for two methods of conducting chemical reactions in calorimeters.

When maltose $C_{12}{H}_{22}{O}_{11}\left(s\right)$ burns in a calorimetric bomb at 298 K yielding carbon dioxide and water, the heat of combustion is $-1350\text{kcal/mol}$, the heat of combustion of maltose at constant
pressure will be:

• A. −2650 kcal/mol
• B. −675 kcal/mol
• C. −1350 kcal/mol
• D. −1100 kcal/mol

Answer 1

The correct option is C −1350 kcal/mol

Point to remember:
$\Rightarrow$ heat of combustion at constant pressure can be calculated by
$q=\Delta E+P\Delta V$ or $q=\Delta E+\Delta n_{g}RT$
where $\Delta E=$ the heat of combustion in calorimeteric bomb.
Reaction:
$C_{12}{H}_{22}{O}_{11}\left(s\right)+12O_2\left(g\right)\to 12C{O}_2\left(g\right)+11H_{2}O\left(l\right)$
On putting values we get
$q=-1350+0=-1350kcal/mol$