Question

The heat of combustion of sucrose, ${{\text{C}}_{{\text{12}}}}{{\text{H}}_{{\text{22}}}}{{\text{O}}_{{\text{11}}}}\left( {\text{s}} \right)$ at constant volume is $- 1348.9{\text{ kcal mo}}{{\text{l}}^{ - 1}}$ at ${25^ \circ }{\text{C}}$, then the heat of reaction at constant pressure, when steam is produced is:
A.$- 1342.334{\text{ kcal}}$
B.$+ 1342.334{\text{ kcal}}$
C.$+ 1250{\text{ kcal}}$
D.None of the above

Answer 1

The amount of heat evolved when one mole of any substance is burned in oxygen at a constant volume is known as the heat of combustion. The amount of heat released or absorbed when a chemical reaction occurs is known as the heat of reaction.

Complete step by step answer:
Step 1: Write the combustion reaction of sucrose is as follows:
Combustion means burning of any substance in presence of oxygen. Thus,
${{\text{C}}_{{\text{12}}}}{{\text{H}}_{{\text{22}}}}{{\text{O}}_{{\text{11}}}}\left( {\text{s}} \right) + {\text{12}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{12C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) + {\text{11}}{{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{g}} \right)$
Use the reaction to calculate the change in number of moles of gases.
Calculate the change in number of moles of gases as follows:
$\Delta {n_{\text{g}}} = {n_{{\text{g}}\left( {{\text{products}}} \right)}} - {n_{{\text{g}}\left( {{\text{reactants}}} \right)}}$
Where, $\Delta {n_{\text{g}}}$ is the change in number of moles of gases,
${n_{{\text{g}}\left( {{\text{products}}} \right)}}$ is the number of moles of gaseous products,
${n_{{\text{g}}\left( {{\text{reactants}}} \right)}}$ is the number of moles of gaseous reactants.
Thus, from the reaction,
$\Delta {n_{\text{g}}} = \left[ {\left( {12 + 11} \right) - 12} \right]{\text{ mol}}$
$\Delta {n_{\text{g}}} = 11{\text{ mol}}$
Thus, the change in number of moles of gases is ${\text{11 mol}}$.
Calculate the heat of the reaction at constant pressure using the equation as follows:
$\Delta H = \Delta E + \Delta {n_{\text{g}}}RT$
Where $\Delta H$ is the heat of the reaction at constant pressure,
$\Delta E$ is the heat of combustion at constant volume,
$\Delta {n_{\text{g}}}$ is the change in number of moles of gases,
$R$ is the universal gas constant,
$T$ is the temperature.
Substitute $- 1348.9{\text{ kcal}}$ for the heat of combustion at constant volume, ${\text{11 mol}}$ for the change in number of moles of gases, $2 \times {10^{ - 3}}{\text{ kcal }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$ for the universal gas constant, ${25^ \circ }{\text{C}} + 273 = 298{\text{ K}}$ for the temperature. Thus,
$\Delta H = \left( { - 1348.9{\text{ kcal}}} \right) + {\text{11 mol}} \times 2 \times {10^{ - 3}}{\text{ kcal }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}} \times 298{\text{ K}}$
$\Delta H = \left( { - 1348.9{\text{ kcal}}} \right) + \left( {6.556{\text{ kcal}}} \right)$
$\Delta H = - 1342.334{\text{ kcal}}$
Thus, the heat of reaction at constant pressure is $- 1342.334{\text{ kcal}}$.
Thus, the correct option is option (A).

Note:
The heat of the reaction has a negative value. The negative sign indicates that heat is released in the reaction. Thus, the reaction is an exothermic reaction.