Combustion of table sugar produces \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) When \(1.46 \mathrm{~g}\) table sugar is combusted in a constant-volume (bomb) calorimeter, \(24.00 \mathrm{~kJ}\) of heat is liberated. a. Assuming that table sugar is pure sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)\), write the balanced equation for the combustion reaction. b. Calculate \(\Delta E\) in \(\mathrm{kJ} / \mathrm{mol} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) for the combustion reaction of sucrose. c. Calculate \(\Delta I I\) in \(\mathrm{kJ} / \mathrm{mol} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) for the combustion reaction of sucrose at \(25^{\circ} \mathrm{C}\).

To write the balanced combustion equation for sucrose, we must remember that combustion reactions involve combining a hydrocarbon (sucrose, in this case) with oxygen gas to produce carbon dioxide and water. The unbalanced equation is: \[C_{12}H_{22}O_{11}(s) + O_{2}(g) \rightarrow CO_{2}(g) + H_{2}O(l)\] Now we need to balance the equation. We have 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms in sucrose. To conserve the number of atoms on both sides, the balanced equation is: \[C_{12}H_{22}O_{11}(s) + 12O_{2}(g) \rightarrow 12CO_{2}(g) + 11H_{2}O(l)\] #step_2#

We are given the mass (1.46 g) of sucrose, so to convert it to moles, we need the molar mass of sucrose, which is: \[M = (12\times12.01)+(22\times1.008)+(11\times16.00) = 342.30\, g/mol\] Now we can solve for the number of moles of sucrose: \[n = \frac{mass}{molar\,mass} = \frac{1.46\,g}{342.30\,g/mol} = 0.00426\,mol\] #step_3#

We are given the heat liberated (24.00 kJ) from the combustion of sucrose in a bomb calorimeter. The heat liberated (\(q_v\)) in a constant volume process is equal to the change in internal energy (\(\Delta E = q_v\)). So, we can calculate the amount of heat in kJ/mol, as follows: \[\Delta E = \frac{q_v}{n} = \frac{24.00\,kJ}{0.00426\,mol} = 5633\,kJ/mol\] #step_4#

To calculate the change in enthalpy (\(\Delta H\)), we will use the following equation: \[\Delta H = \Delta E + \Delta (n \times RT)\] where \(\Delta (n \times RT)\) is the difference between products and reactants in moles multiplied by the ideal gas constant (\(R = 8.314 J/mol \cdot K\)) and the temperature (\(T\)) in Kelvin. In our case, the temperature is given as \(25^{\circ}C\), which is equivalent to 298 K. Let's first determine \(\Delta (n \times RT)\). The balanced equation shows 12 moles of CO2 (gas) are produced and 11 moles of H2O (liquid) and 12 moles of O2 (gas) are consumed: \[\Delta (n \times RT) = [(12 - 12) \times 8.314\,J/mol\cdot K \times 298\,K]\] The term \(\Delta (n \times RT)\) equals 0 because the number of moles of gaseous reactants equals the number of moles of gaseous products. Therefore, the change in enthalpy (\(\Delta H\)) is the same as the change in internal energy (\(\Delta E\)): \[\Delta H = \Delta E = 5633\,kJ/mol\] In summary: a. The balanced equation for the combustion of sucrose is: \[C_{12}H_{22}O_{11}(s) + 12O_{2}(g) \rightarrow 12CO_{2}(g) + 11H_{2}O(l)\] b. The change in internal energy (\(\Delta E\)) for the combustion reaction is 5633 kJ/mol. c. The change in enthalpy (\(\Delta H\)) for the combustion reaction is also 5633 kJ/mol.