Question

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 \(25^{\circ} \mathrm{C}\).

Short answer

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.

Step-by-step solution

Write the balanced combustion equation for sucrose.

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#

Calculate the number of moles of sucrose used.

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#

Calculate \(\Delta E\) (change in internal energy) for the combustion reaction of sucrose.

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#

Calculate \(\Delta H\) (change in enthalpy) for the combustion reaction of sucrose.

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.

Key concepts

balanced chemical equation

In a combustion reaction, a fuel reacts with oxygen to produce heat and light. Specifically, for the combustion of sucrose (table sugar), it involves reacting sucrose with oxygen to form carbon dioxide and water. To ensure that the number of atoms of each element is the same on both sides of the equation, the reaction needs to be balanced.
A balanced equation provides a clear snapshot of the chemical transformation. For sucrose, which has the molecular formula \( C_{12}H_{22}O_{11} \), the combustion reaction begins with identifying the unbalanced reaction:

• \( C_{12}H_{22}O_{11}(s) + O_{2}(g) \rightarrow CO_{2}(g) + H_{2}O(l) \)

The unbalanced equation is adjusted by altering stoichiometric coefficients, ensuring equal numbers of each type of atom on both sides. The balanced chemical equation for this reaction is:

• \( C_{12}H_{22}O_{11}(s) + 12O_{2}(g) \rightarrow 12CO_{2}(g) + 11H_{2}O(l) \)

This equation demonstrates conservation of mass and the stoichiometry required for the reaction. The equation helps us understand that one mole of sucrose requires twelve moles of oxygen to yield twelve moles of carbon dioxide and eleven moles of water.

change in internal energy

The change in internal energy, denoted as \( \Delta E \), reflects the energy change within a system during a reaction. At constant volume, \( \Delta E \) is directly represented by the heat exchanged, known as \( q_v \).
In the exercise, when sucrose combusts in a bomb calorimeter, which maintains a constant volume, the heat liberated provides the change in internal energy. For the combustion of 1.46 g of sucrose, 24.00 kJ of heat is released.
The molar internal energy change can be calculated by converting the mass of sucrose to moles using its molar mass (342.30 g/mol). This finds the moles involved in the reaction:

• \( n = \frac{1.46 \, g}{342.30 \, g/mol} \approx 0.00426 \, mol \)

We then divide the total heat liberated by the number of moles to find \( \Delta E \) per mole of sucrose:

• \( \Delta E = \frac{24.00 \, kJ}{0.00426 \, mol} = 5633 \, kJ/mol \)

This value, 5633 kJ/mol, indicates the energetic change when one mole of sucrose combusts at constant volume, signifying the system's total energy shift during the reaction.

change in enthalpy

Enthalpy change \( \Delta H \) represents the heat change at constant pressure and is vital in understanding the energy absorbed or released in a reaction. Despite maintaining constant volume in the exercise, the relationship between internal energy and enthalpy is essential.
For the combustion of sucrose, \( \Delta H \) can be calculated using \( \Delta H = \Delta E + \Delta (n \times RT) \). Here, \( \Delta (n \times RT) \) considers the change in moles of gas involved, multiplied by the ideal gas constant \( R \) and temperature \( T \).
Since the equation balance results in equal moles of gaseous reactants and products:

• \( \Delta(n \times RT) = [12 - 12] \times 8.314 \, J/mol \cdot K \times 298 \, K = 0 \)

This means that \( \Delta H \) is essentially equivalent to \( \Delta E \) in this reaction, making the enthalpy change 5633 kJ/mol. Thus, when considering constant pressure conditions, the heat released remains unchanged in this specific case, aligning \( \Delta H \) with \( \Delta E \).

constant-volume calorimetry

Constant-volume calorimetry is a method where a reaction occurs in a rigid, sealed container known as a bomb calorimeter. This apparatus helps measure the heat released or absorbed during chemical reactions while maintaining fixed volume conditions.
In this specific exercise, the calorimetry method focuses on the combustion of sucrose, capturing the heat that reflects changes in the system's internal energy. Since volume doesn't change, it simplifies calculations as the heat measured equals the change in internal energy, \( \Delta E \).
Key characteristics of constant-volume calorimetry include:

• A solid, non-expandable container.
• Measuring just energy change, without accounting for the effects of pressure shifts.
• Ideal for reactions that involve gases, where pressure changes do not affect the energy measurement directly.

This method was essential in determining the 24.00 kJ of heat released during sucrose combustion in the exercise. It straightforwardly relates heat with internal energy, offering a practical means to assess the thermodynamic properties of reactions.