Question

A 0.692 -g sample of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), is burned in a constant-volume calorimeter. The temperature rises from \(21.70{ }^{\circ} \mathrm{C}\) to \(25.22{ }^{\circ} \mathrm{C}\). The calorimeter contains \(575 \mathrm{~g}\) water, and the bomb has a heat capacity of \(650 \mathrm{~J} / \mathrm{K}\). Determine \(\Delta_{t} E\) per mole of glucose.

Short answer

The energy change per mole of glucose is approximately \(-2794.40\, \text{kJ/mol}.\)

Step-by-step solution

Calculate the temperature change

To find the temperature change of the system, subtract the initial temperature from the final temperature. The temperatures given are 25.22°C and 21.70°C.\[\Delta T = 25.22\, ^{\circ} \mathrm{C} - 21.70\, ^{\circ} \mathrm{C} = 3.52\, ^{\circ} \mathrm{C}\]

Calculate the heat absorbed by the water

Use the formula for heat, \(q = m \cdot c \cdot \Delta T\), where \(m\) is the mass of the water, \(c\) is the specific heat capacity of water (4.184 J/g·K), and \(\Delta T\) is the temperature change.\[q_{\text{water}} = 575 \text{ g} \times 4.184 \text{ J/g·K} \times 3.52 \text{ K} = 8451.584 \text{ J}\]

Calculate the heat absorbed by the bomb

The heat absorbed by the bomb can be calculated using its heat capacity and the temperature change.\[q_{\text{bomb}} = 650 \text{ J/K} \times 3.52 \text{ K} = 2288 \text{ J}\]

Calculate the total heat released by the combustion

The total heat released, \(q_{\text{total}}\), is the sum of the heat absorbed by the water and the bomb.\[q_{\text{total}} = q_{\text{water}} + q_{\text{bomb}} = 8451.584 \text{ J} + 2288 \text{ J} = 10739.584 \text{ J}\]

Determine the moles of glucose

The molar mass of glucose \(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6\) is calculated as follows:- Carbon: 6 atoms \(\times 12.01\, \text{g/mol} = 72.06\, \text{g/mol}\)- Hydrogen: 12 atoms \(\times 1.01\, \text{g/mol} = 12.12\, \text{g/mol}\)- Oxygen: 6 atoms \(\times 16.00\, \text{g/mol} = 96.00\, \text{g/mol}\)The total molar mass is:\[ \text{Molar mass of glucose} = 72.06 + 12.12 + 96.00 = 180.18\, \text{g/mol} \]Convert the mass of glucose to moles:\[ n = \frac{0.692\, \text{g}}{180.18\, \text{g/mol}} = 0.003842\, \text{mol}\]

Calculate the energy change per mole of glucose

Divide the total heat by the number of moles to find the energy change per mole of glucose:\[\Delta_{t} E = \frac{10739.584 \text{ J}}{0.003842 \text{ mol}} = 2794397.654 \text{ J/mol} \approx -2794.40\, \text{kJ/mol}\]The energy change is negative because the reaction is exothermic, as heat is released when glucose is burned.

Key concepts

Calorimetry

Calorimetry is the science of measuring heat energy transferred in or out of a chemical system. It helps us understand how substances store and release energy during physical or chemical changes. In a calorimeter, the heat exchanges are isolated, allowing us to analyze the temperature changes due to reactions or physical transformations.

In the exercise above, a constant-volume bomb calorimeter is used. Here, glucose is burned, and the goal is to calculate the energy change involved in this process. By measuring how much the water and the bomb heated up, we can back-calculate to find the heat released by the glucose combustion. The system ensures no heat escapes into the surroundings, providing a clear picture of the energy changes occurring due to the reaction.

Calorimetry is a valuable tool in chemistry for studying thermodynamics and calculating reaction enthalpies.

Heat Capacity

Heat capacity is an important concept in thermodynamics that tells us how much heat is required to change the temperature of an object by one degree Celsius or Kelvin. The heat capacity depends on the material's mass and its specific heat capacity. Specific heat capacity is a property that describes how much heat a material can hold per unit mass for a one-degree temperature change.

In the example given, the heat capacity of the bomb is crucial in calculating the total heat exchanged. The bomb has a fixed heat capacity of 650 J/K, meaning that for every degree the temperature increases, it absorbs 650 joules of energy.

• For water, the specific heat capacity is 4.184 J/g·K.
• We multiply this value by the water's mass and the temperature change to find how much heat the water absorbed. This is calculated independently of the bomb's heat absorption.

This separation is crucial to determine the total energy changes happening in the calorimeter.

Temperature Change

The temperature change in a calorimetry experiment helps us quantify the heat transferred during a reaction. By observing how much the temperature rises, we can understand the amount of energy being released or absorbed.

The original exercise provides measurements of temperature increase within the calorimeter. This temperature change, \( \Delta T = 25.22\, ^{\circ} C - 21.70\, ^{\circ} C = 3.52\, ^{\circ} C \), is key to calculating the heat absorbed by each component of the calorimetry setup.

Knowing the initial and final temperatures, the temperature change allows us to use formulas relating heat transfer to mass and specific heat capacity. This direct relationship between temperature change and energy makes calorimetry a powerful technique for determining reaction energetics.

Molar Mass Calculation

Molar mass calculation is essential when converting mass into moles for chemical equations and calculations. The molar mass of a compound is the sum of the masses of its individual atoms found on the periodic table.

In the exercise provided, glucose (\( \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 \)) is combusted. To calculate its molar mass, we consider:

• Carbon: 6 atoms \( \times 12.01\, \text{g/mol} = 72.06\, \text{g/mol} \)
• Hydrogen: 12 atoms \( \times 1.01\, \text{g/mol} = 12.12\, \text{g/mol} \)
• Oxygen: 6 atoms \( \times 16.00\, \text{g/mol} = 96.00\, \text{g/mol} \)

Thus, the molar mass totals 180.18 g/mol. This comprehensive breakdown helps convert the mass of glucose used into moles, allowing the calculation of the energy change per mole of glucose. Because chemical reactions depend on mole ratios, understanding molar mass is critical for these thermodynamic calculations.