Math Problem Statement

A bomb calorimeter is ideally suited for measuring the heat evolved during a combustion reaction. A sample of sucrose (C12H22O11(s)) in a bomb calorimeter causes the temperature of 345 mL of water to rise from 24.9 o C to 28.3 o C. The heat capacitiy of the calorimeter assembly is 4.90 kj/mol. What is the experimental molar enthalpy of combustion of sucrose in kj/mol?

Solution

To determine the experimental molar enthalpy of combustion of sucrose, we need to calculate the heat released by the combustion of sucrose and then relate this to the number of moles of sucrose combusted.

Step 1: Calculate the heat absorbed by the water.

The heat absorbed by the water (qwaterq_{\text{water}}) can be calculated using the formula:

qwater=mwaterCwaterΔTq_{\text{water}} = m_{\text{water}} \cdot C_{\text{water}} \cdot \Delta T

Where:

  • mwaterm_{\text{water}} is the mass of the water,
  • CwaterC_{\text{water}} is the specific heat capacity of water (4.18J/gC4.18 \, \text{J/g}^\circ \text{C}),
  • ΔT\Delta T is the change in temperature.

First, let's calculate ΔT\Delta T:

ΔT=28.3C24.9C=3.4C\Delta T = 28.3^\circ \text{C} - 24.9^\circ \text{C} = 3.4^\circ \text{C}

Given that the density of water is 1g/mL1 \, \text{g/mL}, the mass of the water is:

mwater=345mL×1g/mL=345gm_{\text{water}} = 345 \, \text{mL} \times 1 \, \text{g/mL} = 345 \, \text{g}

Now, calculate the heat absorbed by the water:

qwater=345g×4.18J/gC×3.4Cq_{\text{water}} = 345 \, \text{g} \times 4.18 \, \text{J/g}^\circ \text{C} \times 3.4^\circ \text{C}

qwater=4905.66J=4.91kJq_{\text{water}} = 4905.66 \, \text{J} = 4.91 \, \text{kJ}

Step 2: Calculate the total heat released by the combustion.

The total heat released (qtotalq_{\text{total}}) by the combustion is the sum of the heat absorbed by the water and the heat absorbed by the calorimeter assembly:

qtotal=qwater+Ccalorimeter×ΔTq_{\text{total}} = q_{\text{water}} + C_{\text{calorimeter}} \times \Delta T

Given the heat capacity of the calorimeter assembly:

qcalorimeter=4.90kJ/C×3.4C=16.66kJq_{\text{calorimeter}} = 4.90 \, \text{kJ/}^\circ \text{C} \times 3.4^\circ \text{C} = 16.66 \, \text{kJ}

So the total heat released is:

qtotal=4.91kJ+16.66kJ=21.57kJq_{\text{total}} = 4.91 \, \text{kJ} + 16.66 \, \text{kJ} = 21.57 \, \text{kJ}

Step 3: Determine the molar enthalpy of combustion.

Finally, to find the molar enthalpy of combustion (ΔHcomb\Delta H_{\text{comb}}), we divide the total heat released by the number of moles of sucrose burned.

However, without the mass of sucrose burned, we cannot directly determine this. Please provide the mass of sucrose burned in the experiment to proceed with the calculation.

Would you like to proceed by providing the mass of sucrose or have any other questions?

Here are some related questions:

  1. What is the relationship between heat capacity and specific heat?
  2. How does a bomb calorimeter ensure accurate measurement of heat?
  3. How can we correct for heat losses in a calorimetry experiment?
  4. What is the significance of the molar enthalpy of combustion?
  5. How does the density of water affect the calorimetry calculations?

Tip: In calorimetry, always ensure you account for all sources of heat absorption to get an accurate measure of the energy change.

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Math Problem Analysis

Mathematical Concepts

Calorimetry
Thermodynamics
Heat Transfer

Formulas

q = m * C * ΔT
ΔH = q / n

Theorems

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Suitable Grade Level

College