Problem 24
Question
A calorimeter has a measured heat capacity of \(6.27 \mathrm{~kJ} \cdot\left({ }^{\circ} \mathrm{C}\right)^{-1}\). The combustion of \(1.84 \mathrm{~g}\) of magnesium led to a temperature change from \(21.30^{\circ} \mathrm{C}\) to \(28.56^{\circ} \mathrm{C}\). Calculare the enthalpy change of the reaction \(2 \mathrm{Mg}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{MgO}(\mathrm{s})\).
Step-by-Step Solution
Verified Answer
The enthalpy change for the combustion of magnesium to form MgO is calculated using the heat absorbed by the calorimeter and dividing it by the number of moles of magnesium and then by 2 to match the stoichiometry of the reaction.
1Step 1 - Calculate the heat absorbed by the calorimeter
To find out the heat absorbed during the combustion, use the heat capacity of the calorimeter and the change in temperature. Use the formula: \( Q = C \cdot \Delta T \), where \( Q \) is the heat absorbed, \( C \) is the heat capacity, and \( \Delta T \) is the change in temperature.
2Step 2 - Calculate the change in temperature
Subtract the initial temperature from the final temperature: \( \Delta T = 28.56^\circ C - 21.30^\circ C \).
3Step 3 - Plug in the values to find the heat absorbed
Now, calculate the heat absorbed by the calorimeter using the values from the previous steps: \( Q = 6.27 \, \mathrm{kJ/}^\circ\mathrm{C} \times \Delta T \).
4Step 4 - Convert the mass of magnesium to moles
Calculate the number of moles of magnesium that combusted. The molar mass of magnesium (Mg) is approximately 24.305 g/mol. Use the formula: \( n = \frac{m}{M} \), where \( n \) is the number of moles, \( m \) is the mass of magnesium in grams, and \( M \) is the molar mass of magnesium.
5Step 5 - Calculate the molar enthalpy change
Divide the heat absorbed by the number of moles of magnesium to get the molar enthalpy change for the reaction. The reaction shows the combustion of two moles of magnesium, so it's important to also divide the heat by 2 to find the enthalpy change per mole of the reaction described.
Key Concepts
Heat Capacity of CalorimeterTemperature Change in CombustionMolar Enthalpy ChangeStoichiometry
Heat Capacity of Calorimeter
The heat capacity of a calorimeter is a crucial factor in the study of thermochemical reactions. It represents the amount of heat energy required to raise the temperature of the calorimeter by one degree Celsius. When a substance combusts within the calorimeter, the heat released from this combustion process is absorbed by the calorimeter, causing a rise in its temperature.
To correctly measure the heat change involved in a reaction, it's important to know the calorimeter's heat capacity. This value is constant for a given calorimeter and is often determined through calibration with reactions that have known energy changes. The formula to calculate the heat absorbed by the calorimeter, as provided in the original problem, is given by the equation: \( Q = C \cdot \Delta T \) where \( Q \) is the heat absorbed, \( C \) denotes the heat capacity of the calorimeter, and \( \Delta T \) indicates the temperature change observed during the reaction.
To correctly measure the heat change involved in a reaction, it's important to know the calorimeter's heat capacity. This value is constant for a given calorimeter and is often determined through calibration with reactions that have known energy changes. The formula to calculate the heat absorbed by the calorimeter, as provided in the original problem, is given by the equation: \( Q = C \cdot \Delta T \) where \( Q \) is the heat absorbed, \( C \) denotes the heat capacity of the calorimeter, and \( \Delta T \) indicates the temperature change observed during the reaction.
Temperature Change in Combustion
Temperature change plays a fundamental role in understanding the heat transfer during a reaction. In the combustion process, substances react with oxygen, releasing energy in the form of heat which causes a measurable temperature change in the calorimeter's water reservoir.
To determine the temperature change, \( \Delta T \), the final temperature is subtracted from the initial temperature. This difference reflects the net effect of the reaction's energy on the calorimeter. The larger the temperature change, the more heat has been transferred, assuming the heat capacity of the calorimeter remains constant. In the provided example, the temperature changed from \(21.30^\circ C\) to \(28.56^\circ C\), resulting in a \( \Delta T \) to be used in subsequent calculations.
To determine the temperature change, \( \Delta T \), the final temperature is subtracted from the initial temperature. This difference reflects the net effect of the reaction's energy on the calorimeter. The larger the temperature change, the more heat has been transferred, assuming the heat capacity of the calorimeter remains constant. In the provided example, the temperature changed from \(21.30^\circ C\) to \(28.56^\circ C\), resulting in a \( \Delta T \) to be used in subsequent calculations.
Molar Enthalpy Change
Molar enthalpy change, often expressed as \( \Delta H \), signifies the amount of heat absorbed or released by one mole of substance during a reaction at constant pressure. It is a critical property in thermodynamics and chemical engineering as it helps to quantify the energy changes involved in chemical reactions.
Once the heat absorbed by the calorimeter is known, we can calculate the molar enthalpy change by dividing this heat, \( Q \), by the number of moles of reactant(s) that participated. It is important to adjust for the stoichiometry, or the mole ratio, of the reaction. In our exercise, the reaction involves two moles of magnesium, so the heat calculated for the total reaction must be halved to find the molar enthalpy change per mole of magnesium.
Once the heat absorbed by the calorimeter is known, we can calculate the molar enthalpy change by dividing this heat, \( Q \), by the number of moles of reactant(s) that participated. It is important to adjust for the stoichiometry, or the mole ratio, of the reaction. In our exercise, the reaction involves two moles of magnesium, so the heat calculated for the total reaction must be halved to find the molar enthalpy change per mole of magnesium.
Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It involves using the balanced chemical equation to determine the proportions in which substances react and form products. Understanding stoichiometry is fundamental when calculating the molar enthalpy change because the reaction's stoichiometry will dictate the amount of energy per mole.
In the context of the given exercise, the stoichiometry indicates that two moles of magnesium react with one mole of oxygen to form two moles of magnesium oxide. Thus, when calculating the molar enthalpy change, it's crucial to divide the total heat absorbed by two to accurately represent the enthalpy change for the combustion of one mole of magnesium, according to the stoichiometric coefficients in the balanced chemical equation.
In the context of the given exercise, the stoichiometry indicates that two moles of magnesium react with one mole of oxygen to form two moles of magnesium oxide. Thus, when calculating the molar enthalpy change, it's crucial to divide the total heat absorbed by two to accurately represent the enthalpy change for the combustion of one mole of magnesium, according to the stoichiometric coefficients in the balanced chemical equation.
Other exercises in this chapter
Problem 21
A calorimeter was calibrated with an electric heater, which was used to supply \(22.5 \mathrm{~kJ}\) of energy to the calorimeter; the heat increased the temper
View solution Problem 23
Calculate the amount of heat needed to raise the temperature of \(0.325 \mathrm{~mol}\) of a monatomic idcal gas from a temperature of \(-25^{\circ} \mathrm{C}\
View solution Problem 25
Which gas do you expect to have the higher molar heat capacity, \(\mathrm{NO}\) or \(\mathrm{NO}_{2}\) ? Why?
View solution Problem 29
(a) At its boiling point, the vaporization of \(0.235 \mathrm{~mol} \mathrm{CH}_{4}(1)\) requires \(1.93 \mathrm{~kJ}\) of heat. What is the enthalpy of vaporiz
View solution