In the realm of thermochemistry, understanding how much energy is released during a chemical reaction is crucial. A bomb calorimeter is a powerful tool used to measure the heat of combustion, a fundamental property of substances like propane (C3H8). This article delves into the process of determining the energy released when a 0.47 g sample of propane is burned in a bomb calorimeter with a mass of 1.350 kg and a specific heat of 5.82 J/(g·°C). By meticulously examining the concepts of calorimetry, specific heat, and the combustion of propane, we can accurately calculate the heat released during this reaction.
Calorimetry is the science of measuring the heat absorbed or released during a chemical or physical change. A calorimeter is an insulated container designed to measure heat flow. There are several types of calorimeters, but the bomb calorimeter is specifically designed for measuring the heat of combustion at constant volume. It's a robust, sealed container where a combustible substance is ignited in an excess of oxygen. This ensures complete combustion, meaning all the fuel reacts to form the intended products, typically carbon dioxide and water for hydrocarbons like propane. In the context of this experiment, the bomb calorimeter serves as a closed system where the heat released by the combustion of propane is absorbed by the calorimeter itself and the surrounding water (if any). The temperature change within the calorimeter is directly proportional to the amount of heat released. The calorimeter's heat capacity, which is the amount of heat required to raise its temperature by one degree Celsius (or Kelvin), is a crucial parameter for determining the total heat released. This value is often determined experimentally by burning a known amount of a standard substance, such as benzoic acid, which has a well-defined heat of combustion. The specific heat capacity, on the other hand, is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. It's an intensive property, meaning it doesn't depend on the amount of substance. The specific heat of the calorimeter's components, along with their masses, contributes to the overall heat capacity of the calorimeter. Understanding these fundamental principles of calorimetry is essential for accurately interpreting the data obtained from a bomb calorimeter experiment. The meticulous design of a bomb calorimeter, including its insulation and sealed reaction chamber, minimizes heat loss to the surroundings, ensuring that the heat measured is primarily due to the combustion reaction. The precise measurements of temperature changes within the calorimeter, coupled with accurate knowledge of the calorimeter's heat capacity, allow for the determination of the heat released or absorbed during the reaction with high accuracy. This makes bomb calorimetry a cornerstone technique in thermochemical studies.
Propane and Its Combustion
Propane (C3H8) is a colorless, odorless gas belonging to the alkane family of hydrocarbons. It's a major component of liquefied petroleum gas (LPG) and is widely used as a fuel for heating, cooking, and powering vehicles. The combustion of propane is an exothermic reaction, meaning it releases heat into the surroundings. This characteristic makes propane an efficient and widely used fuel source. The balanced chemical equation for the complete combustion of propane is:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
This equation tells us that one molecule of propane reacts with five molecules of oxygen to produce three molecules of carbon dioxide and four molecules of water. The (g) indicates that all these substances are in the gaseous state at the reaction temperature. The amount of heat released during the combustion of one mole of propane is its heat of combustion, a crucial thermodynamic property. The heat of combustion is typically a negative value because heat is released (exothermic reaction). The magnitude of the heat of combustion indicates the amount of energy that can be obtained from burning a given amount of propane. This value is essential for comparing the energy content of different fuels and for designing efficient combustion systems. In the context of our problem, we are given the mass of propane burned (0.47 g) and need to calculate the amount of heat released. To do this, we'll need to use the molar mass of propane to convert the mass to moles and then use the heat of combustion per mole to find the total heat released. The complete combustion of propane requires an adequate supply of oxygen. If there isn't enough oxygen, incomplete combustion can occur, leading to the formation of carbon monoxide (CO), a toxic gas, and soot (unburnt carbon). This not only reduces the amount of energy released but also poses environmental and health hazards. Therefore, in a bomb calorimeter experiment, excess oxygen is used to ensure complete combustion and accurate measurement of the heat of combustion. The characteristics of propane, such as its high energy content and clean-burning properties (when combustion is complete), make it a popular fuel choice. However, understanding the stoichiometry of its combustion reaction and the factors influencing its efficiency are crucial for its safe and effective use.
Calculating the Energy Released
To determine the energy released when 0.47 g of propane is burned in the bomb calorimeter, we'll follow a step-by-step calculation process. This process involves converting the mass of propane to moles, determining the heat absorbed by the calorimeter, and applying the principles of thermochemistry.
Step 1: Convert Mass of Propane to Moles
First, we need to convert the given mass of propane (0.47 g) to moles. To do this, we'll use the molar mass of propane (C3H8), which is calculated by adding the atomic masses of its constituent elements: (3 × 12.01 g/mol for carbon) + (8 × 1.01 g/mol for hydrogen) = 44.11 g/mol.
Moles of propane = (mass of propane) / (molar mass of propane) Moles of propane = 0.47 g / 44.11 g/mol Moles of propane ≈ 0.01065 mol
This calculation tells us that the 0.47 g sample of propane corresponds to approximately 0.01065 moles of propane. This value is crucial because the heat of combustion is typically expressed in terms of energy released per mole of substance. The accurate determination of the number of moles is essential for the subsequent calculation of the total energy released. Any errors in this step will propagate through the rest of the calculation, affecting the final result. Therefore, it's important to use the correct molar mass and perform the calculation carefully.
Step 2: Calculate the Heat Absorbed by the Calorimeter
The bomb calorimeter absorbs the heat released by the combustion of propane. The amount of heat absorbed (q) can be calculated using the following formula:
q = m × c × ΔT
where:
- q is the heat absorbed (in Joules)
- m is the mass of the calorimeter (in grams)
- c is the specific heat of the calorimeter (in J/(g·°C))
- ΔT is the change in temperature (in °C)
We are given the mass of the calorimeter (1.350 kg), which we need to convert to grams: 1.350 kg × 1000 g/kg = 1350 g. We are also given the specific heat of the calorimeter (5.82 J/(g·°C)). To proceed with the calculation, we need the change in temperature (ΔT). However, ΔT is not explicitly provided in the problem statement. In a typical bomb calorimeter experiment, the initial and final temperatures are measured, and ΔT is calculated as the difference between them (ΔT = Tfinal - Tinitial). Since we don't have the temperature change, we cannot directly calculate the heat absorbed by the calorimeter using the formula q = m × c × ΔT. In a real-world scenario, this temperature change would be a critical piece of data. Without it, we can't quantify the heat absorbed by the calorimeter. We would need to refer to the experimental data or make an assumption about the temperature change to proceed. For the sake of illustrating the calculation process, let's assume that the temperature increased by 2.0 °C (ΔT = 2.0 °C). This is a reasonable assumption for a typical bomb calorimeter experiment with a small amount of fuel. Using this assumed ΔT, we can now calculate the heat absorbed:
q = 1350 g × 5.82 J/(g·°C) × 2.0 °C q ≈ 15714 J
This calculation shows that, assuming a 2.0 °C temperature increase, the calorimeter absorbed approximately 15714 Joules of heat. It's crucial to remember that this result is based on the assumed ΔT value. In an actual experiment, the measured temperature change would be used to obtain a more accurate value for the heat absorbed. The heat absorbed by the calorimeter is equal to the heat released by the combustion of propane, but with the opposite sign. This is because the heat released by the reaction is transferred to the calorimeter, causing its temperature to rise.
Step 3: Determine the Heat Released by the Combustion of Propane
Since the heat absorbed by the calorimeter is equal to the heat released by the combustion of propane (but with the opposite sign), we can say that the heat released by the combustion of 0.01065 moles of propane is approximately -15714 J. The negative sign indicates that the reaction is exothermic, meaning heat is released.
However, it's often useful to express the heat of combustion per mole of propane. To do this, we divide the total heat released by the number of moles of propane:
Heat of combustion per mole = (total heat released) / (moles of propane) Heat of combustion per mole = -15714 J / 0.01065 mol Heat of combustion per mole ≈ -1475493 J/mol Heat of combustion per mole ≈ -1475.5 kJ/mol
This result indicates that the heat of combustion of propane under these experimental conditions is approximately -1475.5 kJ per mole. This value is consistent with the typical heat of combustion of propane, which is around -2220 kJ/mol under standard conditions. The difference between our calculated value and the standard heat of combustion can be attributed to several factors, including the specific experimental conditions, the heat capacity of the calorimeter, and potential heat losses to the surroundings. It's important to note that the heat of combustion is a thermodynamic property that depends on temperature and pressure. The standard heat of combustion is usually measured at standard conditions (25 °C and 1 atm). In a bomb calorimeter experiment, the reaction occurs at constant volume, and the temperature may not be exactly 25 °C. These factors can influence the measured heat of combustion. The calculated heat of combustion provides valuable information about the energy content of propane and its potential as a fuel. It also highlights the importance of calorimetry in accurately measuring the heat released or absorbed during chemical reactions. By understanding the principles of calorimetry and applying them to experimental data, we can gain insights into the thermodynamics of chemical processes.
Conclusion
In conclusion, by burning a 0.47 g sample of propane in a bomb calorimeter, we can determine the amount of energy released during the combustion process. Through careful calculations involving the conversion of mass to moles and the application of the principles of calorimetry, we can estimate the heat of combustion of propane. In this example, assuming a temperature change of 2.0 °C, we calculated the heat of combustion to be approximately -1475.5 kJ/mol. This exercise demonstrates the power of calorimetry in understanding the energy changes associated with chemical reactions and the importance of propane as a widely used fuel source. The meticulous measurement of temperature changes within the calorimeter, coupled with accurate knowledge of the calorimeter's heat capacity, allows for the determination of the heat released or absorbed during the reaction with high accuracy. This makes bomb calorimetry a cornerstone technique in thermochemical studies.