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Chapter 5: Problem 97

How are fuel values calculated from molar enthalpies of combustion?

Short Answer

Expert verified
Based on the step-by-step solution provided, to calculate the fuel value from molar enthalpy of combustion, you need to first understand the concept of molar enthalpy of combustion and fuel value. Then, you need to convert between different units by knowing the molecular weight of the fuel. Finally, you can calculate the fuel value by dividing the molar enthalpy of combustion by the molecular weight and converting the energy per gram to the desired unit (e.g., J/g or Btu/lb).

Step by step solution

01

Understanding Molar Enthalpy of Combustion

Molar enthalpy of combustion is the amount of energy released as heat when one mole of a substance undergoes complete combustion in the presence of oxygen, under constant pressure. Molar enthalpy of combustion can be used to determine the heat energy released during the combustion process.
02

Understanding Fuel Value

Fuel value, also known as heat content or heating value, is defined as the amount of heat energy released during the combustion of a given amount of fuel. It is often expressed in units such as Joules per gram (J/g) or British thermal units per pound (Btu/lb).
03

Units Conversion

Molar enthalpy of combustion is usually expressed in kJ/mol (kilojoules per mole), while the fuel value is expressed in J/g (Joules per gram) or Btu/lb. To convert between these units, we need to know the molecular weight of the fuel. Molecular weight is the mass (grams) of one mole of the substance.
04

Fuel Value Calculation

To calculate the fuel value from molar enthalpy of combustion, follow these steps: 1. Convert molar enthalpy of combustion to energy per gram: Divide the molar enthalpy of combustion (kJ/mol) by the molecular weight (g/mol) of the fuel. This gives you the energy content in kJ/g. 2. Convert the energy per gram to the desired unit: If you want the fuel value in J/g, multiply the energy content obtained in step 1 by 1000 (1 kJ = 1000 J). If you want the fuel value in Btu/lb, multiply the energy content obtained in step 1 by 0.453592 (1 lb = 0.453592 kg), then multiply by 1000 to get the heat content in Btu/lb. Example: Let's consider a fuel with a molar enthalpy of combustion of 1500 kJ/mol and a molecular weight of 30 g/mol. 1. Convert molar enthalpy of combustion to energy per gram: 1500 kJ/mol 梅 30 g/mol = 50 kJ/g. 2. Convert the energy per gram to the desired unit: a. J/g: 50 kJ/g * 1000 = 50,000 J/g. b. Btu/lb: 50 kJ/g * 0.453592 * 1000 = 22,679.6 Btu/lb. The fuel value of the given fuel is 50,000 J/g or 22,679.6 Btu/lb.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fuel Value Calculation
Fuel value calculation is essential in determining the practical utility of a fuel for energy production. This assessment reflects how much heat a fuel can release during the combustion process and is critical for industries reliant on thermal energy, from power generation to heating systems. To start, we consider the molar enthalpy of combustion, which provides us with the heat energy released per mole of fuel burned.

To find the fuel's energy content per unit mass, which is more relatable to everyday usage, we divide the molar enthalpy of combustion by the fuel's molecular weight. The energy content thus obtained will be in kJ/g if using the SI units. For the everyday user, this could further be converted to more conventional units like J/g or Btu/lb, catering to various regional preferences and applications. It鈥檚 crucial to note that the conversion from kJ/g to J/g is a simple multiplication by 1000, as there are 1000 joules in one kilojoule.
Heat Energy
Heat energy is the form of energy transferred from one body to another as a result of a difference in temperature. In the context of combustion, heat energy is released when a fuel reacts with oxygen, and this exothermic reaction can be harnessed for various applications. The amount of energy produced depends on the type of fuel and the conditions under which combustion occurs.

Understanding the heat energy involved in fuel combustion allows for efficient energy management and sustainability practices. For example, using fuels with a high molar enthalpy of combustion can yield more heat per amount of fuel, improving efficiency. Key considerations include not only the total energy released but also the rate at which the fuel releases this energy and its impact on the environment.
Combustion Process
The combustion process is a chemical reaction where a fuel combines with an oxidant, typically oxygen, producing heat and light. For complete combustion to occur, there must be adequate fuel, oxygen, and an ignition source. Incomplete combustion results in inefficient energy production and can release harmful by-products.

For example, the combustion of hydrocarbons, which are commonly found in fossil fuels, results in water and carbon dioxide when combusted completely. However, if insufficient oxygen is present, other products like carbon monoxide or unburnt hydrocarbons might form, which are dangerous pollutants. Understanding the nuances of the combustion process, coupled with efficient fuel value calculations, can lead to better fuel utilization and reduced environmental impact.

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Most popular questions from this chapter

Ammonium nitrate decomposes to \(\mathrm{N}_{2} \mathrm{O}\) and water vapor at temperatures between \(250^{\circ} \mathrm{C}\) and \(300^{\circ} \mathrm{C} .\) Write a balanced chemical reaction describing the decomposition of ammonium nitrate, and calculate the enthalpy of reaction by using the appropriate enthalpies of formation from Appendix 4.

How does the energy required to recycle 1.00 mole of copper compare with that required to recover copper from CuO? The balanced chemical equation for the smelting of copper is: \(\mathrm{CuO}(s)+\mathrm{CO}(g) \rightarrow \mathrm{Cu}(s)+\mathrm{CO}_{2}(g)\) Copper melts at \(1084.5^{\circ} \mathrm{C}\) with \(\Delta H_{\text {fus }}^{\circ}=13.0 \mathrm{kJ} / \mathrm{mol}\) and a molar heat capacity \(c_{\mathrm{P}, \mathrm{Cu}}=24.5 \mathrm{J} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right) .\) In addition, \(\Delta H_{\mathrm{f}, \mathrm{CuO}}^{\circ}=-155 \mathrm{kJ} / \mathrm{mol}\)

Why is it important for Hess's law that enthalpy is a state function?

Adding \(2.00 \mathrm{g}\) of \(\mathrm{Mg}\) metal to \(95.0 \mathrm{mL}\) of \(1.00 \mathrm{MHCl}\) in a coffee-cup calorimeter leads to a temperature increase of \(9.2^{\circ} \mathrm{C}\) a. Write a balanced net ionic equation for the reaction. b. If the molar heat capacity of \(1.00 M \mathrm{HCl}\) is the same as that for water \(\left[c_{\mathrm{P}}=75.3 \mathrm{J} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\right],\) what is \(\Delta H_{\mathrm{rxn}} ?\)

For which of the following reactions does \(\Delta H_{\mathrm{rxn}}^{\circ}\) represent an enthalpy of formation? a. \(C(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)\) b. \(\mathrm{CO}_{2}(g)+\mathrm{C}(s) \rightarrow 2 \mathrm{CO}(g)\) c. \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g)\) d. \(2 \mathrm{H}_{2}(g)+\mathrm{C}(s) \rightarrow \mathrm{CH}_{4}(g)\)
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