Question

From the following data for three prospective fuels, calculate which could provide the most energy per unit volume: $$ \begin{array}{lcc} \hline & \begin{array}{c} \text { Density } \\ \text { at } 20{ }^{\circ} \mathbf{C} \\ \left(\mathrm{g} / \mathrm{cm}^{3}\right) \end{array} & \begin{array}{c} \text { Molar Enthalpy } \\ \text { of Combustion } \\ \text { Fuel } \end{array} & \mathrm{kJ} / \mathrm{mol} \\ \hline \text { Nitroethane, } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NO}_{2}(l) & 1.052 & -1368 \\ \text { Ethanol, } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l) & 0.789 & -1367 \\ \text { Methylhydrazine, } \mathrm{CH}_{6} \mathrm{~N}_{2}(I) & 0.874 & -1305 \\\ \hline \end{array} $$

Short answer

The energy per unit volume for the three prospective fuels is calculated as follows: Nitroethane: -18.2 kJ/cm³, Ethanol: -23.2 kJ/cm³, and Methylhydrazine: -24.6 kJ/cm³. Therefore, Methylhydrazine provides the most energy per unit volume.

Step-by-step solution

Calculate the energy per unit volume for Nitroethane

The energy per unit volume can be calculated by dividing the molar enthalpy of combustion by the density of the fuel, and converting it to kJ/cm³. For Nitroethane, the density is 1.052 g/cm³ and the molar enthalpy of combustion is -1368 kJ/mol. First we need to convert grams to moles. The molecular weight of Nitroethane is: \(C_{2}H_{5}NO_{2} = (2 \times 12.01) + (5 \times 1.008) + (14.01) + (2 \times 16.00) = 75.07 \ \mathrm{g/mol}\) Now, let's calculate the energy per unit volume for Nitroethane: \(\frac{-1368\ \mathrm{kJ/mol}}{75.07\ \mathrm{g/mol}} \times \frac{1.052\ \mathrm{g}}{\mathrm{cm}^3} = -18.2\ \mathrm{kJ/}\mathrm{cm}^3\)

Calculate the energy per unit volume for Ethanol

For Ethanol, the density is 0.789 g/cm³ and the molar enthalpy of combustion is -1367 kJ/mol. First, we need to find the molecular weight: \(C_{2}H_{5}OH = (2 \times 12.01) + (6 \times 1.008) + (16.00) = 46.07\ \mathrm{g/mol}\) Now, let's calculate the energy per unit volume for Ethanol: \(\frac{-1367\ \mathrm{kJ/mol}}{46.07\ \mathrm{g/mol}} \times \frac{0.789\ \mathrm{g}}{\mathrm{cm}^3} = -23.2\ \mathrm{kJ/}\mathrm{cm}^3\)

Calculate the energy per unit volume for Methylhydrazine

For Methylhydrazine, the density is 0.874 g/cm³ and the molar enthalpy of combustion is -1305 kJ/mol. First, we need to find the molecular weight: \(CH_{6}N_{2} = (12.01) + (6 \times 1.008) + (2 \times 14.01) = 46.08\ \mathrm{g/mol}\) Now, let's calculate the energy per unit volume for Methylhydrazine: \(\frac{-1305\ \mathrm{kJ/mol}}{46.08\ \mathrm{g/mol}} \times \frac{0.874\ \mathrm{g}}{\mathrm{cm}^3} = -24.6\ \mathrm{kJ/}\mathrm{cm}^3\)

Compare the energy per unit volume for all fuels

Now that we have calculated the energy per unit volume for all three fuels, we can compare: - Nitroethane: -18.2 kJ/cm³ - Ethanol: -23.2 kJ/cm³ - Methylhydrazine: -24.6 kJ/cm³ Methylhydrazine provides the most energy per unit volume, followed by Ethanol and then Nitroethane.

Key concepts

Molar Enthalpy of Combustion

The molar enthalpy of combustion is a measure of the amount of energy released when one mole of a substance burns in the presence of oxygen. It is denoted usually in kilojoules per mole (kJ/mol). This value is crucial for understanding the energy content of a fuel, which determines how much energy can be obtained from it.
The value is often negative, indicating that energy is released in the process, making the reaction exothermic. When comparing different chemical fuels, the molar enthalpy of combustion allows scientists and engineers to predict which fuel will release the most energy per mole when burned.
In practical applications, one must consider this value to ensure efficient energy production and to choose suitable fuels for specific requirements.
To illustrate, if a fuel has a molar enthalpy of combustion of \(-1368\ kJ/mol\), it means burning one mole of that substance will release 1368 kJ of energy.

Density of Substances

Density is a physical property defined as mass per unit volume, typically expressed in grams per cubic centimeter (\(g/cm^3\)). This property is pivotal in determining how much space a substance will occupy under given conditions. In the context of fuels, density tells us the compactness of the substance, which has immense practical implications.
For energy considerations, density allows us to convert energy measures from per mole (mol) to per volume (cm³), providing a more tangible understanding of how much energy a specified volume of the fuel can release.
Working with density is straightforward. If you have a fuel with a density of \(0.789\ g/cm³\), it means that each cubic centimeter of the fuel weighs 0.789 grams. Thus, combining this information with the molar enthalpy of combustion allows more meaningful comparisons between different fuels in terms of energy release per given volume.

Chemical Fuels Comparison

When comparing chemical fuels, it’s essential to assess both the molar enthalpy of combustion and the density. These parameters help determine the energy efficiency and practicality of a fuel.
In terms of energy release, calculating the energy per unit volume for each fuel gives a clearer picture of how they perform relative to one another. This is a crucial step when deciding which fuel is more advantageous for specific purposes, such as transportation or energy generation.
For example, in comparing Nitroethane, Ethanol, and Methylhydrazine, considering both the energy density and physical properties like density, it becomes evident which fuel delivers the most energy per unit volume. Such analysis shows Methylhydrazine offers the highest energy output per unit volume among the three, hence might be preferred where volume efficiency is vital.
Ultimately, understanding these attributes helps engineers and decision-makers optimize fuel choices according to usage needs and environmental considerations.