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In a pound-for-pound comparison, which would you prefer as a rocket fuel, hydrogen \(\left(\mathrm{H}_{2}\right)\) or dimethythydrazine \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NNH}_{2} ? \mathrm{H}_{2}(\mathrm{~g})+(1 / 2) \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell)\) \(\Delta \mathrm{H}=-68.3 \mathrm{~K} \mathrm{cal} / \mathrm{mole}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NNH}_{2}+4 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+\) \(4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{CO}_{2}(\mathrm{~g}), \Delta \mathrm{H}=-404.9 \mathrm{~K}\) cal/mole.

Short Answer

Expert verified
In a pound-for-pound comparison, hydrogen \(\left(\mathrm{H}_{2}\right)\) is preferred as a rocket fuel over dimethylhydrazine \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NNH}_{2}\), as hydrogen releases more energy per gram \(-33.88\,Kcal/g\) than dimethylhydrazine \(-6.74\,Kcal/g\).

Step by step solution

01

Calculate the molar mass of hydrogen and dimethylhydrazine

To make a pound-for-pound comparison, we need to calculate the energy released per gram of each fuel. To do so, we will first determine the molar mass of hydrogen (H₂) and dimethylhydrazine [(CH₃)₂NNH₂]. The molar mass of hydrogen (H₂) is \(2 \times 1.008\,g/mol = 2.016\,g/mol\). The molar mass of dimethylhydrazine [(CH₃)₂NNH₂] = \((2 \times 12.01) + (6 \times 1.008) + 14.01 + 2 + 1.008 = 60.096\,g/mol\).
02

Calculate the energy released (ΔH) per gram of fuel

Now that we have the molar mass of each fuel, we can determine how much energy is released per gram of each fuel by dividing the given enthalpy change by the molar mass: Energy released per gram of hydrogen (H₂) = \(-68.3\,Kcal/mol \div 2.016\,g/mol = -33.88\,Kcal/g\). Energy released per gram of dimethylhydrazine [(CH₃)₂NNH₂] = \(-404.9 \,Kcal/mol \div 60.096\,g/mol = -6.74\,Kcal/g\).
03

Compare the energy released per gram

The energy released per gram of hydrogen and dimethylhydrazine are now calculated. We can compare these values to determine which fuel is more preferred based on a pound-for-pound comparison. Energy released per gram of hydrogen (H₂) is \(-33.88\,Kcal/g\). Energy released per gram of dimethylhydrazine [(CH₃)₂NNH₂] is \(-6.74\,Kcal/g\). As hydrogen releases more energy per gram than dimethylhydrazine, it is preferred as a rocket fuel in a pound-for-pound comparison.

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Key Concepts

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

Molar Mass Calculation
Calculating the molar mass of substances is a crucial step in chemistry, especially when comparing different fuels. Molar mass is the mass of one mole of a substance. For a molecule, you find this by summing the atomic masses of all atoms in the molecule.
In the exercise provided, we calculate the molar mass of two potential rocket fuels: hydrogen (\(\mathrm{H}_{2}\)) and dimethylhydrazine (\((\mathrm{CH}_3)_2 \mathrm{NNH}_2\)).
Let's break it down:
  • Hydrogen (\(\mathrm{H}_2\)): Since each hydrogen atom has an atomic mass of approximately 1.008 g/mol, two hydrogen atoms result in a molar mass of \(2 \times 1.008 = 2.016 \text{ g/mol}\).
  • Dimethylhydrazine (\((\mathrm{CH}_3)_2 \mathrm{NNH}_2\)): This compound consists of: two carbon atoms (\(2 \times 12.01\)), eight hydrogen atoms (\(8 \times 1.008\)), and two nitrogen atoms (14.01 for N). Adding them gives \(60.096 \text{ g/mol}\).
This calculation provides the basis for comparing the efficiency of the fuels based on their enthalpy change.
Enthalpy Change
Enthalpy change (\(\Delta H\)) is a measure of heat energy released or absorbed during a chemical reaction. When comparing fuels, enthalpy change helps determine which fuel releases more energy per mole burned.
In this case, for each reaction:
  • Hydrogen: The enthalpy change is given as \(-68.3 \text{ Kcal/mol}\), indicating that this amount of energy is released when one mole of hydrogen reacts.
  • Dimethylhydrazine: Here, the reaction releases \(-404.9 \text{ Kcal/mol}\). This reaction has more complex chemistry, resulting in different products and more energy release per mole.
Understanding enthalpy change is key because it tells us how effective a fuel is in terms of energy produced.
Energy Released per Gram
Energy released per gram is a crucial factor for fuel efficiency. It's more useful than molar enthalpy when comparing different fuels by weight. You calculate it by dividing the \(\Delta H\) for the fuel by its molar mass.
For these fuels:
  • Hydrogen has an energy release of \(-33.88 \text{ Kcal/g}\), which means it releases this energy amount per gram burned.
  • Dimethylhydrazine releases \(-6.74 \text{ Kcal/g}\).
This metric helps us evaluate which fuel is more efficient on a pound-for-pound basis. Because weight is often a limiting factor in rockets, energy per gram becomes quite important.
Comparison Analysis
In a fuel comparison analysis, we focus on which fuel provides more energy for weight carried. This is particularly important in scenarios like space travel where weight efficiency is critical.
From the calculations:
  • Hydrogen releases \(-33.88 \text{ Kcal/g}\), making it much more energy-dense per unit of weight compared to dimethylhydrazine.
  • Dimethylhydrazine, though releasing substantial energy per mole, is less energy-efficient per gram at \(-6.74 \text{ Kcal/g}\).
The analysis shows hydrogen as the superior choice when considering energy released per weight. Its higher energy release per gram means it can provide more thrust for the same weight compared to dimethylhydrazine, which is why it is the preferred rocket fuel in this comparison.

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Most popular questions from this chapter

\(500 \mathrm{~g}\) of ice at \(0^{\circ} \mathrm{C}\) is added to \(500 \mathrm{~g}\) of water at \(64^{\circ} \mathrm{C}\). When the temperature of the mixture is \(0^{\circ} \mathrm{C}\), what weight of ice is still present? Heat of fusion of \(\mathrm{H}_{2} \mathrm{O}=80 \mathrm{cal} / \mathrm{g}\).

Calculate \(\Delta \mathrm{H}^{\circ} \mathrm{r}\) for the combustion of methane, \(\mathrm{CH}_{4}\). The balanced reaction is \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\ell)\) \(\Delta \mathrm{H}^{\circ} \mathrm{f}\) in \(\mathrm{Kcal} / \mathrm{mole}=-17.89,0,-94.05,2(-68.32)\)

Given that \(\Delta \mathrm{H}^{\circ} \mathrm{CO} 2(\mathrm{~g})=-94.0, \Delta \mathrm{H}^{\circ} \mathrm{CO}(\mathrm{g})=-26,4, \Delta \mathrm{H}^{\circ} \mathrm{H} 2 \mathrm{O}(\ell)\) \(=68.4\) and \(\Delta \mathrm{H}^{\circ} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})=-57.8 \mathrm{Kcal} / \mathrm{mole}\), determine the heats of reaction of (1) \(\mathrm{CO}(\mathrm{g})+(1 / 2) \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}),(2)\) \(\mathrm{H}_{2}(\mathrm{~g})+(1 / 2) \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell)\) and (3) \(\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

In the reaction, \(\mathrm{CaCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{CaCO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) at \(950^{\circ} \mathrm{C}\) and \(\mathrm{CO}_{2}\) pressure of \(1 \mathrm{~atm}\), the \(\Delta \mathrm{H}\) is found to be 176 \(\mathrm{kJ} /\) mole. Assuming that the volume of the solid phase changes little by comparison with the volume of gas generated, calculate the \(\Delta \mathrm{E}\) for this reaction.

It is known that the heat of vaporization of water is 5 times as great as the heat of fusion. Explain this fact.

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