Problem 136 - Chemistry The Science in Context

Question

The payload of a rocket includes a fuel and oxygen for combustion of the fuel. Reactions 1 and 2 describe the combustion of dimethylhydrazine and hydrogen, respectively. Pound for pound, which is the better rocket fuel, dimethylhydrazine or hydrogen? (1) \(\quad\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NNH}_{2}(\ell)+4 \mathrm{O}_{2}(g) \rightarrow\) \(\mathrm{N}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=-1694 \mathrm{kJ}\) (2) \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=-242 \mathrm{kJ}\)

Step-by-Step Solution

Answer

Answer: Hydrogen is a better rocket fuel on a pound-for-pound basis.

Step 1: Calculate energy per mole for both fuels

First, using the given \(\Delta H_{\mathrm{rxn}}^{\circ}\) values, we will calculate the energy released per mole of fuel burnt: 1. For reaction (1) with dimethylhydrazine: \(\Delta H_{\mathrm{rxn1}}^{\circ} = -1694 \ \mathrm{kJ/mol \ of \ fuel}\) 2. For reaction (2) with hydrogen: \(\Delta H_{\mathrm{rxn2}}^{\circ} = -242 \ \mathrm{kJ/mol \ of \ fuel}\)

Step 2: Calculate molar masses of the fuels

We will now calculate the molar mass of each fuel: 1. For dimethylhydrazine: Molar mass = \(2 (12.01 \ \mathrm{g/mol}) + 2 (1.01 \ \mathrm{g/mol})+1(14.01 \ \mathrm{g/mol})+4(1.01 \ \mathrm{g/mol}) = 46.14 \ \mathrm{g/mol}\) 2. For hydrogen: Molar mass = \(2 (1.01 \ \mathrm{g/mol}) = 2.02 \ \mathrm{g/mol}\)

Step 3: Calculate energy per unit mass for both fuels

We will now calculate the energy released per unit mass for each fuel: 1. For dimethylhydrazine: Energy per unit mass = \(\frac{-1694 \ \mathrm{kJ/mol}}{46.14 \ \mathrm{g/mol}} = -36.7 \ \mathrm{kJ/g}\) 2. For hydrogen: Energy per unit mass = \(\frac{-242 \ \mathrm{kJ/mol}}{2.02 \ \mathrm{g/mol}} = -119.8 \ \mathrm{kJ/g}\)

Step 4: Compare the energy per unit mass of both fuels

Comparing the energy released per unit mass of each fuel, hydrogen has a higher energy value (-119.8 kJ/g) than dimethylhydrazine (-36.7 kJ/g). This means that hydrogen is a better rocket fuel on a pound-for-pound basis.

Key Concepts

Enthalpy Change

Understanding enthalpy change is crucial in determining the suitability of rocket fuels. Enthalpy change, denoted as \(\Delta H_{\text{rxn}}^\circ\), represents the heat absorbed or released during a chemical reaction at constant pressure. It is typically measured in kilojoules per mole (\(\text{kJ/mol}\)), providing insights into the energy dynamics of the reaction.

In our case, we have two reactions: combustion of dimethylhydrazine and hydrogen. For dimethylhydrazine (Reaction 1), the enthalpy change is \(-1694 \text{kJ/mol}\), indicating that energy is released during the reaction. Similarly, hydrogen's combustion (Reaction 2) has an enthalpy change of \(-242 \text{kJ/mol}\). This negative sign signifies exothermic reactions, meaning heat is emitted when the fuel combusts.

By comparing these enthalpy changes, one can gauge how much energy each fuel releases per mole when it burns. However, to evaluate which is a better fuel, one must consider not only the magnitude of \(\Delta H_{\text{rxn}}^\circ\) but also the molar mass and energy per unit mass.

Molar Mass Calculation

Calculated molar masses provide vital information for determining how much energy a substance can release per unit of its mass. The molar mass is the weight of one mole of a substance, usually expressed in grams per mole (\(\text{g/mol}\)).

For dimethylhydrazine, we add the individual atomic weights: two carbon (C) atoms, six hydrogen (H) atoms, and two nitrogen (N) atoms. Thus, its molar mass is \(2(12.01) + 6(1.01) + 2(14.01) = 46.14 \text{g/mol}\).

• Carbon: \(2 \times 12.01\)
• Hydrogen: \(6 \times 1.01\)
• Nitrogen: \(2 \times 14.01\)

For hydrogen, the molar mass is determined by two hydrogen atoms: \(2(1.01) = 2.02 \text{g/mol}\).

This data is essential in converting energy release values from per mole to per gram, which allows us to make fair comparisons of fuel efficiency by weight.

Energy per Unit Mass

To choose the best rocket fuel, calculating the energy released per unit mass provides a helpful metric. This is especially useful because a good rocket fuel should release a significant amount of energy while minimal by weight.

"Energy per unit mass" is determined by dividing the enthalpy change by the molar mass of the fuel. For dimethylhydrazine, this calculation is \(\frac{-1694 \text{kJ/mol}}{46.14 \text{g/mol}} = -36.7 \text{kJ/g}\).

Similarly, for hydrogen, the calculation is \(\frac{-242 \text{kJ/mol}}{2.02 \text{g/mol}} = -119.8 \text{kJ/g}\).

• Dimethylhydrazine: \(-1694 \text{kJ/mol} / 46.14 \text{g/mol} = -36.7 \text{kJ/g}\)
• Hydrogen: \(-242 \text{kJ/mol} / 2.02 \text{g/mol} = -119.8 \text{kJ/g}\)

This approach shows that hydrogen, despite its lower enthalpy change per mole, provides more energy per gram, making it more efficient as a rocket fuel on a weight basis.

Combustion Reaction

Combustion reactions play a critical role in energy production for rockets. These reactions involve rapid oxidation, releasing energy in the form of heat. The process converts reactants, often fuels and oxygen, into products like water and carbon dioxide.

For the combustion of dimethylhydrazine (Reaction 1), the reaction is: \((\text{CH}_3)_2 \text{NNH}_2(\ell)+4 \text{O}_2(g) \rightarrow \text{N}_2(g)+4 \text{H}_2 \text{O}(g)+2 \text{CO}_2(g)\). Here, dimethylhydrazine burns using oxygen to produce nitrogen, water, and carbon dioxide with an enthalpy change of \(-1694 \text{kJ/mol}\).

In hydrogen's combustion (Reaction 2), the formula is simpler: \(\text{H}_2(g)+\frac{1}{2} \text{O}_2(g) \rightarrow \text{H}_2 \text{O}(g)\), leading to an enthalpy change of \(-242 \text{kJ/mol}\).

Understanding these reactions allows for evaluating how each fuel's chemical properties contribute to energy release. Efficient combustion means more energy for propulsion, a crucial criterion while selecting fuels for rocket engines.