Question

Hydrazine, a component of rocket fuel, undergoes combustion to yield \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) : $$\mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ (a) Draw an electron-dot structure for hydrazine, predict the geometry about each nitrogen atom, and tell the hybridization of each nitrogen. (b) Use the following information to set up a Hess's law cycle, and then calculate \(\Delta H^{\circ}\) for the combustion reaction. You will need to use fractional coefficients for some equations. $$\begin{aligned}&2 \mathrm{NH}_{3}(g)+3 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow 4 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) \\ &\Delta H^{\circ}=-1011.2 \mathrm{~kJ} \\ &\mathrm{~N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) \\ &\Delta H^{\circ}=-317.2 \mathrm{~kJ} \\ &4 \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{~N}_{2} \mathrm{H}_{4}(l)+2 \mathrm{H}_{2} \mathrm{O}(l) \\ &\Delta H^{\circ}=-286.0 \mathrm{~kJ} \\ &\mathrm{H}_{2} \mathrm{O}(I) \quad \Delta H_{\mathrm{f}}^{\circ}=-285.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$$ (c) How much heat is released on combustion of \(100.0 \mathrm{~g}\) of hydrazine?

Short answer

The combustion of 100 g of hydrazine releases approximately -1674.7 kJ. The electron geometry of hydrazine is tetrahedral with \(\text{sp}^3\) hybridization.

Step-by-step solution

Draw Electron-dot Structure for Hydrazine

Hydrazine, \(\mathrm{N}_{2}\mathrm{H}_{4}\), consists of two nitrogen atoms connected by a single bond with each nitrogen attached to two hydrogens. Nitrogen atoms each have one lone pair. This gives us an electron-dot structure: H-N-N-H with each nitrogen having two H atoms and one lone pair of electrons.

Predict Geometry and Hybridization

Each nitrogen atom in hydrazine is bonded to three other atoms (two hydrogens and one nitrogen) and has one lone pair. The electron geometry is tetrahedral, but the molecular geometry around each nitrogen atom is 'trigonal pyramidal.' Each nitrogen atom has \(\mathrm{sp^3}\) hybridization to accommodate the three sigma bonds and one lone pair.

Set Up Hess's Law Cycle

We have three reactions to combine:1. \(2\mathrm{NH}_{3}(g) + 3\mathrm{N}_{2}\mathrm{O}(g) \rightarrow 4\mathrm{N}_{2}(g) + 3\mathrm{H}_{2}\mathrm{O}(l)\) with \(\Delta H^{\circ} = -1011.2 \mathrm{~kJ}\)2. \(\mathrm{N}_{2}\mathrm{O}(g) + 3\mathrm{H}_{2}(g) \rightarrow \mathrm{N}_{2}\mathrm{H}_{4}(l) + \mathrm{H}_{2}\mathrm{O}(l)\) with \(\Delta H^{\circ} = -317.2 \mathrm{~kJ}\)3. \(4\mathrm{NH}_{3}(g) + \mathrm{O}_{2}(g) \rightarrow 2\mathrm{N}_{2}\mathrm{H}_{4}(l) + 2\mathrm{H}_{2}\mathrm{O}(l)\) with \(\Delta H^{\circ} = -286.0 \mathrm{~kJ}\).These reactions are combined to form the desired combustion reaction of hydrazine, relating hydrazine formation and decomposition as needed.

Calculate \(\Delta H^{\circ}\) for Combustion Reaction

The target reaction is \(\mathrm{N}_{2}\mathrm{H}_{4}(l) + \mathrm{O}_{2}(g) \rightarrow \mathrm{N}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(l)\). From Hess's Law, combine reactions 2 and 3 while reversing reaction 1:- Use reaction 3 as is: \(\Delta H^{\circ} = -286.0 \mathrm{~kJ}\)- Reverse reaction 2: \(\mathrm{N}_{2}\mathrm{H}_{4}(l) + \mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{N}_{2}\mathrm{O}(g) + 3\mathrm{H}_{2}(g)\), resulting in \(\Delta H^{\circ} = +317.2 \mathrm{~kJ}\)- Reverse reaction 1: \(\mathrm{N}_{2}(g) + 1.5\mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{NH}_{3}(g) + 1.5\mathrm{N}_{2}\mathrm{O}(g)\) for 0.5 mol, resulting in \(\Delta H^{\circ} = +505.6 \mathrm{~kJ}\).Finally, add these \(\Delta H\) values: \(-286.0 + 317.2 + 505.6\), giving \(\Delta H^{\circ} = 536.8 \mathrm{~kJ/mol}\).

Calculate Heat Released for 100 g Hydrazine

First, determine moles of hydrazine in 100 g:\(\text{Molar mass of } \mathrm{N}_{2}\mathrm{H}_{4} = 32.05 \mathrm{~g/mol}\), hence 100 g corresponds to \(\frac{100}{32.05} \approx 3.12\) moles.Calculate total heat released using the calculated \(\Delta H^{\circ}\):\(-536.8 \mathrm{~kJ/mol} \times 3.12 \mathrm{~mol} = -1674.7 \mathrm{~kJ}\).

Key concepts

Chemical Bonding

Chemical bonding is the process by which two or more atoms join together to form a stable molecule. In the case of hydrazine (\(\mathrm{N}_{2}\mathrm{H}_{4}\)), the chemical bonds involved are covalent bonds. This type of bond occurs when atoms share pairs of electrons.

• In hydrazine, each nitrogen atom is bonded to two hydrogen atoms and one nitrogen atom.
• These covalent bonds are formed through the sharing of electrons, ensuring that each atom achieves a stable electron configuration.

A Lewis Structure, or electron-dot structure, helps us visualize the bonding in molecules. For hydrazine, the Lewis Structure involves:

• Two nitrogen atoms bonded together by a single covalent bond.
• Each nitrogen bonded to two hydrogen atoms.
• Each nitrogen atom containing a lone pair of electrons.

Being familiar with chemical bonds is important because it helps us understand the properties and reactions of molecules like hydrazine, which is used as rocket fuel because of its ability to release large amounts of energy upon combustion.

Hybridization

Hybridization is a concept in chemistry where atomic orbitals merge to form new hybrid orbitals. These hybrid orbitals influence the geometry of the molecule, specific to the central atom being scrutinized.In hydrazine, the nitrogen atoms undergo \(\mathrm{sp^3}\) hybridization:

• This hybridization involves one 's' orbital and three 'p' orbitals mixing to form four equivalent \(\mathrm{sp^3}\) hybrid orbitals for each nitrogen atom.
• The geometry formed as a result of \(\mathrm{sp^3}\) hybridization is tetrahedral. However, due to the presence of lone pair electrons, the actual molecular geometry around each nitrogen is trigonal pyramidal.

Understanding hybridization allows chemists to predict molecular shapes and bond angles accurately, as well as to infer reactivity and interaction with other molecules. This knowledge is beneficial when analyzing reaction mechanisms and the effectiveness of molecules such as hydrazine in fuel applications.

Enthalpy Change

Enthalpy change, often represented as \(\Delta H^{\circ}\), is the heat change at constant pressure during a chemical reaction. In practical terms, it informs us whether a reaction absorbs heat (endothermic) or releases heat (exothermic).For hydrazine's combustion, we utilized Hess's Law to simplify the calculation of the net enthalpy change by considering multiple related chemical reactions:

• Hess's Law states that the total enthalpy change in a series of reactions is the sum of the enthalpy changes for each individual reaction.
• In the combustion of hydrazine, the target reaction and auxiliary reactions were combined to determine the change in enthalpy.

Through the cycle derived from the associated reactions:

• The calculated net \(\Delta H^{\circ}\) for the combustion of hydrazine is \(536.8 \mathrm{~kJ/mol}\).
• This means that the reaction releases \(536.8 \mathrm{~kJ/mol}\) of energy, classifying it as an exothermic process.

Enthalpy changes are critically important in designing and optimizing energy-related processes, such as fuel efficiency and energy safety, which are key considerations for chemical propellants like hydrazine.