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Hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}\), is a colorless liquid used as a rocket fuel. What is the enthalpy change for the process in which hydrazine is formed from its elements? $$ \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l) $$ Use the following reactions and enthalpy changes: \(\mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) ; \Delta H=-622.2 \mathrm{~kJ}\) \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) ; \Delta H=-285.8 \mathrm{~kJ}\)

Short Answer

Expert verified
Use Hess's law to combine reactions.

Step by step solution

01

Understand the target reaction

The target reaction for which we need to determine the enthalpy change is: \( \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}\mathrm{H}_{4}(l) \). We aim to find \( \Delta H \) for this process.
02

Analyze the provided reactions

The first given reaction is \( \mathrm{N}_{2} \mathrm{H}_{4}(l) + \mathrm{O}_{2}(g) \rightarrow \mathrm{N}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(l) \) with \( \Delta H = -622.2 \mathrm{~kJ} \). The second given reaction is \( \mathrm{H}_{2}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l) \) with \( \Delta H = -285.8 \mathrm{~kJ} \).

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Key Concepts

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

Thermochemistry
Thermochemistry is a branch of chemistry that studies the heat involved in chemical reactions and physical transformations. It focuses on understanding how energy, in the form of heat, is transferred between systems and their surroundings. In chemical reactions, energy changes are particularly important, as they determine whether a reaction releases or absorbs heat. A key concept in thermochemistry is enthalpy, which is a measure of the total heat content of a system at constant pressure.

When analyzing reactions, we often measure these energy changes through enthalpy changes (ΔH). These changes help in understanding how much heat is needed or released during a reaction. Whenever a reaction occurs, bonds are broken and formed, and this process involves energy changes that are crucial for predicting reaction behavior.
  • Exothermic reactions release heat, resulting in a negative ΔH.
  • Endothermic reactions absorb heat, leading to a positive ΔH.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products. This transformation includes the breaking and forming of chemical bonds, and it encompasses the rearrangement of atoms. Every chemical reaction is accompanied by a change in energy, as old bonds break and new bonds form.

In the example of hydrazine formation, the reaction \[ \mathrm{N}_{2}(g) + 2 \mathrm{H}_{2}(g) \rightarrow \mathrm{N}_{2}\mathrm{H}_{4}(l) \] involves combining nitrogen and hydrogen gases to create liquid hydrazine. To understand this reaction, it's essential to consider the given reactions and their enthalpy changes. In thermochemistry, knowing the ΔH of reactions helps predict whether the formation of a product will require or release energy.
  • Reactants: The starting substances, in this case, nitrogen gas and hydrogen gas.
  • Products: The substances formed, like hydrazine.
  • Energy change: Important for determining the enthalpy of formation.
Hydrazine Formation
Hydrazine (\( \mathrm{N}_{2} \mathrm{H}_{4} \)) is a compound with significant uses, such as rocket fuel. Its formation from elemental nitrogen and hydrogen is an example of a chemical reaction where understanding the enthalpy change is vital.
The enthalpy change for hydrazine's formation can be determined using reactions where hydrazine participates, combined with Hess's Law. The two given reactions describe the formation and decomposition of compounds related to hydrazine. This is crucial because the direct formation of a compound can often be linked back to simpler known reactions involving the compound.
  • Hydrazine: A liquid fuel, characterized by its high energy content.
  • Enthalpy determination: Helps in assessing the energy needed or released during hydrazine formation.
Hess's Law
Hess's Law is a powerful principle in thermochemistry stating that the total enthalpy change of a chemical reaction is the same, no matter how many steps the reaction is carried out in. This law is very useful because it allows chemists to calculate the enthalpy change of a reaction even when it cannot be measured directly.

For hydrazine formation, the enthalpy change can be found by using the given reactions and manipulating them to obtain the target reaction. According to Hess's Law, by reversing and combining the provided reactions, we can determine the ΔH of the hydrazine formation reaction. The steps to apply Hess's Law involve ensuring all intermediates cancel out, leaving only the overall intended reaction.
  • Utilize known enthalpy changes: Modify and combine them to calculate unknown ΔH.
  • Strategically align the reactions: Ensure that intermediary substances cancel out, achieving the target reaction.
  • Achieve simplicity: A single overall ΔH value regardless of complex reaction pathways.

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Most popular questions from this chapter

Calcium oxide (quicklime) reacts with water to produce calcium hydroxide (slaked lime). $$ \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s) ; \Delta H=-65.2 \mathrm{~kJ} $$ The heat released by this reaction is sufficient to ignite paper. How much heat is released when \(24.5 \mathrm{~g}\) of calcium oxide reacts?

The sugar arabinose, \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5}\), is burned completely in oxygen in a calorimeter. $$ \mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5}(s)+5 \mathrm{O}_{2}(g) \longrightarrow 5 \mathrm{CO}_{2}(g)+5 \mathrm{H}_{2} \mathrm{O}(l) $$ Burning a \(0.548-\mathrm{g}\) sample caused the temperature to rise from \(20.00^{\circ} \mathrm{C}\) to \(20.54^{\circ} \mathrm{C}\). The heat capacity of the calorimeter and its contents is \(15.8 \mathrm{~kJ} /{ }^{\circ} \mathrm{C} .\) Calculate \(\Delta H\) for the combustion reaction per mole of arabinose.

You have two samples of different metals, metal \(\mathrm{A}\) and metal \(\mathrm{B}\), each having the same mass. You heat both metals to \(95^{\circ} \mathrm{C}\) and then place each one into separate beakers containing the same quantity of water at \(25^{\circ} \mathrm{C}\). a. You measure the temperatures of the water in the two beakers when each metal has cooled by \(10^{\circ} \mathrm{C}\) and find that the temperature of the water with metal \(\mathrm{A}\) is higher than the temperature of the water with metal \(\mathrm{B}\). Which metal has the greater specific heat? Explain. b. After waiting a period of time, the temperature of the water in each beaker rises to a maximum value. In which beaker does the water rise to the higher value, the one with metal \(\mathrm{A}\) or the one with metal \(\mathrm{B} ?\) Explain.

The process of dissolving ammonium nitrate, \(\mathrm{NH}_{4} \mathrm{NO}_{3}\), in water is an endothermic process. What is the sign of \(q ?\) If you were to add some ammonium nitrate to water in a flask, would you expect the flask to feel warm or cool?

Compounds with carbon-carbon double bonds, such as ethylene, \(\mathrm{C}_{2} \mathrm{H}_{4}\), add hydrogen in a reaction called hydrogenation. $$ \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g) $$ Calculate the enthalpy change for this reaction, using the following combustion data: $$ \begin{gathered} \mathrm{C}_{2} \mathrm{H}_{4}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H=-1411 \mathrm{~kJ} \end{gathered} $$ $$ \begin{gathered} \mathrm{C}_{2} \mathrm{H}_{6}(g)+\frac{7}{2} \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H=-1560 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) ; \Delta H=-286 \mathrm{~kJ} \end{gathered} $$

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